Malla Reddy College of Engineering

MALLA REDDY COLLEGE OF ENGINEERING & TECHNOLOGY (An Autonomous Institution – UGC, Govt.of India) Recognizes under 2(f) and 12(B) of UGC ACT 1956 (Affiliated to JNTUH, Hyderabad, Approved by AICTE –Accredited by NBA & NAAC-“A” Grade-ISO 9001:2015 Certified)

Mathematics-I

B.Tech – I Year – I Semester

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MATHEMATICS - I

(R18A0021) Mathematics -I

Course Objectives: To learn 1. The concept of rank of a matrix which is used to know the consistency of system of linear equations and also to find the eigen vectors of a given matrix. 2. Finding maxima and minima of functions of several variables. 3. Applications of first order ordinary differential equations. ( Newton’s law of cooling, Natural growth and decay) 4. How to solve first order linear, non linear partial differential equations and also method of separation of variables technique to solve typical second order partial differential equations. 5. Solving differential equations using Laplace Transforms.

UNIT I: Matrices

Introduction, types of matrices-symmetric, skew-symmetric, Hermitian, skew-Hermitian, orthogonal, unitary matrices. Rank of a matrix - echelon form, normal form, consistency of system of linear equations (Homogeneous and Non-Homogeneous). Eigen values and Eigen vectors and their properties (without proof), Cayley-Hamilton theorem (without proof), Diagonalisation. UNIT II:Functions of Several Variables Limit continuity, partial derivatives and total derivative. Jacobian-Functional dependence and independence. Maxima and minima and saddle points, method of Lagrange multipliers, Taylor’s theorem for two variables. UNIT III: Ordinary Differential Equations First order ordinary differential equations: Exact, equations reducible to exact form. Applications of first order differential equations - Newton’s law of cooling, law of natural growth and decay. Linear differential equations of second and higher order with constant coefficients: Non-homogeneous term of the type f(x) = eax, sinax, cosax, xn, eax V and xn V. Method of variation of parameters. UNIT IV: Partial Differential Equations

Introduction, formation of partial differential equation by elimination of arbitrary constants and arbitrary functions, solutions of first order Lagrange’s linear equation and non-linear equations, Charpit’s method, Method of separation of variables for second order equations and applications of PDE to one dimensional (Heat equation). UNIT V: Laplace Transforms Definition of Laplace transform, domain of the function and Kernel for the Laplace transforms, Existence of Laplace transform, Laplace transform of standard functions, first shifting Theorem, Laplace transform of functions when they are multiplied or divided by “t”, Laplace transforms of derivatives and integrals of functions, Unit step function, Periodic function. Inverse Laplace transform by Partial fractions, Inverse Laplace transforms of functions when they are multiplied or divided by ”s”, Inverse Laplace Transforms of derivatives and integrals of functions, Convolution theorem, Solving ordinary differential equations by Laplace transforms.

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TEXT BOOKS: i) Higher Engineering Mathematics by B V Ramana ., Tata McGraw Hill. ii) Higher Engineering Mathematics by B.S. Grewal, Khanna Publishers. iii) Advanced Engineering Mathematics by Kreyszig, John Wiley & Sons.

REFERENCE BOOKS: i)Advanced Engineering Mathematics by R.K Jain & S R K Iyenger, Narosa Publishers. ii)Advanced Engineering Mathematics by Michael Green Berg, Pearson Publishers . iii)Engineering Mathematics by N.P Bali and Manish Goyal.

Course Outcomes: After learning the concepts of this paper the student will be able to 1.Analyze the solution of the system of linear equations and to find the Eigen values and Eigen vectors of a matrix. 2.Find the extreme values of functions of two variables with / without constraints. 3.Solve first and higher order differential equations. 4.Solve first order linear and non-linear partial differential equations. 5.Solve differential equations with initial conditions using Laplac

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UNIT-I

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MATRICES Matrix : A system of mn numbers real (or) complex arranged in the form of an ordered set of ‘m’ rows, each row consisting of an ordered set of ‘n’ numbers between [ ] (or) ( ) (or) || || is called a matrix of order m xn.

a11 a12...... a1n  a a ...... a   21 12 2n  Eg: ......  = [aij ]mxn where 1≤i≤m, 1≤j≤n.   ......  a a ...... a   m1 m2 mn  m x n Order of the Matrix: The number of rows and columns represents the order of the matrix. It is denoted by mxn, where m is number of rows and n is number of columns. Types of Matrices: Row Matrix: A Matrix having only one row is called a “Row Matrix”.

Eg: 1 2 3  x31 Column Matrix: A Matrix having only one column is called a “Column Matrix” . 1  Eg:   1  2    x13

Null Matrix: A= a such that aij=0  i and j. Then A is c ij mxn alled a “Zero Matrix”. It is denoted by Omxn. 0 0 0  Eg: O2x3=  0 0 0 Rectangular Matrix: If A= a , and m  ij mxn n then the matrix A is called a “Rectangular Matrix” 1 1 2  Eg :   is a 2x3 matrix 2 3 4 Square Matrix: If A= a ij mxn and m = n then A is called a “Square Matrix”. 1 1  Eg :   is a 2x2 matrix 2 2 Lower Triangular Matrix: A square Matrix Aa  is said to be lower Triangular nXn ij nxn of aij = 0 if i

aij = 0 if i.>j. i.e. all the elements below the principle diagonal are zeros.

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1 3 8    Eg: 0 4  5 is an Upper triangular matrix 0 0 2   Triangle Matrix: A square matrix which is either lower triangular or upper triangular is called a triangle matrix.

Principal Diagonal of a Matrix: In a square matrix, the set of all aij, for which i = j are called principal diagonal elements. The line joining the principal diagonal elements is called principal diagonal. Note: Principal diagonal exists only in a square matrix.

Diagonal elements in a matrix: A= [aij]nxn, the elements aij of A for which i = j.

i.e. a11, a22….ann are called the diagonal elements of A 1 2 3  Eg: A=  diagonal elements are 1, 5, 9 4 5 6 7 8 9 

Diagonal Matrix: A Square Matrix is said to be diagonal matrix, if aij = 0 for i  j i.e. all the elements except the principal diagonal elements are zeros. Note: 1. Diagonal matrix is both lower and upper triangular.

2. If d1, d2…….dn are the diagonal elements in a diagonal matrix it can be represented as diag d, d, ., d 12 n  3 0 0  Eg : A = diag (3,1,-2)=   0 1 0  0 0  2 Scalar Matrix: A diagonal matrix whose leading diagonal elements are equal is called a

2 0 0 “Scalar Matrix”. Eg : A=   0 2 0 0 0 2 

Unit/Identity Matrix: If A = a such that aij=1 for i = j, and aij=0 for ij then A is ij nxn I called a “Identity Matrix” or Unit matrix. It is denoted by n 1 0 0 1 0   Eg: I2 =   ; I3=0 1 0 0 1 0 0 1  Trace of Matrix: The sum of all the diagonal elements of a square matrix A is called Trace of a matrix A, and is denoted by Trace A or tr A.

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a h g    Eg : A = h b f  then tr A = a+b+c g f c  Singular & Non Singular Matrices: A square matrix A is said to be “Singular” if the determinant of │A│= 0, Otherwise A is said to be “Non-singular”. Note: 1. Only non-singular matrices possess inverse. 2. The product of non-singular matrices is also non-singular. Inverse of a Matrix: Let A be a non-singular matrix of order n if there exist a matrix B such that AB=BA=I then B is called the inverse of A and is denoted by A-1. If inverse of a matrix exist, it is said to be invertible. Note: 1. The necessary and sufficient condition for a square matrix to posses inverse is that

2 .Every Invertible matrix has unique inverse. || ≠ . 3. If A, B are two invertible square matrices then AB is also invertible and

AB1  B11 A  

AdjA 4. A1 where detA  0, det A Theorem: The inverse of a Matrix if exists is Unique. Note: 1. (A-1)-1 = A 2. I-1 = I Theorem: If A, B are invertible matrices of the same order, then (i). (AB)-1 = B-1A-1 (ii). (A1)-1 = (A-1)1 Sub Matrix: - A matrix obtained by deleting some of the rows or columns or both from the given matrix is called a sub matrix of the given matrix.

1 5 6 7  Eg: Let A =   . Then 8 9 10 is a sub matrix of A obtained by deleting first 8 9 10 5     43 5  x32 3 4 5 1 row and 4th column of A. Minor of a Matrix: Let A be an mxn matrix. The determinant of a square sub matrix of A is called a minor of the matrix. Note: If the order of the square sub matrix is ‘t’ then its determinant is called a minor of order ‘t’.

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2 1 1 3 1 2 Eg: A =   be a 4x3 matrix 1 2 3   5 6 7 

Here B = is a sub-matrix of order ‘2’ B = 2-3 [= - ] 1 is a minor of order ‘2

And C = is a sub-matrix of order ‘3’ det C = 2(7-12)-1(21-10)+(18-5)[ ] = -9 Properties of trace of a matrix: Let A and B be two square matrices and  be any scalar 1) tr (  A) =  (tr A) ; 2) tr(A+B) = trA + trB ; 3) tr (AB) = tr(BA) Idempotent Matrix: A square matrix A Such that A2=A then A is called “Idempotent Matrix”.

Eg: A = Involutary[ Matrix:] A square matrix A such that A2 = I then A is called an Involutary Matrix.

Eg: A = Nilpotent[ Matrix:] A square matrix A is said to be Nilpotent if there exists a + ve integer n such that An = 0 here the least n is called the Index of the Nilpotent Matrix.

Eg: A = Transpose[ of ] a Matrix: The matrix obtained by interchanging rows and columns of the given matrix A is called as transpose of the given matrix A. It is denoted by AT or A1

Eg: A = Then AT = Properties[ of ] transpose of[ a matrix:] If A and B are two matrices and AT , BT are their transposes then

T T T T 1) AAT   ; 2)ABAB  TT  ; 3) KA   KAT ; 4) AB   BTT A

Symmetric Matrix: A square matrix A is said to be symmetric if AAT 

T If A a  then A  a  nxn where aaij ji ij nxn  ji  a h g  Eg:   is a symmetric matrix h b f  g f c 

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Skew-Symmetric Matrix: A square matrix A is said to be Skew symmetric If AAT  .

If A a  then AT   a  where a   a ij nxn  ji  nxn ij ji .

 0 a  b Eg :   is a skew – symmetric matrix  a 0 c   b  c 0  Note: All the principle diagonal elements of a skew symmetric matrix are always zero.

Since aij = -aij  aij = 0 Theorem: Every square matrix can be expressed uniquely as the sum of symmetric and skew symmetric matrices. 1 1 Proof: Let A be a square matrix, A = AA  = AAAATT    = 2 2 11 1 1 AAAATT     = P + Q, where P = AA T  ; Q = AA T  22 2 2 Thus every square matrix can be expressed as a sum of two matrices.

T TTTTT1 1TT 1 1 T Consider PAAAAAA          = AA T  =P, since PP , 2 2 2   2 P is symmetric

T TTTTT1 1TT 1 1 Consider QAAAAAA          = - AA T  = - Q 2 2 2   2

SinceQQT  , Q is Skew-symmetric. To prove the representation is unique: Let A= R+S 1  be the representation, where R is symmetric and S is skew symmetric. i.e. RRSSTT,  

Consider ARSRSRSTTT  T      2  1 1   2  AASSAAQ TT  2      2 Therefore every square matrix can be expressed as a sum of a symmetric and a skew symmetric matrix

Ex. Express the given matrix A as a sum of a symmetric and skew symmetric matrices 2 4 9 where A=  14 7 13 9 5 11 

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2 14 3 Solution: T  A 4 7 5 9 3 11

4 10 12   2 5 6  1 A ATT 10 14 18   P  A  A    5 7 9  ; P is symmetric  2   12 18 22   6 9 11 

0 18 6   0 9 3  1 A AT 18 0 8   Q  A  A    9 0 4  ; Q is skew  symmetric  2   6  8 0    3  4 0 

2 5 6 0 9 3 Now A =P+Q = 5 7 9 +  9 0 4 6 9 11 3 4 0 Orthogonal Matrix: A square matrix A is said to be an Orthogonal Matrix if AAT=ATA=I, Similarly we can prove that A=A-1; Hence A is an orthogonal matrix. Note: 1. If A, B are orthogonal matrices, then AB and BA are orthogonal matrices. 2. Inverse and transpose of an orthogonal matrix is also an orthogonal matrix. Result: If A, B are orthogonal matrices, each of order n then AB and BA are orthogonal matrices. Result: The inverse of an orthogonal matrix is orthogonal and its transpose is also orthogonal

Solved Problems : cos sin   1. Show that A =   is orthogonal.  sin  cos 

cos sin   T cos  sin   Sol: Given A =   then A =    sin  cos  sin  cos 

T cos sin   cos  sin   Consider A.A =      sin  cos  sin  cos 

cos2   sin 2   cos sin   cos sin   1 0 = = I  2 2     sin  cos  cos sin  cos   sin   0 1 A is orthogonal matrix.

1 2 2  1 2. Prove that the matrix 2 1 2  is orthogonal. 3   2 2 1

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1 2 2  1 2 2  1   T 1 Sol: Given A = 2 1 2 Then A = 2 1 2  3   3   2 2 1 2 2 1

1 2 2  1 2 2  9 0 0 1 0 0 T 1   1     Consider A .A = 2 1 2  2 1 2 =0 9 0 =0 1 0 = I 9     9     2 2 1 2 2 1 0 0 9 0 0 1 A.AT = I T ⇒Similarly A .A = I Hence A is orthogonal matrix

0 2b c 3. Determine the values of a, b, c when   is orthogonal. a b  c a  b c Sol: - For orthogonal matrix AAT = I

0 2b c 0 a a  T So, AA =   2b b  b  I a b  c   a  b c c  c c 

4b 2  c 2 2b 2  c 2  2b 2  c 2  1 0 0  2 2 2 2 2 2 2 2  2b  c a  b  c a  b  c  = I =   0 1 0   2 2 2 2 2 2 2 2   2b  c a  b  c a  b  c  0 0 1  Solving 2b2-c2 =0, a2-b2-c2 =0 We get c =  2b a2 =b2+2b2 =3b2

 a =  3b From the diagonal elements of I 1 4b2+c2= 1  4b2+2b2=1 (since c2=2b2)  b =  6 1 1 1 a =  3b = ; b = ; c =  2b =  2 6 3

2 3 1 4. Is matrix  Orthogonal? 4 3 1 3 1 9

2 3 1 2 4 3  Sol:- Given A= 4 3 1      3 3 1 3 1 9 119

2 3 1   2 4 3  14 0 0   4 3 1   3 3 1  =  I       0 26 0 3 3 1 9   1 1 9  0 0 91

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      I3  Matrix is not orthogonal. Complex matrix: A matrix whose elements are complex numbers is called a complex matrix. Conjugate of a complex matrix: A matrix obtained from A on replacing its elements by the corresponding conjugate complex numbers is called conjugate of a complex matrix. It is denoted by A

If A aij, A a ij , where aij is the conjugate of aij . mxn mxn 2 3i 5 2 3i 5 Eg: If A= then A = 6 7ii  5  6 7ii  5  Note: If A and B be the conjugate matrices of A and B respectively, then (i) A  =A (ii) AB = A + B (iii)KA  = K A Transpose conjugate of a complex matrix: Transpose of conjugate of complex matrix is called transposed conjugate of complex matrix. It is denoted by A or A* .

Note: If A and B be the transposed conjugates of A and B respectively, then

  (i) A  = A (ii) ABAB   

(iii) KA   KA (iv) AB   A B

Hermitian Matrix: A square matrix A is said to be Hermitian Matrix iff AA  . 4 1 3i 4 1 3i 4 1 3i θ  Eg: A= then A = and A = 1 3i 7 1 3i 7 1 3i 7 Note: 1. In Hermitian matrix the principal diagonal elements are real. 2. The Hermitian matrix over the field of Real numbers is nothing but real symmetric matrix.

3. In Hermitian matrix A= a , a a i, j . ij nxn ij ji Skew-Hermitian Matrix: A square matrix A is said to be Skew-Hermitian Matrix iff AA  .

32ii 32ii T  3i  2  i Eg: Let A= then A = and A     2 ii  2 ii 2  i i 

T  A  =-A  A is skew-Hermitian matrix. Note: 1. In Skew-Hermitian matrix the principal diagonal elements are either Zero or Purely Imaginary.

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2. The Skew- Hermitian matrix over the field of Real numbers is nothing but real Skew - Symmetric matrix.

3. In Skew-Hermitian matrix A= a , a  a  i, j . ij nxn ij ji Unitary Matrix: A Square matrix A is said to be unitary matrix iff

   1 AA A A  I or A  A 1   4  2  4i Eg: B    6 2  4i  4  Theorem1: Every square matrix can be uniquely expressed as a sum of Hermitian and skew – Hermitian Matrices.

1 1 1 AAAAAAAA(2 )  (  )  (    ) Proof: - Let A be a square matrix write 2 2 2 11 A()(). A  A  A  A i eA  P  Q 22 11 Let PAAQAA  ;    22 11 Consider PAAAAAAP()()()            22

I.e. PPP  , is Hermitian matrix.

 1 1 1 QAAAAAAQ()()             2 2 2

Ie QQQ  , is skew – Hermitian matrix. Thus every square matrix can be expressed as a sum of Hermitian & Skew Hermitian matrices.

To prove such representation is unique: Let A = R+S------(1) be another representation of A where R is Hermitian matrix & S is skew – Hermitian matrix. RRSS ;  

Consider ARSRSRS()          . Ie ARA        (2)

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1 1   2  A  A  2 R ie R  A  A   P 2 1 1   2  A  A  2 S ie S  A  A   Q 2 Thus every square matrix can be uniquely expressed as a sum of Hermitian & skew Hermitian matrices.

Solved Problems :

3 7 4ii  2  5 1) If A=  then show that A is Hermitian and iA is skew- 7 4ii  2 3  2  5ii 3  4 Hermitian.

3 7 4ii  2  5 Sol: Given A=  then 7 4ii  2 3  2  5ii 3  4

3 7 4ii  2  5 3 7 4ii  2  5  And T  A7  4 i  2 3  i A  7  4 i  2 3  i 2  5ii 3  4 2  5ii 3  4

T AA  Hence A is Hermitian matrix. Let B= iA

3i 4 7 i  5  2 i i.e B=  then 4  7i  2 i  1  3 i 5 2i 1 3 i 4 i

3i 4  7 i  5  2 i  B 4  7 i 2 i  1  3 i 5 2i 1  3 i  4 i

3i  4  7 i 5  2 i 3i 4  7i  5  2i T   B4  7 i 2 i 1  3 i      )1(  4  7i  2i 1 3i   B 5  2i  1  3 i  4 i  5  2i 1 3i 4i  T  B  =-B B= iA is a skew Hermitian matrix. 2). If A and B are Hermitian matrices, prove that AB-BA is a skew-Hermitian matrix. Sol: Given A and B are Hermitan matrices

T T  AA   And BB   ------(1)

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T T Now AB BA   AB  BA 

T AB BA 

TTTTTT AB   BA   B   A   A   B  BA AB (By (1))  AB  BA  Hence AB-BA is a skew-Hemitian matrix.

a ic  b  id 2 2 2 2 3). Show that A=  is unitary if and only if a +b +c +d =1 b id a ic a ic  b  id Sol: Given A=  b id a ic a ic  b  id Then A   b id a ic

 T  a  ic b  id  Hence A  A      b  id a  ic 

 a  ic  b  id  a  ic b  id   AA     b  id a  ic  b  id a  ic  a2 b 2  c 2  d 2 0 = 2 2 2 2 0 a b  c  d

AA I if and only if a2 b 2  c 2  d 2 1

0 1 2i 1 4)Given that A= , show that I  A I  A  is a unitary matrix. 1 2i 0

1 0  0 1 2i Sol: we have I  A       0 1 1 2i 0 

1 1 2i   And 1 2i 1

1 0  0 1 2i I  A       0 1 1 2i 0 

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1 1 2i =  1 2i 1

1 1  1 1 2i (I  A)  2   1 4i 1 1 2i 1 

1 1 1 2i   6 1 2i 1

Let B  I  A I  A 1

1  1 1 2i 1 1 2i 1 1 1(  2i)(1 i)2 1 2i 1 2i  B        6 1 2i 1 1 2i 1  6  1 2i 1 2i (1 2i)(1 i)2 1

1   4  2  4i B    6 2  4i  4 

1 4  2  4i T 1 4 2 4i Now B   and B    6 2 4i 4 6 2  4i  4

T 1 4  2  4i   4 2  4i BB     36 2 4i 4  2  4i  4 

1 36 0   1 0        I 36 0 36   0 1 

T BB   1 i.e. B is unitary matrix.

I  A I  A 1 is a unitary matrix.

5) Show that the inverse of a unitary matrix is unitary.

Sol: Let A be a unitary matrix. Then AA  I

1 i.e AA   I 1

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1 AAI  1

 AAI11 

Thus A1 is unitary. Rank of a Matrix: Let A be mxn matrix. If A is a null matrix, we define its rank to be ‘0’. If A is a non-zero matrix, we say that ‘r’ is the rank of A if i. Every (r+1)th order minor of A is ‘0’ (zero) & ii. At least one rth order minor of A which is not zero. It is denoted by  (A) and read as rank of A. Note: 1. Rank of a matrix is unique. 2. Every matrix will have a rank. 3. If A is a matrix of order mxn, then Rank of A ≤ min (m,n) 4. If ρ (A) = r then every minor of A of order r+1, or minor is zero.

5. Rank of the Identity matrix In is n. 6. If A is a matrix of order n and A is non-singular then ρ (A) = n 7. If A is a singular matrix of order n then  (A) < n Important Note: 1. The rank of a matrix is ≤ r if all minors of (r+1)th order are zero. 2. The rank of a matrix is ≥ r, if there is at least one minor of order ‘r’ which is not equal to zero.

1 2 3 1. Find the rank of the given matrix   3 4 4 7 10 12

1 2 3 Sol: Given matrix A =   3 4 4 7 10 12 det A = 1(48-40)-2(36-28)+3(30-28) = 8-16+6 = -2 ≠ 0 We have minor of order 3 ρ (A) =3

1 2 3 4 2. Find the rank of the matrix∴   5 6 7 8  8 7 0 5 Sol: Given the matrix is of order 3x4 Its Rank ≤ min(3,4) = 3

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Highest order of the minor will be 3.

1 2 3  Let us consider the minor   5 6 7  8 7 0  Determinant of minor is 1(-49)-2(-56) + 3(35-48) = -49+112-39 = 24 ≠ 0. Hence rank of the given matrix is ‘3’. Elementary Transformations on a Matrix: th th i). Interchange of i row and j row is denoted by Ri ↔ Rj th (ii). If i row is multiplied with k then it is denoted by Ri kRi (iii). If all the elements of ith row are multiplied with k and added to the corresponding th elements of j row then it is denoted by Rj  Rj +kRi

Note: 1. The corresponding column transformations will be denoted by writing ‘c’. i.e ci ↔cj, ci  k cj cj  cj + kci 2. The elementary operations on a matrix do not change its rank. Equivalance of Matrices: If B is obtained from A after a finite number of elementary transformations on A, then B is said to be equivalent to A.It is denoted as B~A. Note : 1. If A and B are two equivalent matrices, then rank A = rank B. 2. If A and B have the same size and the same rank, then the two matrices are equivalent. Elementary Matrix or E-Matrix: A matrix is obtained from a unit matrix by a single elementary transformation is called elementary matrix or E-matrix. Notations: We use the following notations to denote the E-Matrices.

th th 1) Eij  Matrix obtained by interchange of i and j rows (columns).

2) Eik   Matrix obtained by multiplying ith row (column) by a non- zero number k.

th 3) Eij k   Matrix obtained by adding k times of j row (column) to ith row (column). Echelon form of a matrix: A matrix is said to be in Echelon form, if (i) Zero rows, if any exists, they should be below the non-zero row. (ii) The first non-zero entry in each non-zero row is equal to ‘1’. (iii) The number of zeros before the first non-zero element in a row is less than the number of such zeros in the next row. Note : 1. The number of non-zero rows in echelon form of A is the rank of ‘A’. 1. The rank of the transpose of a matrix is the same as that of original matrix. 2. The condition (ii) is optional.

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 01 0 0 Eg: 1.   is a row echelon form.  10 0 0  00 11     00 00 

1  3 1 2.   is a row echelon form.  10 1   00 0 

Solved Problems :

2 3 7 1. Find the rank of the matrix A =   by reducing it to Echelon form. 3  2 4  1  3 1

2 3 7 Sol: Given A =   Applying row transformations on A. 3  42  1  3 1 1  3 1   R1 ↔ R3 A ~ 3  2 4 2 3 7

1  3 1    R2 → R2 –3R1; R3→ R3 -2R1 ~0 7 7 0 9 9

1  3 1   R2 → R2/7,R3→ R3/9 ~0 1 1  0 1 1 

1  3 1   R3 → R3 –R2 ~0 1 1  0 0 0  This is the Echelon form of matrix A. The rank of a matrix A= Number of non – zero rows =2 4 4  3 1  1 1 1 0  2. For what values of k the matrix   has rank ‘3’. k 2 2  2   9 9 k 3  Sol: The given matrix is of the order 4x4 If its rank is 3  det A =0

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4 4  3 1  1 1 1 0  A =   k 2 2  2   9 9 k 3 

Applying R2 → 4R2-R1, R3 →4R3 – kR1, R4 → 4R4 – 9R1 4 4  3 1  0 0 1 1  We get A ~   0 8  4k 8  3k 8  k   0 0 4k  27 3  Since Rank A = 3  det A =0 0 1 1  4 8  4k 8  3k 8  k  0 0 4k  27 3

 1[(8-4k) 3]-1(8-4k) (4k+27)] = 0  (8-4k) (3-4k-27) = 0  (8-4k)(-24-4k) =0  (2-k)(6+k)=0  k =2 or k = -6

2 1 3 5 3). Find the rank of the matrix using echelon form 4 2 1 3 A   8 4 7 13  8 4 3 1

2 1 3 5 Sol: Given 4 2 1 3 A   8 4 7 13  8 4 3 1 2 1 3 5 0 0 5 7 By applying RRRRRRRRR2 2 2 1 ; 3  3  4 1 ; 4  4  4 1  ~  0 0 5 7  0 0 15 21

2 1 3 5  RR12R3 0 0 5 7 RRR1,, 2  3   ~  1  1  3 0 0 5 7  0 0 5 7

2 1 3 5  RRRRRR ,   0 0 5 7 3 3 2 4 4 2  ~  0 0 0 0  0 0 0 0

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A is in echelon form Rank of A = 2

1 2 0 1 ⇒4). Find the rank of the matrix∴ A =  by reducing into echelon form. 2 1 1 0 3 3 1 1  1 1 1 1

1 2 0 1 Sol: By applying RRRRRRRRR 2 ;   3 ;   A ~  2 2 1 3 3 1 4 4 1 0 3 1 2 0 3 1 2  0 3 1 2

1 2 0 1 RRR A ~  3 3 2 0 3 1 2 0 0 0 0  0 3 1 2

1 2 0 1 RR A ~  34 0 3 1 2 0 3 1 2  0 0 0 0

1 2 0 1 RRR A ~  3 3 2 0 3 1 2 0000  0000 Clear it is in echelon form, rank of A = 2 Normal form/Canonical form of a Matrix: Every non-zero Matrix can be reduced to any one of the following forms.

IIrr0     ;IIrr 0  ;   ;   Known as normal forms or canonical forms by using Elementary 0 0   0  row or column or both transformations where Ir is the unit matrix of order ‘r’ and ‘O’ is the null matrix. Note: 1.In this form “the rank of a matrix is equal to the order of an identity matrix. 2. Normal form another name is “canonical form”

Solved Problems : 1 2 3 4    1. By reducing the matrix 2 1 4 3  into normal form, find its rank. 3 0 5 10

1 2 3 4    Sol: Given A =2 1 4 3  3 0 5 10

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1 2 3 4    R2 → R2 – 2R1; R3 → R3 – 3R1 A ~ 0  3  2 5  0  6  4  22

1 2 3 4    R3 → R3/-2 A ~ 0  3  2  5 0 3 2 11 

1 2 3 4    R3 → R3+R2 A ~ 0  3  2  5 0 0 0 6 

0 0 0 0    c2→ c2 - 2c1, c3→c3-3c1, c4→c4-4c1 A ~ 0  3  2  5 0 0 0 6 

1 0 0 0   c3 → 3 c3 -2c2, c4→3c4-5c2 A ~ 0  3 0 0 0 0 0 18

1 0 0 0   c2→ c2 /-3, c4→c4/18 A ~ 0 1 0 0  0 0 0 1 

1 0 0 0   c4 ↔ c3 A~ 0 1 0 0  0 0 1 0 

This is in normal form [I3 0], Hence Rank of A is ‘3’.

1 1 1 1 2). Find the rank of the matrix A∴ =  by reducing into canonical form or 1 2 3 4 2 3 5 5  3 4 5 8 normal form.

1 1 1 1 Sol: Given A =  1 2 3 4 2 3 5 5  3 4 5 8

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1 1 1 1 By applying RRRRRRRRR ,   2 ,   3 0 1 2 5 2 2 1 3 3 1 4 4 1  ~  0 1 3 7  0 7 8 5 1 1 1 1 RRRRRR ,7   0 1 2 5 3 3 2 4 4 2  ~  0 0 1 2  0 0 6 30 1 1 1 1 RRR6 0 1 2 5 4 4 3  ~  0 0 1 2  0 0 0 18

1 1 1 1 R R  4 0 1 2 5 4  ~  18 0 0 1 2  0 0 0 1

1 0 0 0 Apply CCCCCCCCC ,,     0 1 2 5 2 2 1 3 3 1 4 4 1  ~  0 0 1 2  0 0 0 1

1 0 0 0 CCCCCC 2 ;   5 0 1 0 0 3 3 2 4 4 2  ~  0 0 1 2  0 0 0 1 1 0 0 0 CCC2 0 1 0 0 4 4 3  ~  0 0 1 0  0 0 0 1

Clearly it is in the normal form I4  Rank of A = 4 3). Define the rank of the matrix and find the rank of the following matrix 2 1 3 5 4 2 1 3  8 4 7 13  8 4 3 1 2 1 3 5  Sol: Let A= 4 2 1 3 8 4 7 13  8 4 3 1 2 1 3 5 RRR2 22 1 0 0 5 7 RRR3 34 1 A ~  0 0 5 7 RRR4 4 4 1  0 0 15 21

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2 1 3 5 RRR 3 3 2 0 0 5 7 RRR3 A ~  4 4 2 0 0 0 0  0 0 0 0 It is in echelon form. So, rank of matrix = no. of non zero rows in echelon form. Rank( A ) 2

2 1 3 4 4). Reduce the matrix A to normal form and hence find its rank 0 3 4 1 A   2 3 7 5  2 5 11 6

2 1 3 4 0 3 4 1 Sol: Given A   2 3 7 5  2 5 11 6 1 1 3 4 1 0 3 4 1 CC11 A ~  2 1 3 7 5  1 5 11 6 1 1 3 4 RRR  3 3 2 0 3 4 1 A ~  RRR4 4 1 0 2 4 1  0 4 8 2

1 1 3 4  RRR 0 1 0 0 2 2 3 A ~  0 2 4 1  0 4 8 2

RRR2 1 0 0 0 3 3 2 0 1 0 0 RRR4 A ~  4 4 2 0 0 4 1  0 0 8 2

1000  RRR2 0 1 0 0 4 4 3 A ~  0 0 4 1  0 0 0 0

1000 0 1 0 0 CCC44 4 3 A ~  0 0 4 1  0 0 0 0

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1 0 0 0 1  I 0 0 1 0 0 3 CC33 A ~ A ~  3 0 0 1 0 00  0000 ⇒ This is in normal form. Thus Rank of matrix = Order of identify matrix. Rank (A ) 3

0 1 2 2 5). Reduce the matrix A =  into canonical form and then find its rank. 4 0 2 6 2 1 3 1

1 0 2 2 Sol: ApplyCC  12 A ~ 0 4 2 6 1 2 3 1

1 0 2 2 RRR  3 3 1 A ~ 0 4 2 6 0 2 1 3

1 0 0 0 CCCCCC 2 ;   2  3 3 1 4 4 1 A ~ 0 4 2 6 0 2 1 3

R 1 0 0 0 R  2  2 2 A ~ 0 2 1 3 0 0 0 0

1000 CC  23 A ~ 0 1 2 3 0 0 0 0 1 0 0 0 CCCCCC 2 ;   3  3 3 2 4 4 2 A ~ 0 1 0 0 0000

I2 0 Which is in the normal form ,   (A) = 2 00 Note: .If A is an mxn matrix of rank r, there exists non-singular matrices P and Q such that

 I r 0 PAQ =   0 0  Suppose we want to find P and Q we have procedure.

Let order of matrix ‘A’ is ‘3 i.e. A = I3 A I3 1 0 0 1 0 0     A = 0 1 0  A 0 1 0  0 0 1 0 0 1

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Now we go on applying elementary row operations and column operations on the matrix A

 I r 0 (L.H.S) until it is reduced to the normal form   0 0  Every row operations will also be applied to the pre-factor of on R.H.S Every column operation will also be applied to the post –factor of on R.H.S.

Solved Problems : 1 0  2   1. Find the non-singular matrices P and Q is of the normal form where A =2 3  4 3 3  6 

Sol: Write A = I3 A I3 1 0  2 1 0 0 1 0 0       ~2 3  4 = 0 1 0  A 0 1 0  3 3  6  0 0 1 0 0 1

1 0  2 1 0 0  1 0 0       R2 → R2-2R1, R3→R3-3R1 ~0 3 0  =  2 1 0 A 0 1 0  0 3 0   3 0 1 0 0 1

1 0  2 1 0 0  1 0 0 1       R3 → R3-R2, R2 → /3 R2 ~0 1 0    3/2 3/1 0 A 0 1 0  0 0 0  1 1 1  0 0 1

1 0 0 1 0 0  1 0 2        c3 → c3-2c1 ~ 0 1 0  =  3/2 3/1 0 A 0 1 0 0 0 0 1 1 1  0 0 1

1 0 0  1 0 2      ~ = PAQ where P = 3/2 3/1 0 Q = 0 1 0 1 1 1  0 0 1 [ ]     2. Find the non-singular matrices P and Q such that the normal form of A is P A Q.

1 3 6 1 Where A =   . Hence find its rank. 1 4 5 1  1 5 4 3 

Sol: we write A = I3A I4

1 3 6 1 1 0 0 1 0 0 0     ~   =   A 0 1 0 0 1 4 5 1 0 1 0      0 0 01  1 5 4 3  0 0 1   0 0 10 

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1 0 0 0 1 3 6 1  1 0 0    0 1 0 0  R2 →R2- R1, R3→R3-R1 ~ 0 1 1 2 = 1 1 0 A       0 0 01  0 2  2 4 1 0 1    00 10 

1 3 6 1  1 0 0 0 1 0 0    R3 →R3-2R2 ~   = A 0 1 0 0 0 1 1 2  11 0      0 0 01  0 0 0 0  1  2 1   0 0 10 

Applying c2→c2-3c1,c3 →c3-6c1, and c4→c4+c1, we get.

1 0 0 0  1 0 0 1  3  16    ~ 0 1 1 2 =   A 0 1 0 0    11 0      0 0 01  0 0 0 0 1  2 1   0 0 10 

Applying c3→c3+c2 and c4→c4-2c2, we get. 1  3  9 7 1 0 0 0 1 0 0    ~ =   A 0 1 1  2  10 0 0 1 1 0       0 0 1 0  0 0 0 0  1  2 1   0 0 0 1 

1  3  9 7  I 2 0 1 0 0     = P A Q Where P = Q= 0 1 1  2    11 0   0 0    0 0 1 0  1  2 1   0 0 0 1 

 I 2 0 Here A ~  , Hence ρ(A) =2 0 0  ∴ System of linear equations: In this chapter we shall apply the theory of matrices to study the existence and nature of solutions for a system of m linear equations in ‘n’ unknowns.

The system of m linear equations in ‘n’ unknowns x1, x 2, x 3  xn given by

a11 x 1 a 12 x 2 ......  a 1nn x  b 1 a x a x ......  a x  b 21 1 22 2 2nn 2 } ------(1) ......

am1 x 1 a m 2 x 2 ......  a mn x n  b m

The above set of equations can be written in the Matrix form as A X = B

a a a x 11 12 1n 1 b1   a a a x2 b 21 22 2n  2       .  . (2)         . .   an12 a n a nn xn bm A-Coefficient Matrix; X-Set of unknowns; B-Constant Matrix

Homogeneous Linear Equations: If b 12bb ...... m  0 then B = 0

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Hence eqn (2) Reduces to AX = 0 which are known as homogeneous linear equations Non-Homogeneous Linear equations:

It at least one of b1, b2----bm is non zero. Then B  0, the system Reduces to AX = B is known as Non-Homogeneous Linear equations.

Solutions: A set of numbers x12, x xn which satisfy all the equations in the system is known as solution of the system. Consistent: If the system possesses a solution then the system of equations is said to be consistent. Inconsistent: If the system has no solution then the system of equations is said to be Inconsistent. Augmented Matrix: A matrix which is obtained by attaching the elements of B as the last column in the coefficient matrix A is called Augmented Matrix. It is denoted by [A|B]

a11 a 12 a 13 a 1n : b 1 A B C  a a a   a: b   21 22 23 2n 2 am1 a m 2 a m 3 a mn : b 3

1. If (AB / ) (A) , then the system of equations AX = B is consistent (solution exits). a). If (AB / ) (A) = r = n (no. of unknowns) system is consistent and have a unique solution b). If (AB / ) (A) = r < n (no. of unknowns) then the system of equations AX = B will have an infinite no. of solutions. In this case (n-r) variables can be assigned arbitrary values.

2. If (AB / ) (A) then the system of equations AX=B is inconsistent (no solution). In case of homogeneous system AX = 0, the system is always consistent.

(or) x1 = 0, x2 = 0, ------, xn = 0 is always the solution of the system known as the” zero solution “. Non-trivial solution: If (A / B) (A) = r < n (no. of unknowns) then the system of equations AX = 0 will have an infinite no. of non zero (non trivial) solutions. In this case (n-r) variables can be assigned arbitrary values. Also we use some direct methods for solving the system of equations. Note: The direct methods are Cramer’s rule, Matrix Inversion, Gaussian Elimination, Gauss Jordan, Factorization Tridiagonal system. These methods will give a unique solution. Procedure to solve AX = B (Non Homogeneous equations)

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Let us first consider n equations in n unknowns ie. m=n then the system will be of the form a11x1+a12x2+a13x3+…..+a1nxn =b1 a21x1+a22x2+a23x3+…..+a2nxn =b2 ………………………………….. an1x1+an2x2+an3x3+…..+annxn =bn The above system can be written as AX = B ------(1) Where A is an n n matrix. Solving AX = B using Echelon form: Consider the system of m equations in n unknowns given by a11x1+a12x2+a13x3+…..+a1nxn =b1 a21x1+a22x2+a23x3+…..+a2nxn =b2 ………………………………….. am1x1+am2x2+am3x3+…..+amnxn =bm We know this system can be we write as AX = B

a11 a 12 a 1n b 1  a a a b The augmented matrix of the above system is [A / B] =21 22 2n 2   am12 a m a mn b m The system AX = B is consistent if ρ (A) = ρ[A/B] i). ρ (A) = ρ [A/B] = r < n (no. of unknowns).Then there is infinite no. of solutions. ii). ρ (A) = ρ [A/B] = number of unknowns then the system will have unique solution. iii). ρ (A) ≠ ρ [A/B] the system has no solution. Solved Problems : 1). Show that the equations x+y+z = 4, 2x+5y-2z =3, x+7y-7z =5 are not consistent. 1 1 1  x 4       Sol: Write given equations is of the form AX = B i.e; 2 5  2 y  3 1 7  7  z  5

1 1 1 4   Consider the Augment matrix is [A /B]  [A/B] = 2 5  2 3 1 7  7 5

1 1 1 4    Applying R2 →R2-2R1 and R3 → R3-R1, we get [A/B] ~ 0 3  4  5 0 6  8 1 

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1 1 1 4    Applying R3→ R3-2R2, we get [A/B] ~ 0 3  4  5 0 0 0 11 ρ (A) =2 and ρ (A/B) =3

∴The given system is inconsistent as ρ (A) ≠ ρ[A/B]. 2). Show that the equations given below are consistent and hence solve them x-3y-8z = -10, 3x+y-4z =0, 2x+5y+6z =3 1  3  8 x 10       Sol: Matrix notation is 3 1  4  y  0  2 5 6  z  3 

1  3  8 10 Augmented matrix [A/B] is [A/B] =   3 1  4 0  2 5 6 3 

1  3  8 10  R2 → R2-3R1 R3 → R3 -2R1 ~  0 10 20 03  0 11 22 23

1  3  8 10  R2 →1/10 R2 ~  0 1 2 3  0 11 22 23 

1  3  8 10  R3 →R3-11R2 ~  0 1 2 3  0 0 0 10 This is the Echelon form of [A/B] ρ (A) =2, ρ (A/B) =3

ρ (A) ≠ ρ[A/B]. ∴ The given system is inconsistent. 3). Find whether the following equations are consistent, if so solve them. x+y+2z=4, 2x-y+3z=9, 3x-y-z=2 1 1 2  x 4       Sol: We write the given equations in the form AX=B i.e; 2 1 3  y  9 3 1 1 z  2

1 1 2 4    The Augmented matrix [A/B] = 2 1 3 9  3 1 1 2   

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1 1 2 4    Applying R2→ R2-2R1 and R3→R3-3R1, we get [A/B] ~ 0  3 1 1  0  4  7 10 

1 1 2 4    Applying R3 →3R3-4R2, we get [A/B] ~ 0  3 1 1  0 0 17  34 

this matrix is in Echelon form. ρ(A) = 3 and ρ(A/B) = 3 Since ρ (A) =ρ [A/B]. The system of equations is consistent.

Here the number of unknowns∴ is 3 Since ρ (A) =ρ [A/B] = number of unknowns  The system of equations has a unique solution 1 1 2  x 4        We have 0  3 1  y  1  0 0 17  z   34        -17z = -34  z = 2 -3y-z =1  -3y =z+1  -3y =3  y= -1 and x+y+2z =4  x=4-y-2z =4+1-4=1  x=1, y=-1, z=2 is the solution. 4). Show that the equations x+y+z=6, x+2y+3z=14, x+4y+7z=30 are consistent and solve them. 1 1 1 x 6        Sol: We write the given equations in the form AX=B i.e. 1 2 3  y  14 1 4 7  z  30

1 1 1 6  The Augmented matrix [A/B] =   1 2 3 14  1 4 7 30 

1 1 1 6  Applying R2→ R2-R1 and R3→R3-R1, we get [A/B] ~   0 1 2 8  0 3 6 24 

1 1 1 6  Applying R3 →R3-3R2, we get [A/B] ~   0 1 2 8  0 0 0 0 

This matrix is in Echelon form. ρ(A) = 2 and ρ(A/B) = 2 Since ρ (A) =ρ [A/B].

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The system of equations is consistent. Here the no. of unknowns are 3 Since rank of A is less than the no. of unknowns, the system of equations will have infinite number of solutions in terms of n-r=3-2=1 arbitrary constant. 1 1 1 x 6       The given system of equations reduced form is 0 1 2  y  8 0 0 0 z  0       x+y+z=6……… (1), y+2z=8…….(2) Let z=k, put z=k in (2) we get y=8-2k Put z=k y=8-2k in (1), we get x=6-y-z=6-8+2k=-2+k x=-2+k, y=8-2k, z=k is the solution, where k is an arbitrary constant.

∴5). Show that x23 y  z  ;3x y  2 z  1; 2x 2 y  3 z  2; x  y  z   1are consistent and solve them

1 2 1   3  x 3 1 2   1   y    Sol: The above system in matrix notation is 2 2 3   2  z  31   1 1 1 4 3  1  4 1 AXB

1 2 1 3 3 1 2 1 The Augmented matrix is AB    2 2 3 2  1 1 1 1

RRR2 23 1 1 2 1 3 0 7 5 8 RRR3 32 1 ~   RRR 0 6 5 0 4 4 1  0 3 2 4 1 2 1 3 0 1 0 4 RRR2 2 3 ~  0 6 5 4  0 3 2 4 1 2 1 3 RRR3 3 3 1 0 1 0 4 ~  RRR4 43 2 0 0 5 20  0 0 2 8

1 2 1 3 R R  3 0 1 0 4 3 ~  5 0 0 1 4  0 0 2 8

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1 2  1 3   RRR2 0 1 0 4 4 4 3 ~  0 0 1 4  0 0 0 0 ρ (A) = 3 = ρ (A/B

∴ρ (A) =ρ (A/B) = No. of unknowns = 3 ∴ The given system has unique solution. The systems of equations equivalent to given system are x2 y  z  3  y   4; z  4 x8  4  3 y  4; z  4

x 3  4   1 x  1, y  4, z  4. 6). Solve x y  z 3;3 x  5 y  2 z  8;5 x  3 y  4 z  14

1 1 1  x   3  Sol: -       3 5 2  y   8  5 3 4  z   14 

1 1 1 3 Augmented Matrix is AB    3 5 2 8 5 3 4 14

1 1 1 3 RRR3 2 2 1 ~ 0 8  1  1 RRR5  3 3 1 0 8  1  1  1 1 1 3  RRR ~ 0 8  1  1 3 3 2  0 0 0 0 A   AB  2  Number of unknowns (3)  The system has infinite number of solutions.

x y z3, 8 y  z 1  8 y  z 1 1k1 k  24  1  k  8 k 23  7 k Let z k  y  and x 3   k   8 8 8 8

23 7 23 xk8 8   8  X  y 11 k   X    where k is any real number.  8 8   8  zk01    7). Find whether the following system of equations is consistent. If so solve them. xyz222,32 xyz  5,253 xyz  4, xyz  460.

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1 2 2   2  x 3 2 1   5   y    Sol: In Matrix form it is 2 5 3   4   z   1 4 6   0  AX B

1 2 2 2 RRR3 0 8  5  1 2 2 1 ~  RRR3 32 1 0 9  1  8  RRR 0 2 4 2 4 4 1  1 2 2 2 0 1 4 7 ~  0 9  1  8  RRR 0 2 4 2 2 2 3  1 2 2 2 0 1 4 7 ~  RRR3 39 2 0 0 37 55  RRR2 0 0 12 16 4 4 2 

1 2 2 2  1 ~ 0 1 4 7 RR44 4 0 0 37 55  0 0 3 4 1 2 2 2 0 1 4 7 RRR37 3 ~ is in echelon form 4 4 3 0 0 37 55  0 0 0 17 A 34 and   AB      A     AB . The given system is in consistent.

∴ 8). Discuss for what values of , the simultaneous equations x+y+z = 6, x+2y+3z =10, x+2y+z = have (i). No solution (ii). A unique solution (iii). An infinite number of solutions. 1 1 1 x 6        Sol: The matrix form of given system of Equations is A X = 1 2 3  y  10 = B 1 2  z   

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1 1 1 6    The augmented matrix is [A/B] = 1 2 3 10 1 2   

1 1 1 6    R2 → R2 – R1, R3 → R3-R1 [A/B] ~0 1 2 4  0 1  1   6

1 1 1 6    R3 →R3 – R2 ~0 1 2 4  0 0   3  10 Case (i): let ≠ 3 the rank of A = 3 and rank [A/B] = 3 Here the no. of unknowns is ‘3’ ρ (A) =ρ (A/B)= No. of unknowns

The system has unique solution if ≠3∴ and for any value of ‘’.

Case (ii): Suppose =3 and ≠10. We have ρ (A) = 2, ρ (A/B) = 3 The system has no solution. Case (iii): Let =3 and =10. We have ρ (A) = 2 ρ (A/B) = 2 Here ρ (A) = ρ (A/B) ≠ No. of unknowns =3 The system has infinitely many solutions. 9). Find the values of a and b for which the equations x+y+z=3; x+2y+2z=6;x+ay+3z=b have (i) No solution (ii) A unique solution (iii) Infinite no of solutions. 1 1 1  x   3  Sol: The above system in matrix notation is       1 2 2  y   6  13a   z   b 

1 1 1 3 Augmented matrix  AB   1 2 2 6 13ab

1 1 1 3 RRR 2 2 1 ~ 0 1 1 3 RRR  3 3 1 0ab 1 2 2 

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1 1 1 3 RRR 3 3 2  ~ 0 1 1 3 0ab 3 0 9  1 1 1 3  For a  3 & b  9 ~ 0 1 1 3 0 0 0 0

A  AB 23  It has infinite no of solutions.     ⇒ 1 1 1 3  For a 3 & b  any value ~ 0 1 1 3 0ab 3 0 9  A    AB   3 It has a unique solution.

⇒ 1 1 1 3  For a 3& b  9 ~ 0 1 1 3 0 0 0b  9 A 23  AB  Inconsistent  no solution     10). Solve the following system⇒ completely. x y  z 1; x  2 y  4 x  ; x  4 y  10 z  2

1 1 1  x   1        1 2 4  y    Sol: The above system in matrix notation is 2 1 4 10  z    AXB

1 1 1 1 Augmented Matrix is  AB   1 2 4  2 1 4 10 

1 1 1 1 RRR2 2 1 ~ 0 1 3  1 RRR  3 3 1 0 3 9 2  1  1 1 1 1  ~ 0 1 3  1 2 RRR3 0 0 0 3 2 3 3 2  Here A23 and AB The given system of equations is consistent if     22320   2   20⇒   210      2,1   Case (i): When   1

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A   AB  2  Number of unknowns.  The system has infinite number of solutions. 1111 The equivalent matrix is  0 1 3 0 0000

The equivalent systems of equations are x y  z 1; y  3 z  0 Let z  k y3 k and x  ( 3 k ) K 1 x 2 k 1 x 1 2 k xk  1 2   1   2          where k is any arbitrary constant. X y   0  3 k   X   0   k   3  zk  0   0   1  Case (ii): When   2

A AB 2  no. of unknowns.      The system has infinite number of solutions. 1111 The equivalent matrix is ~  0 1 3 1 0000 The system of equations equivalent to the given system is x + y + z = 1; y + 3z = 1

Let z k  y 1  3 k and x  (1  3 K )  k  1  x  2 k xk  0 2   0   2          X  y   1  3 k   X   1   k   3  zk  0   0   1          where k is any arbitrary constant. 11). Show that the equations 345xyzaxyzbxyzc   ;456    ;567    don’t have a solution unless a c2. b solve equations when a=b=c= -1. 3 4 5  xa    4 5 6  yb    Sol: The Matrix notation is       5 6 7  zc    AXB

3 4 5 a Augment Matrix is  AB   4 5 6 b 5 6 7 c

3 4 5 a RRR34 2 2 1 ~ 0 1  2 3ba  4 RRR35  3 3 1 0 2  4 3ca  5 

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3 4 5 a RRR2 ~ 0 1  2 3ba  4 3 3 2  0 0 0 3a 6 b 3 c Here A 23 and  AB 

The given system of equations is consistent if 3a 6 b  3 c  0  3 a  3 c  6 b  a  c  2 b Thus the equations don’t have a solution unless a c 2 b , when a  b  c   1

3 4 5 1 The equivalent matrix is  0 1 2 1 0 0 0 0

A   AB  2  No. of unknowns.  The system has infinite number of solutions. The system of equations equivalent to the given system 3x 4 y  5 z  1; y 2 z  1 y 2 z  1

Let z k y1 2 k and 3 x  4 8 k 5 k  1 x 1 k

xk  1   1   1          X y    1  2 k     1   k   2  zk  0   0   1 

Linearly dependent set of vectors: A set { x1,x2,------xr} of r vectors is said to be a linearly dependent set, if there exist r scalars k1,k2---kr not all zero, such that k1x1+k2x2+------

-+krxr = 0

Linearly independent set of vectors: A set { x1, x2,------xr} of r vectors is said to be a linearly independent set, if k1x1+k2x2+------+krxr =0 then k1 =0, k2 = 0------kr =0. Linear combination of vectors:

A vector x which can be expressed in the form x = k1x1+k2x2+------+knxn is said to be a linear combination of x1,x2------xn here k1,k2------kn are any scalars. Linear dependence and independence of Vectors: Solved Problems : 1). Show that the vectors (1, 2, 3), (3,-2, 1), (1,-6,-5) from a linearly dependent set. 1  3   1  X  2X   2X   6 Sol: The Given Vector 1   2   3   3  1  5

The Vectors X1, X2, X3 from a square matrix.

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1 3 1 1 3 1  Let A2  2  6 Then A  2  2  6 3 1 5 3 1 5  = 1(10+6)-2(15-1) + 3(-18+2) = 16+32-48 = 0 The given vectors are linearly dependent  |A| = 0

2). Show that the Vector X1=(2,2,1), X2=(1,4,-1) and X3=(4,6,-3) are linearly dependent.

Sol: Given Vectors X1=(2,-2,1) X2=(1,4,-1) and X3=(4,6,-3) The Vectors X1, X2, X3 form a square matrix. 2 1 4 2 1 4  Let A 2 4 6 Then A   2 4 6 1 1  3 1  1  3  = 2(-12+6) + 2(-3+4)+1(6-16) = -20≠ 0 The given vectors are linearly dependent |A|≠0

Consistency of system of Homogeneous linear equations: A system of m homogeneous linear equations in n unknowns, namely

a11 x 1 a 12 x 2  a 13 x 3 ..  a 1nn x  0   a21 x 1 a 22 x 2  a 23 x 3 ..  a 2nn x  0    (1)     a x a x  a x ..  a x  0 m1 1 m 2 2 m 3 3 mn n  0  x1    a11 a12 a13....a1n  0   x     2  i.e. a21 a22 a23....a2n     Here A is called Co –efficient matrix. ...      .. am1 am2 am3....amn      xn   0. 

Note: 1. Here x1= x2 = ------xn = 0 is called trivial solution or zero solution of AX = 0 2. A zero solution always linearly dependent.

Theorem: The number of linearly independent solutions of the linear system AX = 0 is (n-r), r being the rank of the matrix A and n being the number of variables. Note: 1.if A is a non-singular matrix then the linear system AX = 0 has only the zero solution.

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2. The system AX =0 possesses a non-zero solution if and only if A is a singular matrix. Working rule for finding the solutions of the equation AX = 0 (i). Rank of A = No. of unknowns i.e. r = n The given system has zero solution.

∴(ii). Rank of A < No of unknowns (r

Solved Problems : 1). Solve the system of equations x+3y-2z = 0, 2x-y+4z = 0, x-11y+14z = 0 1 3  2  x 0       Sol: We write the given system is AX = 0 i.e. 2 1 4 y  0 1 11 4  z  0

1 3  2    R2 →R2 -2R1; R3→R3-R1 A ~ 0  7 8  0 14 16

1 3  2   R3 →R3 -2R2 A ~0  7 8  0 0 0  The Rank of the A = 2 i.e. ρ (A) = 2 < No. of unknowns = 3 We have infinite No. of solution Above matrix can we write as x+3y-2z = 0 -7y+8z = 0, 0 = 0 Let z = k then y=8/7k & x= -10/7 k Giving different values to k, we get infinite no. of values of x,y,z.

2). Find all the non-trivial solution 2x y  30;32 z  x  y  z  0; x  450. y  z 

2 1 3  x   0        Sol: In Matrix form it is 3 2 1  y   0  1 4 5  z   0  AX  0

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2 1 3 0  The Augumented matrixAO   3 2 1 0  1 4 5 0 1 4 5 0 RR ~ 3 2 1 0 13  2 1 3 0  1 4 5 0 RRR2 23 1  ~ 0 14 14 0 RRR3 32 1 0 7 7 0 1 4 5 0 RRR2  ~014140  it is echelon form. 3 3 2  0 0 0 0 

The Rank of the A = 2 i.e. ρ(A) = 2 < No. of unknowns = 3 Hence the system has non trivial solutions. From echelon form, reduced equations are x4 y  5 z  0 and 14 y  14 z  0 Let z k then y  k and x 4 k  5 k  0  x   k . xk    1  Thus, the solution set is       X y    k   k 1.  K zk    1 

3). Show that the only real number for which the system x+2y+3z = x, 3x+y+2z= y, 2x+3y+z = z, has non-zero solution is 6 and solve them. 1  2 3  x 0 Sol: Above system can we expressed as AX = 0 i.e.       3 1  2  y  0 2 3 1  z  0 Given system of equations possess a non –zero solution i.e. ρ (A) < no. of unknowns.  For this we must have det A = 0 1  2 3 3 1  2  0 2 3 1 

666

RRRR1 1 2 3 3 1 2  0 2 3 1 

1 1 1 (6  ) 3 1  2  0 2 3 1 

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→ − , → − − | − − − | =  (6-)[(-2- )(-1- )+1] =0 − −  (6- ) (2+3 +3) = 0  = 6 only real values. When = 6, the given system becomes  5 2 3  x 0       3  5 2  y  0 2 3  5 z  0

 5 2 3  x 0 R2 → 5R2+3R1, R3→5R3+2R1 ~       0 19 19  y  0 0 19 19 z  0

 5 2 3  x 0 R3 →R3+R2 ~       0 19 19 y  0 0 0 0  z  0 -5x+2y+3z = 0 and -19y+19z = 0  y = z Let z = k  y = k and x = k. The solution is x = y = z = k.

Eigen Values∴ and Eigen vectors: Let A= a be a square Matrix. Suppose the linear transformation Y = AX transforms X ij nxn into a scalar multiple of itself i.e. AX = Y =  X, Then the unknown scalar  is known as an “Eigen value” of the Matrix A and the corresponding non-zero vector X is known as “Eigen Vector” of A. Corresponding to Eigen value  . Thus the Eigen values (or) characteristic values (or) proper values (or) latent roots are scalars  which satisfy the equation. AX =  X for X  0, AX IX 0  ( A  I ) X  0

Which represents a system of ‘n’ homogeneous equations in ‘n’ variables x1, x2, ----, xn this system of equations has non-trivial solutions If the coefficient matrix (A-  I) is singular i.e.

a11 a 12 a 1n

a21 a 22  a 2n AI 0         0       a a a n12 n nn

n n n12 n th Expansion of the determinant is ( 1)  KKK12          n is the n degree of a polynomial Pn () which is known as “Characteristic Polynomial”. Of A

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n n n12 n ( 1)  KKK12          n =0 is known as “Characteristic Equation”. Thus the Eigen values of a square Matrix A are the roots of the characteristic equation.

Eg: Let A= X = AX = [ ] = [ ] - − - [ ][ ] [ ] = . [ ] = . Here Characteristic − vector of A is and Characteristic root of A is “1”. Eigen Value: The roots of the characteristic[ ] equation are called Eigen values or characteristic − roots or latent roots or proper values. Eigen Vector: Let A= a be a Matrix of order n. A non-zero vector X is said to be a ij nxn characteristic vector (or) Eigen vector of A if there exists a scalar  such that AX =  X. Method of finding the Eigen vectors of a matrix.

Let A=[aij] be a nxn matrix. Let X be an eigen vector of A corresponding to the eigen value . Then by definition AX = X. AX = IX

⇒ AX –IX = 0 ⇒ (A-I)X = 0 ------(1) This⇒ is a homogeneous system of n equations in n unknowns. Will have a non-zero solution X if and only |A-I| = 0  A-I is called characteristic matrix of A  |A-I| is a polynomial in of degree n and is called the characteristic polynomial of A  |A-I|=0 is called the characteristic equation  Solving characteristic equation of A, we get the roots , These are

called the characteristic roots or eigen values of the matrix.휆,휆 , 휆, … … . 휆,  Corresponding to each one of these n eigen values, we can find the characteristic vectors. Procedure to find Eigen values and Eigen vectors

Let A = … … … . . […….…….…….] 푖 푖 … … . ℎ푖푖 푖 푖 − λI

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a11  a 12 a 1n a a  a i. e ., A I 21 22 2n   an12 a n a nn   Then the characteristic polynomial is A  I

a11   a12 ... a1n a a   ... a say   A  I  21 22 2n ......

an1 an2 ... ann   The characteristic equation is |A- we solve the we get n roots, these are called eigen values or latentλI| =values or proper values.∅λ = |A − λI| = , Let each one of these eigen values say their eigen vector X corresponding the given value is obtained by solving Homogeneous system

a11  a 12 a 1n   x 1  0 a a  a   x  0 21 22 2n   2    and determining the non-trivial solution.           an12 a n a nn    x n  0

Solved Problems

1. Find the eigen values and the corresponding eigen vectors of − [ ] Sol: − = [ ] Characteristic matrix = − λ − Characteristic equation is [ − 휆=0] = [ ] − λ | − I| − λ − ⟹ | | = − λ − λ − λ + = ⟹ + λ − λ + = ⟹ λ − λ + = ⟹ λ − λ − = Consider the system⟹ λ = , are eigen values of A − λ − Eigen vector corresponding to [ ] = − λ 훌 =

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Put in the above system, we get − λ = = − ⟹ − = − − − − = − − − Let x = 1  ℎ =

x1   1  Eigen vector is     x2   1 

[Eigen] 푖 Vector 푖 corresponding 푖 to , 푖 푖 휆 = 훌 = − λ = in the above system, we get = − from (1) and (2) we have x = 2x ⟹ − = −1 − − 2 − = − − −

Let x2   x1  2 22    Eigen vector       1 

[ ] 푖 푖 푖 푖 푖 휆 = 2. Find the eigen values and the corresponding eigen vectors of matrix [ ] Sol: Let A = [ ] The characteristic equation is |A-I|=0 i.e. |A-I| = − λ | | = − λ − λ ⟹ − λ − λ − + [ − − λ] = ⟹ − λ — λ − = ⟹ λ − [– λ − − ] = ⟹ λ − [−λ + λ − ] =  =1, 2, 3 ⟹ λ − λ − λ − = The eigen values of A is 1, 2, 3.

푖푖 푖 ℎ 푖 − λIX =

− λ ⟹ [ − λ ] [ ] = [] DEPARTMENT OF HUMANITIES − λ & SCIENCES ©MRCET (EAMCET CODE: MLRD) 44

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훌 =

[ ] [ ] = [] + = = + = = −, = = 훼 ⇒ = −훼 = , = 훼

−훼 − [ ] = [ ] = 훼 [ ] 훼 − [ ] 푖 푖 훌 =

[ ] [ ] = [] = = 푖 푖. = 훼

[ ] = [훼] = 훼 [] 푖 푖 [ ] 훌 =

− [ − ] [ ] = [] − − + = − = − = ℎ 푖 = , = =∝ x1=∝ ,  =1  , =∝ x   0   0  2      x3   1

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푖 푖 [] 3 6 2  3. Find the Eigen values and Eigen vectors of the matrix is 6 7 4 2 4 3 8 6 2  Sol: Let A  6 7  4 2 4 3 8 6 2 Consider characteristic equation is AI 0 i.e. 6 7   4  0 2 4 3 

8 7   3   16663      822427       0 2 8   217   3   16 6186  8224142      0 2 8      10   566102102           0 82 80  40  3 10  3  5  36  60 20 4  0  32 18   45   0 2    18   45  0  0 (OR )  2 18  45  0  0,   3,   15 Eigen Values

= , , Case (i): If

− [− −] = −

− [− −][ ] = [] −

8x1  6 x 2  2 x 3  0     (1)

6x1  7 x 2  4 x 3  0     (2)

2x1 4 x 2  3 x 3  0     (3)

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Consider (2) & (3)

x1 x 2 x 3 6 7 4 2 4 3 xx x 12  3  k 21 16  18  8 24  14 xx x 12  3  k 5 10 10

x1  k, x 2  2 k , x 3  2 k

xk1    1  Eigen Vector is x  22 k  k   2      xk3  22    Case (ii): If

8= 6 2   0      6 7   4 x   0  2 4 3   0      5 6 2  x1   0  6 4  4  x    0    2    2 4 0  x3   0 

5x1  6 x 2  2 x 3  0       (1)

6x1  4 x 2  4 x 3  0      (2)

2xx12 4  0  0        (3) Consider (2) & (3)

x1 x 2 x 3 6 4 4 2 4 0

xx x 12  3  k 0 16 0  8 24  8 xx x 12  3  k 16 8 16 xx x 12  3  k 2 1 2

x1  k,  x23  k , x  2 k 2

x1  2 k , x 2   k , x 3  2 k

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xk1  22    x    k   k  1  Eigen Vector is 2      xk3  22   

Case (iii): If

7 6= 2  6  8  4x  0 2 4 12

7 6 2  x1   0  6  8  4  x    0    2    2 4 12  x3   0 

 7x1   6 x 2   2 x 3  0     (1)

6x1  8 x 2  4 x 3  0      (2)

2x1 4 x 2  12 x 3  0      (3) Consider (2) & (3)

x1 x 2 x 3

6  8  4 2 4 12 xx x 12  3  k 96 16 72  8 24  16

xx x 12  3  k 80 80 40 xxx 12 k,,  k3  k 2 2 1

x1 2 k , x 2  2 k , x 3  k

xk1  22    Eigen Vector is x   22 k      k 2      xk3    1  4. Find the Eigen values and the corresponding Eigen vectors of the matrix. 2 2 3  2 1 6 1  2  0

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2 2 3 Sol: Let  A 2 1 6 1  2  0 2  2  3 The characteristic equation of A is AI 0 i.e. 2 1  6  0 12  

2     1   122263221             0

32   21   45  0  3   3   5   0   3,  3,5 The Eigen values are -3,-3, and 5 Case (i): If

2 = 3 − 2  3  x1   0  We get 2 1 3  6  x    0    2    1  2 0  3  x3   0 

1 2  3 0   The augment matrix of the system is 2 4 6 0 1 2 3 0

1 2  3 0  RRRR2,  Performing 2 1 3 1 , we get 0 0 0 0 0 0 0 0

Hence we have x12 x 2  3 xy  0  x 1   2 x 2  3 x 3

Thus taking x2 k 1 and x 3 k 2 , we get x1 2 k 1  3 k 2 ; x 2  k 1 ; x 3  k 2

x1  23    Hence x  k 10  k   2  1   2   x3  01   

2 3 So and are the Eigen vectors corresponding to  3 1 0 0 1 Case (ii): If

7= 2 3 x1   0       We get 2 4  6x  0  2         1  2  5 x3   0 

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 7x1  2 x 2  3 x 3  0     (1)

2x1 4 x 2  6 x 3  0     (2)

x1 2 x 2  5 x 3  0     (3)

Consider (2) & (3)

x1 x 2 x 3 2 4 6 1  2  5 xx x 12  3  k 20 12  10  6  4  4 3 xx x 12  3  k 8 16 8 3 xx x 12  3  k 1 2 1 3

x1  1  Eigen vector is x  2  k 23    x3  1  5. Find the Eigen values and Eigen vectors of the matrix A and it’s inverse where 1 3 4  A  0 2 5 0 0 3 1 3 4  Sol: Given A  0 2 5 0 0 3

The characteristic equation of “A” is given by AI 0

1  3 4 0 2  5  0 0 0 3 

1   2    3     0  1,2,3i . e . EigenValues are 1,2,3

Note: In upper le (or) Lower lar of a square matrix the Eigen values of a diagonal matrix are just the diagonal elements of the matrix. Case (i): If

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A I  x 0

∴ 1  3 4  x1   0  0 2  5  x    0    2    0 0 3   x3   0 

0 3 4  x1   0  0 1 5  x   0    2    0 0 2  x3   0 

340;x2  x 3  x 2  50;20 x 3  x 3   x 1  k 1 ; x 2  0; x 3  0

xk11    1  X x   00     k 21      x3  00    Case (ii): If

1=  3 4  x1   0  0 2  5  x    0    2    0 0 3   x   0    3   

1 3 4  x1   0  0 0 5  x   0    2    0 0 1  x   0    3   

1 3 4  x1   0  0 0 5  x   0    2    0 0 1  x3   0 

 x 3 x  4 x  0; 5 x  0; x  0 1 2 3 2 3

x13 k  4(0)  0 x 1 3 k  0 x 1 3 k

xk1  33    X x    k   k 1  2      x3  00    Case (iii): If

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1  3 4  x1   0  0 2  5  x    0    2    0 0 3   x3   0 

2 3 4  x1   0  0  1 5  x    0    2    0 0 0  x   0    3   

 2x1  3 x 2  4 x 3  0;  x 2  5 x 3  0; x 3  0 Let x k 3  x2 5 x 3  0  x 2  5 k

and2 x1  3 x 2  4 k  0   2 x 1  15 k  4 k  0 19  2x  19 x  0  x  k 112

19 19 xk1 22    X x 55 k   k   2         xk3   1  1 1 1 1 1 Note: Eigen Values of A-1 are , ,ie . 1, , and the Eigen vectors of A-1 are same as 1  2  3 23 Eigen vectors of the matrix A 6. Determine the Eigen values and Eigen vectors of

2 1 84 B23 A  A  I where A   2 22

1 84 Sol :  Given that B23 A  A   A   2 2 2 

2 8 4   8  4   56  40  we have A A. A        2 2   2 2   20 4 

BAAI232 1  2 56 40 1  8 4   1 0  23        20 4 2  2 2   0 1  112 80   4 2   3 0          40 8   1 1   0 3  111 78   39 6

Characteristic equation of B is BI 0

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111 78 0 39 6  i.e. 2 105  2376  0  33   72   0  33or 72 Eigen Values of B are 33 and 72. Case (i): If = 33 111 78 0 39 6 

78 78 x1  0      39 39 x2  0 

39x1  78 x 2  0  x 1  2 x 2 xx 12k() say 21 x 2 1 k x 1 2  Case (ii): If = 72 111 78 0 39 6  111 72 78 X 0 39 6 72

39 78 X 0 39 78

39 78 x1  0      39 78 x2  0 

39x1  78 x 2  0  x 1  2 x 2 xx 12   k() say 21

x1 2 k x2 1

Properties of Eigen Values: Theorem 1: The sum of the eigen values of a square matrix is equal to its trace and product of the eigen values is equal to its determinant. Proof: Characteristic equation of A is |A-I|=0

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a11  a 12 a 1n a a  a i.e. 21 22 2n   an12 a n a nn   푖 ℎ푖

a11  a 22     ann     a 12 (a polynomial of degree n – 2)+ a13 (a polynomial of degree n -2) + … = 0

n n n1  ( )1   (a11  a22 .... ann )  a polynomial of deg ree(n  )2  0

( )1 nn  ( )1 n1(Trace A)n1 a polynomial of deg ree(n  )2 in   0

s λ 푛+, λ … . . λ ℎ ℎ푖 푖 − 푛 ℎ = − = ℎ |A − λI| = − λ + ⋯ . +a 휆 = ℎ || = − − − − − λ + a λ + a λ + … . + a = − ℎ = = − = || = det Theorem 2: If is an eigen value ℎ of A corresponding to the eigen vector X, then is eigen value An corresponding to the eigen vector X. Proof: Since is an eigen value of A corresponding to the eigen value X, we have AX= ------(1) Pre multiply (1) by A, A(AX) = A( X) (AA)X = (AX) A2X= ( X) A2X= 2X 2 is eigen value of A2 with X itself as the corresponding eigen vector. Thus the theorm is true for n=2 Let we assume it is true for n = k i.e. AKX = KX------(2) Premultiplying (2) by A, we get A(AkX) = A( KX) (AAK)X= K(AX)= K( X)

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AK+1X= K+1X K+1 is eigen value of K+1 with X itself as the corresponding eigen vector. Thus, by Mathematical induction. n is an eigen value of An. Theorem 3: A Square matrix A and its transpose AT have the same eigen values. Proof: We have (A- I)T = AT- IT = AT- I |(A- I)T|=|AT- I| (or) |A- I|=|AT- I|  AT  A 

|A- I|=0 if and only if |AT- I|=0 Hence the theorem. Theorem 4: If A and B are n-rowed square matrices and If A is invertible show that A-1B and B A-1 have same eigen values. Proof: Given A is invertble i.e, A-1 exist We know that if A and P are the square matrices of order n such that P is non-singular then A and P-1 AP have the same eigen values. Taking A=B A-1 and P=A, we have B A-1 and A-1 (B A-1 ) A have the same eigen values ie.,B A-1 and (A-1 B)( A-1 A) have the same eigen values ie.,B A-1 and (A-1 B)I have the same eigen values ie.,B A-1 and A-1 B have the same eigen values

Theorem 5: If 1,2 ,...... n are the eigen values of a matrix A then k 1, k 2, ….. k n are the eigen value of the matrix KA, where K is a non-zero scalar. Proof: Let A be a square matrix of order n. Then |KA- KI| = |K (A- I)| = Kn |A- I| Since K≠0, therefore |KA- KI| = 0 if and only if AI  0

.ei ., K is aneigen valueof KAif  is aneigen valueof A

Thus k 1, k 2 … k n are the eigen values of the matrix KA if

1, 2 … n are the eigen values of the matrix A Theorem 6: If is an eigen values of the matrix A then +k is an eigen value of the matrix A+KI Proof: Let be an eigen value of A and X the corresponding eigen vector. Then by definition

AX= X Now (A+KI) X  AX  IKX  X  KX

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= (+K) X +k is an eigen value of the matrix A+KI.

Theorem 7: If 1, 2 … n are the eigen values of A, then 1 – K, 2 – K, … n – K, aretheeigen valuesof thematrix(A  KI), where K is a non  zero scalar Proof: Since The characteristic polynomial of A is

|A – I| = ( 1 – ) ( 2 – ) … ( n – )------1 Thus the characteristic polynomial of A-KI is |(A – KI) – I| = |A – (k+ )I|

= [ 1- (+K)][ 2-( +K)]...... [ n-( +K) ].

= [ (1-K)-)][ (2-K)-]...... [( n-K)- ]. Which shows that the eigen values of A-KI are Theorem 8: If are the eigen values of A, find the eigen values of the matrix

Proof: First we will find the eigen values of the matrix A- Since are the eigen values of A The characteristics polynomial is | A- ( ( The characteristic polynomial of the matrix (A- |A- -KI| = |A-( +K)I| = [ ( +K)] = [ K)] Which shows that eigen values of (A- I) are We know that if the eigen values of A are then the eigen values of A2 are

2 2 2 2 Thus eigen values of (A  I) are(1   (,) 2   ,) .....(n  ) Theorem 9: If is an eigen value of a non-singular matrix A corresponding to the eigen vector X, then –1 is an eigen value of A–1 and corresponding eigen vector X itself. Proof: Since A is non-singular and product of the eigen values is equal to |A|, it follows that none of the eigen values of A is 0. If s an eigen value of the non-singular matrix A and X is the corresponding eigen vector ≠0 and AX= .

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premultiplying this with A–1, we get A–1(AX) = A–1( X)

1  1  1 ()A A X  A X  IX  A X X = AXAXX1  1  1 (  )0

Hence  1 is an eigen value of A1

Theorem 10: If is an eigen value of a non-singular matrix A, then is an Eigen value of || the matrix AdjA. Proof: Since is an eigen value of a non-singular matrix, therefore ≠ 0 Also is an eigen value of A implies that there exists a non-zero vector X such that AX = -----(1)

(adjA)A  IA 

A A  X  (adj A) X or (adj A) X  X   Since X is a non – zero vector, therefore the relation (1) A it is clear that is an eigen value of the matrix Adj A 

Theorem 11: If is an eigen value of an orthogonal matrix A, then is also an Eigen value A Proof: We know that if is an eigen value of a matrix A, then is an eigen value of A–1

Since A is an orthogonal matrix, therefore A–1 = A1 is an eigen value of ∴ But the matrices A and A1 have the same eigen values, since the determinants |A- I| and | A1 - I| are same. Hence is also an eigen value of A. 2 Theorem 12: If is eigen value of A then prove that the eigen value of B = a0A +a1A+a2I is

2 a0 +a1 +a2 Proof: If X be the eigen vector corresponding to the eigen value , then AX = X --- (1) Premultiplying by A on both sides

This shows that 2 is an eigen value of A2

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2 We have B = a0A +a1A+a2I 2 BX = a0A +a1A+a2I) X 2 = a0A X+a1AX+a2 X 2 2 = a0 X+a1 X+a2X = (a0 +a1 +a2 ) X

2 (a0 +a1 +a2 ) is an eigen value of B and the corresponding eigen vector of B is X. Theorem 13: Suppose that A and P be square matrices of order n such that P is non singular. Then A and P-1AP have the same eigen values. Proof: Consider the characteristic equation of P-1AP

1 It is | ( P-1AP)-I| = | P-1AP- P-1IP| (I  P P) = | P-1 (A-I) P| = | P-1 | |A-I| |P| = |A-I| since |P-1 | |P| = 1 Thus the characteristic polynomials of P-1AP and A are same. Hence the eigen values of P-1AP and A are same. Corollary 1: If A and B are square matrices such that A is non-singular, then A-1B and BA-1 have the same eigen values. Corollary 2: If A and B are non-singular matrices of the same order, then AB and BA have the same eigen values. Theorem 14: The eigen values of a triangular matrix are just the diagonal elements of the matrix.

a11 a12 ...... a1n

0 a22 ...... a2n Proof: Let A = ...... be a triangular matrix of order n

0 0...... ann The characteristic equation of A is |A- I|=0 i.e.,

i.e, (a11- ) (a22- ) ….. (ann- )=0

a11 , a22 ,…. ann

Hence the eigen values of A are a11 , a22 ,…. ann which are just the diagonal elements of A. Note: Similarly we can show that the eigen values of a diagonal matrix are just the diagonal elements of the matrix. Theorem 15: The eigen values of a real symmetric matrix are always real.

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Proof: Let be an eigen value of a real symmetric matrix A and Let X be the corresponding eigen vector then AX= Take the conjugate Taking the transpose Since Post multiplying by X, we get ------(2)

TT Premultiplying (1) with , we get X AX  X X ------(3)

T (1) – (3) gives  XX 0 but    0

     is real. Hence the result follows Theorem 16: For a real symmetric matrix, the eigen vectors corresponding to two distinct eigen values are orthogonal.

Proof: Let 1, 2 be eigen values of a symmetric matrix A and let X1, X2 be the corresponding eigen vectors.

Let 1 ≠ 2. We want to show that X1 is orthogonal to X2 (i.e.,

Sice X1, X2 are eigen values of A corresponding to the eigen values 1, 2 we have

AX1= 1X1 ----- (1) AX2 = 2 X2 ------(2) Premultiply (1) by

Taking transpose to above, we have

T T T T T T T  X 2 A X 2  1X1 X 2  ------(3) s Premultiplying (2) by Hence from (3) and (4) we get

()12

Note: If is an eigen value of A and f(A) is any polynomial in A, then the eigen value of f(A) is f() . Theorem17: The Eigen values of a Hermitian matrix are real. Proof: Let A be Hermitian matrix. If X be the Eigen vector corresponding to the eigen value

 of A, then AX =  X ------(1)

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Pre multiplying both sides of (1) by X  ,we get X  AX  X  X ------(2) Taking conjugate transpose of both sides of (2)

 We get X AX   X X 

    i.e X  A X    X  X   ABC  C B A and KA   KA 

 (or) XAXXXXXAA    ,   ------(3)     From (2) and (3), we have XXXX i.e    XX 00       XX  0 

 Hence  is real. Note: The Eigen values of a real symmetric are all real Corollary: The Eigen values of a skew-Hermitian matrix are either purely imaginary (or) Zero Proof: Let A be the skew-Hermitian matrix If X be the Eigen vector corresponding to the Eigen value  of A, then AX X() or iA  X  i  X From this it follows that i is an Eigen value of iA Which is Hermitian (since A is skew-hermitian) AA   Now iA   Ai   iA  i A   iA Hence i is real. Therefore  must be either Zero or purely imaginary. Hence the Eigen values of skew-Hermitian matrix are purely imaginary or zero Theorem 18: The Eigen values of an unitary matrix have absolute value l. Proof: Let A be a square unitary matrix whose Eigen value is  with corresponding eigen vector X  AX X 1 

TT  AXX  XAX T  2 

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T Since A is unitary, we have AAI  3 

TT T (1) and (2) given XA AX    X X

TT i.e XXXX  From (3) T  XX10 

T Since XX 0 ,we must have 10   1

Since  =

We must have  =1

Note 1: From the above theorem, we have “The characteristic root of an orthogonal matrix is of unit modulus”. 2. The only real eigen values of unitary matrix and orthogonal matrix can be  1 Theorem 19: Prove that transpose of a unitary matrix is unitary. Proof: Let A be a unitary matrix, then AAAAI..

 where A is the transposed conjugate of A.

TT AA    A A    I T

TT  AAAAI TT  

  AAAAITTTT   

Hence AT is a unitary matrix.

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Solved Problems: 1 2 3  32 1. For the matrix A  0 3 2 find the Eigen values of 3AAAI 5  6  2 0 0 2 1 2 3 Sol: The Characteristic equation of A is AI 0 i.e. 0 3 2 0 0 0 2 

1  3     2     0

⇒ Eigen values are 1, 3,-2. If  is the Eigen value of A. and F (A) is the polynomial in A then the Eigen value of f (A) is f (  ) Let f A  3 A32  5 A  6 A  2 I Eigen Value of f (A) are f (1), f (-2), f (3)

∴f (1) = 3+5-6+2 = 4 f (-2) = 3(-8)+5(4)-6(-2)+2 = -24+20+12+2 = 10 f (3) = 3(27)+5(9)+6(3)+2 = 81+45-18+2 = 110 The Eigen values of f (a) are f (  ) = 4, 10,110

2. Find the eigen values and eigen vectors of the matrix A and its inverse, where

A =

Sol: Given A =

The characteristic equation of A is given by |A-I| = 0

Charecetstic roots are 1, 2, 3. Case (i): If

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and x1   1     X  0   0 is the solution where  is arbritary constant 0 0 1    X  0 is the eigen vector corresponding to  1 0 Case (i): If

5x3  0  x3  0

 x1 3x2  0  x1  3x2

Let x2  k

x1  3k 3k 3     X   k   k1  0  0 is the solution where k is arbitrary constant 3    X  1 is the eigen vector corresponding to   2 0 Case (iii): If  2 3 4 x 0 =   1    For   ,3 becomes 0 1 5x   0   2     0 0 0x3  0

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X = is the solution, where k/2 is arbitrary constant.

19    X  10 is the eigen vector corresponding to   3  2 

Eigen values of A –1are 1, .

We know Eigen vectors of , are same as eigen vectors of A. − 32ii 3. Find the eigen values of A=  2 ii  32ii Sol: we have A=  2 ii 

32ii T  3i  2  i So A =  and A    2 ii 2  i  i 

 A = AT Thus A is a skew-Hermitian matrix. The characteristic equation of A is AI 0

3i    2  i  AT   0  2  i  i  

2  2i  8  0   4ii , 2 are the Eigen values of A

 1 3   i  4.Find the eigen values of A   2 2  3 1  i   2 2 

 1 3   1 3   i   i 2 2 T   Now A    and A    2 2  3 1 3 1   i   i  2 2   2 2 

T 10 We can see that AAI.  01 Thus A is a unitary matrix

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 The characterstic equation is AI 0

1 3 i    2 2  0 3 1 i   2 2 3 1  3 1 Which gives    i and  i and 2 2 2 2 Hence above  values are Eigen values of A.

Cayley-Hamilton Theorem: Every Square Matrix satisfies its own characteristic equation To find Inverse of matrix: If A is non-singular Matrix, then A-1 exists, Pre multiplying (1) above by A-1 we have a Ann1 a A  2      a I  a A  1  0 0 1nn 1 , 1 1nn  1  2 A a0 A  a 1 A     an 1 I an To find the powers of A: - Let K be a +ve integer such that Kn

K K1 k n k-n a01 A a A      an A  0 Pre multiplying (1) by A we get , 1 K k11 K  K  n A a12 A  a A     an A a0 Solved Problems :

1. S.T the matrix A = satisfies its characteristic equation and hence find A-1

Sol: Characteristic equation of A is det (A-I) = 0



C2 C2+C3

→

By Cayley Hamilton theorem, we have A3-A2+A-I = 0 – ⇒ − + − = 1  2 2 1 0 0 1 2  2   2   3   A  1  2 3 A  1 1 2 A   2 2 1 0 1 2 1 0 1 1 1 0 

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1 2  2 1 0 0 1  2 2 1 0 0 3 2         = A  A  A  I   2 2 1  1 1 2  1  2 3  0 1 0 1 1 0  1 0 1 0 1 2 0 0 1 Multiplying with A–1 we get A2 – A + I =A–1 1 0 0 1  2 2 1 0 0 1 2  2 1         A  1 1 2  1  2 3  0 1 0   2 2 1 1 0 1 0 1 2 0 0 1 1 1 0  2. Using Cayley - Hamilton Theorem find the inverse and A4 of the matrix

A = − [− − ] - Sol: Let A = - −- - [ ] 7   2  2 The characteristic equation is given by |A-I|=0 .ei .,  6 1  2  0 6 2 1 

By Cayley Hamilton theorem we have A3-5A2+7A- – ⇒ − + − = 3I=0…..(1) Multiply with A-1 we get

− =25 [8 − 8 + ] 79 26  26 2   3   A   24  7 8  A   78  25 26   24 8  7  78 26  25

 3  2 2  1 A1   6 5  2 3    6  2 5  multiplying (1) with A,we get, A4-5A3+7A2-3A= 0 A4 = 5A3-7A2+3A

 395 130 130  175 56  56  21 6  6  241 80 80   390 125 130   168  49 56   18  3 6            240  79 80   390 130 125  168 56  69  18 6  3          240 80  79

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2 1 2 3.  Verify Cayley-Hamilton theorem hence find A-1 If A  5 3 3 1 0 2

2 1 2 Sol: - Given that  A  5 3 3 1 0 2 The characteristic equation of A is |A-I|=0 i.e. =0 −

| − 2 | 2    632−    −  − 11053203       2    2    6  1  5   7   2  3 x   0  22  2  12  3  2 6  5  7 6 2  0  32 3   7   1  0 32 3   7   1  0     (1) According to Cayley Hamilton theorem. Square matrix ‘A’ satisfies equation (1) Substitute A in place of 

2 1 2 Now  A  5 3 3 1 0 2

2 1 2   2 1 2  7 5 3 A2  5 3 3   5 3 3        22 14 13 1 0  2    1 0  2  0 1 2

7 5 3   2 1 2  36 22 23 AAA32. 22 14 13   5 3 3        101 64 60 0 1 2    1 0  2  7  3  7

32 Now AAAI3  7   0

36 22 23   21 15  9   14  7  14    1 0 0   0 0 0            101 64 60    66 42 39   35 21 21   0 1 0    0 0 0   0 737    03  6   7014   001000     Cayley –Hamilton theorem is verified. To find A-1 32 AAAI3  7   0 -1 ⇒Multiply A , we get

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AAAAI1 33 2  7    0 AAIA21 3  7   0 AAAI12  37  7 5 3   6  3  6    7 0 0  1       A 22 14 13    15  9 9    0  7 0  0 1 2   3 0 6   0 0 7 

6 2 3 1  A 7 2 4 3 1 1 Check A.A-1= I

2 1 2   6 2 3  1 0 0 -1 A.A = 5 3 3   7 2 4       0 1 0 I     1 0  2   3  1 1  0 0 1

12 4. Using Cayley – Hamilton theorem, find A8, if A   21 12 Sol: Given A   21

Characteristic equation of A is AI 0

1   1   4  0     2 5  0     (1) Substitute A in place of  AIAI225  0   5 find A8 AAAAA8 5 6  5( 2 )( 2 )( 2 )

 5 5III  5  5   625I AI8 625

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Diagonalization of a Matrix by similarity transformation: Similar Matrix: A matrix A is said to be similar to the Matrix B if there Exists a non- singular matrix P such that B= P1 AP . This transformation of A to B is known as “Similarity Transformation”

Diagonalization of a Matrix: Let A be a square Matrix. If there exists a non-singular Matrix P and a diagonal Matrix D such that P-1AP=D, then the Matrix A is said to be diagonalizable and D is said to be “Diagonal” form (or) canonical diagonal form of the Matrix A Modal Matrix:The modal matrix which diagonalizes A is called the modal Matrix of A and is obtained by grouping the Eigen vectors of A into a Square Matrix. Spectral Matrix: The resulting diagonal Matrix D is known as Spectral Matrix. In this spectral Matrix D whose principal diagonal elements are the Eigen values of the Matrix. Calculation of powers of a matrix: We can obtain the power of a matrx by using diagonalization Let A be the square matrix then a non-singular matrix P can be found such that D = P-1AP D2= (P–1AP) (P–1AP) = P–1A (PP–1) AP = P–1A2P (since PP–1=I) Simlarly D3 = P–1A3P In general Dn = P–1AnP……..(1) To obtain An, Premultiply (1) by P and post multiply by P–1

Then PDnP–1 = P(P–1AnP)P–1 = (PP–1)An (PP–1) = An  An  PDn P1

n 1 0 0 0 n n Hence A = P 02 0 0 1 P  n 0 0 0 n Diagonalization of a matrix: Theorem: If a square matrix A of order n has n linearly independent eigen vectors

(X1,X2…Xn) corresponding to the n eigen values 1,2….n respectively then a matrix P can be found such that P-1AP is a diagonal matrix.

Note: 1. If X1,X2…Xn are not linearly independent this result is not true.

2. Suppose A is a real symmetric matrix with n pair wise distinct eigen values 12,  n then the corresponding eigen vectors X1,X2…Xn are pairwise orthogonal.

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Hence if P = (e1,e2…en)

Where e1 = (X1 / ||X1||), e2 = (X2 / ||X2||)….en = (Xn)/ ||Xn|| then P will be an orthogonal matrix. i.e, PTP=PPT=I Hence P–1 = PT

− Solved∴ Problems= : 1. Determine the modal matrix P of A= . Verify that P-1AP is a diagonal matrix.

Sol: The characteristic equation of A is |A-I| = 0 i.e.

which gives (-5) (+3)2= 0 Thus the eigen values are =5, =-3 and =-3

When =5

By solving above we get X1 =

Similarly, for the given eigen value = -3 we can have two linearly independent eigen vectors

X2 =

P = [ ]

P = = modal matrix of A Now[ det P=1(- −1) -2(2)+3(0-1)] = -8 −

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= = diag [5,-3,-3]. [ ] −= diag [5,-3,-3]. − − ∴ 2. Find a matrix P which transform the matrix A = to diagonal form. Hence − calculate A4. [ ]

Sol: Characteristic equation of A is given by |A-I| = 0 i.e.

=> (1-

=> 9 [ − − − ] − − [ − − ] = => − − − = Thus =the , eigen= ,values= of A are 1, 2, 3. If x1, x2, x3 be the components of an eigen vector corresponding to the eigen value , we have

[A-I] X =

Case (i): If

= i.e, 0.x1+0.x2+0.x3=0 and x1+x2+x3=0

x3=0 and x1+x2+x3=0 x3=0, x1=-x2 x1=1, x2=-1, x3=0 Eigen vector is [1,-1,0]T Also every non-zero multiple of this vector is an eigen vector corresponding to =1 For =2, =3 we can obtain eigen vector [-2,1,2]T and [-1,1,2]T

P =

The Matrix P is called modal matrix of A  0 2 1 1 P-1=   2 2 0  2    2  2 1

 1 0 1 1 0 1 1  2 1  2   NowP1 AP  1 1 01 2 1 1 1 1   1    1 1 2 2 3  0 2 2   2

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A4 = PD4P-1  1 1  2 1 1 0 0 0 1    2   1 1 1 0 16 0 1 1 0       0 2 2 0 0 81 2  2 1  

3. Determine the modal matrix P for and hence diagonalize A [ ] Sol: Given that [ ] 1  1 3 The characteristic equation of A is AI 0 i.e. 1 5 1 0 3 1 1 

1   5    1    111      33135       0 1  5  5   2 1   2   3  1  15 3    0 1  4  6  2   2   3  14 3    0 4 6 2 4  6  2   3 2  42 9  0 32 7   9   9   36  0 32 7  36  0   2,3,6 The Eigen Values are -2, 3, and 6 Case (i): If

 [A-= − I]X = 1  1 3  x1   0  1 5 1  x   0    2    3 1 1   x   0  ⇒   3   

3 1 3  x1   0  1 7 1  x   0    2    3 1 3  x   0  ⇒   3    3x  x  3 x  0     (1) 1 2 3 x17 x 2  x 3  0     (2) 3x x  3 x  0     (3) 1 2 3

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From (2) & (3)

x1 x 2 x 3 1 7 1 3 1 3

xx x 12  3  k 1 21 3 3 21 1 xx x 12  3  k 20 0 20

x1  20 k , x 2  0, x 3  20 k

xk1  20   1  X x   0 k   20 k  0  2      xk3  20   1  Case (ii): If

 [A-= I]X = 2 1 3  x1   0  1 2 1  x   0    2    3 1 2  x   0  ⇒   3   

 2x1  x 2  3 x 3  0     (1)

x12 x 2  x 3  0      (2)

3x1 x 2  2 x 3  0     (3) Consider (1) & (2)

x1 x 2 x 3 213 1 2 1

xx x 12  3  k 1 6  2  3  4  1 xx x 12  3  k 5  5  5

xk1  51    X  x   5 k    5 k   1  2      xk  51    3      Case (iii): If = 6  [A-

I]X =

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1  1 3  x1   0  1 5  1  x    0    2          3 1 1   x3   0 

5 1 3  x1   0  1  11 1  x    0    2    3 1 5  x3   0 

 5x1  x 2  3 x 3  0     (1)

x1 x 2  x 3 0    (2)

3x1 x 2  5 x 3  0    (3) Consider (2) & (3)

x1 x 2 x 3 1 1 1 3 1 5 xx x 12  3  k 5 1  5  3  1  3 xx x 12  3  k 4 8 4 xxx 12  3  k 1 2 1

x1  1  X  x   2  k 2    x3  1 

1 1 1   0 1 2 1 1 1   1(  1  2)  1(0  2)  1(0  1)  (  1)(  3)  1(  2)  1  3  2  1  6

3 2 1   0 2 2 3 2 1 3 2 1  Adj( ) 0 2 2 3 2 1 Adj  1   3 0 311 0 1 22 Cofactor of  2  2 2 1  1 1 6 3 3 3 1 2 1 1 1 1 6 3 6

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DA 1

11 220 1 1 3   1 1 1  D 1  1 1 1 5 1   0  1 2  3 3 3     1 1 1 6 3 6 3 1 1   1 1 1  11 220  1  0  3 1  1  3 1  2  3 1  1 1  1  0  1 1  5  1 1  10  1 3 3 3  1 1 1 6 3 6 3  0  1 3  1  1 3  2  1 11  220 2 3 6  200  1 1 1 0 3 12   0 3 0  3 3 3     1 1 1 6 3 6 2 3 6   0 0 6 

Find (a) A8 (b) A4 . 푰 푨 = [ ] Sol: Given −that A= [ ] 1  1 1 − The Characteristic equation of A is AI 0 i.e. 0 2 1 0 4 4 3 

1   2    3    41041042       0 2 1    6  2   3     4  4  8  4x  0 2 1      5   2  4  4   0 2 5   2   3  5  2  2   4  4   0  32 6   11   6  0  1,2,3 The Eigen values are 1, 2, and 3 Case (i): If

[ = − I]X = => − [ − ] = => − − [ ][ ] = [] x1+x−2=0, x 1+x 2=0, -4x1+4x 2+2x3=0

Let x3 = k, x2 + k=0, x2 = -k

k 4x  4( k ) 2 K  0 4 x  2 k  0 4 x  2 k  x 1 1 1 1 2

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x k  1 1  1 22 k x   k 12  k    2    2 xk 12   3    1 X 2 1  2 Case (ii): If

=> [ = − I]X = => − [ − ] = => − − − [ ][ ] = []  x  x  x 0     (1) − 1 2 3 x3 0    (2)

4x1  4 x 2  x 3  0    (3) Consider (1) & (3)

x1 x 2 x 3 1 1 1 4 4 1

xx x 12  3  k 1 4  1  4  4  4 xx x 12  3  k 3 3 0

x12   k; x   k xk 1 1      1 x     k   1   k  2        X 1 2  x3  00    0 Case (iii): If

=> [ = − I]X = => − [ − ] = => − − − [ − ][ ] = [] −

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 20x  x  x  1 2 3 xx23   0 4xx  4  0 12 Let x k and x k 13

2x1  x 2 x 3 0 2 k x 2 k 0 x 2  k

xk1    1  X  x    k   1  k 32     

xk3    1  1 1 1 p  X X X  2 1 1 1 2 3   2 0 1

1 1 0 1   4  3  1 2 2 1 1 0 0   1 0 0  1     D AP 0 2 0    0 2 0  0 0 3   0 0 3 

18 0 0 1 0 0 88 D 0 2 0 0 256 0 8 0 0 3 0 0 6561

(a ). A8 PD 8 P 1 1 1 1   1 0 0   1 1 0        2 1 1   0 256 0   4  3  1  2 0 1   0 0 6561   2 2 1  12099 12355 6305  12100 12356 6305 13120 13120 6561

14 0 0 1 0 0 44 (bD ). 0 2 0 0 16 0 4 0 0 3 0 0 81

1 1 1   1 0 0   1 1 0  4 4 -1       A =PD P 2 1 1   0 16 0   4  3  1  2 0 1   0 0 81   2 2 1 

1 16 81   1 1 0  1  64  162 1  48  162 0  16  81 4      2  64  162 2  48  162 0  16  81 A 2 16 81   4  3  1    2 0 81   2 2 1  2 0  162  2  0  162 0  0  81

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99 115 65 4  A  100 116 65  160 160 81

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UNIT-II

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FUNCTIONS OF SEVERAL VARIABLES

Introduction: We know that is a function where ‘ ’ is dependent variable and ‘ ’ is independent variable. We are going to expand the idea of functions to include functions for more than = one independent variable. In day to day life we deal with things which depend on two or more quantity. For example, the area of the room which is a rectangle consists of two variables: length(say a) and breadth (say b) is given by . Similarly the volume of the rectangular parallelepiped consists of three variables a, b, c i.e., length, breadth, height is given by V = a b c. = In this chapter we say that z is a function of two variables x , y and write where ‘ ’ is dependent variable and ‘ are independent variables. = , & Limit of a function of two variables: A function f(x, y) is said to tend to the limit l as (x, y) tends to (a,b) i.e., x if corresponding to any given positive number a positive number → → for all points (x,y) whenever , ∈ ∃ . 훿 ℎ ℎ |, − | <∈ In other words the variable value (x ,y) approaches a finite fixed value l when the variable | − | 훿 | − | 훿 value (x, y) approaches a fixed value (a, b) i.e.,

→lim , = , lim→, , = Continuity of a function→ of two variables at a point: A function f(x,y) is continuous at a point (a, b) if, corresponding to any given positive number a positive number for all points (x,y) whenever 0<

2 2 (x-a) +(y-b)∈ ∃< 훿 ℎ ℎ |, − , | <∈ Note: Every differentiable function is always continuous, but converse need not be true. 훿 Solved Problems:

1. Evaluate →lim + + → Sol. = →lim + + →lim {lim→ [ + +]} → = = →lim + = (or) = →lim + + →lim {lim→ [ + +]} → = →lim +

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= = 2. If f(x ,y) = show that − + →lim {→lim , } ≠ →lim {→lim , } Sol. = − = →lim {lim→ , } →lim {lim→ +} = →lim = − = → lim {lim→ , } →lim {lim→ +} − = -1 →lim Hence the result follows.

3. Discuss the continuity of the function

f(x,y) = + , , ≠ , { Sol. Let us consider, the ,limit of= ,the function for testing the continuity along the line y = mx. Now →lim , = →lim + → = → = →lim + Which is different for the different m selected. does not+ exist.

∴Consider→lim , → = = →lim , = → lim + →lim = = , f(x,y) is continuous for given values of x and y but it is not continuous at (0,0) →lim , = → lim + →lim = = ,

Partial∴ Differentiation:

Let be a function of two variables x and y. Then , if it exists , +∆,−, is saidz = to f bex, ypartial derivative or partial differential coefficient→lim of z or f(x,y)∆ , w.r.t.x . It is denoted by the symbol or or . i.e, for the partial derivative of w.r.t. ‘x’ , ‘y’ is kept constant.

Similarly, the partial derivative of = , w.r.t. ‘y’ , ‘x’ is kept constant and is defined as and is denoted by = ,or or . ,+∆−, Higher→lim order∆ Partial Derivatives:

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In general the first order partial derivatives and are also functions of x and y and they can be differentiated repeatedly to get higher order partial derivatives, So = , = , = , = , ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ = , = , = , and so on. ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ The∂ chain∂ rule∂ of Partial∂ ∂ Differentiation: ∂ ∂ ∂ ∂ ∂ Let where u = and v = g(x,y) . Then

= , = ∅x, y and = ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ Total differential∂ ∂ coefficient:∂ + ∂ ∂ ∂ ∂ ∂ + ∂ ∂ Let where x = and y = g(t)

Substitutingz = fx, yx and y in z =∅ f(x,y)t , z becomes a function of a single variable t. Then the derivative of z w.r.t. ‘t’ i.e, is called the total differential coefficient or total derivative of z.

∂ ∂ ∂ ∂ ∂ ∴Note:∂ = In ∂ the ∂ differential+ ∂ ∂ form , this result can be written as ∂ ∂ Here, du is called the total differential of u. = ∂ . dx + ∂ . dy Solved Problems:

1. If U = log( , prove that = 훛 훛 훛 − Sol: Given that U =+ log( + − 훛 + 훛 + 훛 퐔 ++ = ( y and z are constant) x + y + z − xyz − ∴ = + + − ( x and z are constant) − = + + − ( y and x are constant) − + + − − − − ∴ + + = + + − + + + − + + + −

x + y + z − xy − yz − zx = x= + y + zx +……..(1) y + z − xy − yz − zx ∂U ∂U ∂U ∂ ∂ ∂ ++ ⟹ + +U = ∂ ∂ ∂ ∂ ∂ ∂ ∂U ∂U ∂U Now ∂ + ∂ + ∂ = ∂ + ∂ + ∂ ∂ + ∂ + ∂ ∂ ∂ ∂ ∂ + ∂ + ∂ ++ [from ]

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= + ∂ ∂ ∂ = -∂ ++ - + ∂ ++ ∂ ++ = -++ ++ − ++ 2. If = e show that ++at x = y = z , = 훛 − Sol: Given that = e 훛훛 −퐥 Taking logarithmx ony zboth sides, we get x log x + y log y + z log z = log e z log z = 1 - x log x – y log y

⟹ ′ = - Differentiating partially w. r. t , ′x , we get ∂ z.= - + . log z ∂ x. +……..(1) . log x ∂ +l ⇒Similarly∂ + l= - ……..(2) ∂ +l When x =∂ y = z+ , wel have = -1 and = -1 Now differentiating (2) partially w.r.t, ‘x’, we get

= = ∂ ∂ ∂ ∂ +l ∂ ∂ = -∂ (1+∂ log y)∂ [− +l ] = …….(3) − ∂ +l ∂ When x = y = z from[− (3) + , logwe zhave ∂] +l ∂

= (-1) ∂ +l ∂ ∂ ∂ = - +l = - since ∂ = (since − log e =1) = - +l = l+l − 3. If lu = f( y-z− ,z-x,x log x-y)ex prove that = 0 Sol: Let r = y - z , s = z - x , t = x – y .Then + u =+ f (r,s,t) Now ∂ ∂ ∂ ∂ =, , ∂ ,= , ∂ = − = − =1, = = -1 , = = 0 ∂ ∂ ∂ and ∂ ∂ ∂

∴ By chain rule of partial differentiation , we have DEPARTMENT OF HUMANITIES & SCIENCES ©MRCET (EAMCET CODE: MLRD) 83

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= + + = = …(1) + − + − + = + + = = …(2) and + − + − −

∂u ∂u ∂r ∂u ∂s ∂u ∂t = + + ∂z = ∂r ∂z ∂s ∂z ∂t ∂z = …(3) ∂ ∂ ∂ ∂ ∂ (1) + (2) + (3) gives ∂ − + ∂ + ∂ − ∂ + ∂ + = 0 ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ + ∂ + ∂ = − ∂ + ∂ ∂ − ∂ + − ∂ + ∂ Jacobian : Let u = u (x , y) , v = v(x , y) are two functions of the independent variables x , y. The jacobian of ( u , v ) w .r .t (x , y ) or the jacobian transformation is given by the determinant ∂ ∂ (or) ∂ ∂ |∂ ∂| The determinant∂ value∂ is denoted by J ( ) or

Similarly if u = u(x, y, z ) , v = v (x, y , z) , w = w(x, y , z) ,then the Jacobian of u , v , w w.r.to x , y , z is given by

J ( ) = =

Properties of Jacobians  vu ),(  yx ),( 1. If J  and J 1  then JJ 1 1  yx ),(  vu ),(

2. If u,v are functions of r, s and r, s are functions of x ,y , then = , , , Solved Problems: , , . ,  zyx ),,( 1. If x + y2 = u , y + z2 = v , z + x2 = w find  vu ,,( w) Sol : Given x + y2 = u , y + z2 = v , z + x2 = w

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We have = =

= 1(1-0) – 2y(0 – 4xz) + 0 = 1 – 2y(-4xz) = 1 + 8xyz

 = =

2. Show that the functions u = x + y + z , v = x2 + y2 + z2 -2xy – 2yz -2xz and w = x3 + y3 + z3 -3xyz are functionally related. Sol: Given u = x + y + z v = x2 + y2 + z2 -2xy – 2yz -2xz w = x3 + y3 + z3 -3xyz we have

c  c  c 1 1 2 c2  c2  c3 0 0 1  6 2x  2y 2y  2z z  y  x x2  yz  y2  xz y2  xz z2  xy z2  xy

=6[2(x - y) (y2 + xy – xz -z2 )-2(y - z)(x2 + xz – yz - y2)] =6[2(x - y)( y – z)(x + y + z) – 2(y – z)(x – y)(x + y + z)]

=0 Hence there is a relation between u,v,w.

3. If x + y + z = u , y + z = uv , z = uvw then evaluate

Sol: x + y + z = u y + z = uv z = uvw y = uv – uvw = uv (1 – w) x = u – uv = u (1 – v)

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R2  R2  R3

= uv [ u –uv +uv] = u2v

4. If u = x2 – y2 , v =2xy where x = r cos , y = r sin S.T = 4r3

Sol: Given u = x2 – y2 , v = 2xy =r2cos2 – r2sin2 = 2rcos r sin = r2 (cos2 – sin2 = r2 sin2 = r2 cos2

= =

= (2r)(2r)

= 4r2 [rcos22 + r sin22 ] =4r2(r)[ cos22 + sin22 ] =4r3

5. If u = , v = , w = find

Sol: Given u = , v = , w =

We have

2 ux = yz(-1/x ) = , uy = , uz =

= , xz(-1/y2) = ,

= , = , = xy (-1/z2) =

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= . .

= 1[-1(1-1) -1(-1-1) + (1+1) ] = 0 -1(-2) + (2) =2 + 2 =4

6. If x = er sec , y = er tan P.T . = 1

Sol: Given x = er sec , y = er tan

= , =

= er sec = x , = ersec tan = er tan = y , = er sec2 x2 – y2 = e2r (sec2 - tan2 )  2r = log (x2 – y2 )  r = ½ log (x2 – y2 ) 1 1 x r  x)2(  x 2 x2  y2 (x2  y2 ) 1 1  y ry  2 2  y)2(  2 2 2 x  y (x  y ) = = =

 = , = sin-1( )

1  1   y  x  y 2   y 2  x  x x2  y2 1 x2

= (1/x) =

= = e2r sec2 - y er sec tan

= e2r sec [sec2 - tan2 ] = e2r sec

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x  y r,  (x2  y 2 ) (x2  y 2 )   yx ),(  y 1 x x2  y 2 x2  y 2

=[ - ]

= = =

. = 1

Functional Dependence: Two functions u and v are functionally dependent if their Jacobian i.e,

J ( ) = = = 0

If the Jacobian of u, v is not equal to zero then those functions u, v are functionally independent.

Solved Problems :

1. If , . Find . Hence prove that u and v are + − − , functionally = − dependent. = 퐚 Find + 퐚the relation between, them. Sol : Given u = and + − − = , − = v =, tan = x +and tan = y ∂ + ∂ + ∂ ∂ ∴ ∂ − ∂ − ∂ + ∂ + = ∂ ∂ = + + = - = 0 ∂, ∂ ∂ − − ∂, ∂ ∂ − − ∴ | | | | ∂ ∂ + + u and v are functionally dependent .

∴Now v = = = − − − + − tan isx +the tan functionaly tan relation − between tan u andu v. − 2. ∴Determine v = tan whetheru the following functions are functionally dependent or not. If they are functionally dependent , find a relation between them. i) u = , v = ii) u = , v = + − Sol: i) Jacobian = = = ∂, u u e sin y e cos y ∂, | | | | v v e cos y − e sin y DEPARTMENT OF HUMANITIES & SCIENCES ©MRCET (EAMCET CODE: MLRD) 88

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= ( - = u,v aree functionallysin y − cos independenty − ≠ . ii) ∴u = , v = + − J( ) = = − = - = 0 , ∂, , ∂, − − − ∴ | | u and v are functionally− dependent − ,

∴ = = + + + Now v = − − − is the functional relation between u and v . + 3. Show that∴ v =the − functions u= xy + yz + zx , v = and w= x + y + z are functionally related .find the relation between them. + + Sol: We have u = xy + yz + zx , v = , w= x + y + z x + y + z ∂ ∂ ∂ = ∂ ∂ ∂ = ∂,, ∂ ∂ ∂ y + z z + x x + y | | ∴ ∂,, ∂ ∂ ∂ | x y z | | | ∂ ∂ ∂ ∂ ∂ ∂ = 2 ( Applying ) + + + | | → + = 2(x + y + z) | | = 2(x + y + z) (0) ( since ) = 0 RandR are identical Hence u,v and w are functionally dependent . That is , the functional relationship exists between them. Now = w = x + is y +the z functional x + yrelation+ z +between xy + u yz,v +andzx w. = v + u 4. ∴Verify w = if uv += 2xu – y + 3z , v = 2x – y – z , w = 2x – y + z are functionally dependent and if so , find the relation between them. Sol: Given u = 2x – y + 3z , v = 2x – y – z , w = 2x – y + z The functions u,v,w are functionally dependent if and only if J( = 0 ,, ,

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∂ ∂ ∂ Now J( = ∂ ∂ ∂ = = 2(-1) = (-2)(0) =0 ,, ∂ ∂ ∂ − | | , ∂ ∂ ∂ | − −| | −| | | ∂ ∂ ∂ − ∂ ∂ ∂ Now u + v -2w = (2x-y +3z) + (2x y z) 2 (2x y + z ) ∴ u, v, w are functionally dependent– – – – = (4x – 2y + 2z) – (4x – 2y +2z) = 0 Hence u + v -2w = 0 is the functional relationship between u, v and w. 5. Show that the functions u = x+y+z, v = x2+y2+z2-2xy-2yz-2zx and w = x3+y3+z3-3xyz are functionally related.

Sol. Given u = x+y+z, v = x2+y2+z2-2xy-2yz-2zx and w = x3+y3+z3-3xyz

Now = = ,, | | ,, |−− −− −−| | | − − − = 6

|−− −− −− | − − − = 6 C1 and C2 | − − −− | → − −++ −++ − → − = 12 ,, − − ,, = 12 (x-y)(y-z) ∴ | − + + − + + | = 12 (x-y)(y-z) (0) [C and C are identical] | 1 2 | = 0 ++ ++ Hence the functional relationship exists between u, v, w. 6. Prove that u = , v = are functionally dependent and find the relation between − them. + +

Sol. We are given u = , v = − = + = + = ( + ).−( − ). + − + = + = + = + ∴ ( + ).−−( − ). − + − + − =2 y + = and + + ( + ).−. − =2 x + = + [ ] ( + ).−. − [ + ] +

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Thus = = − , + + , | | | − − | = + + + − − = + + = 0 ( − )− − u,v are functionally dependent. + u2 + v2 = + = = 1 ∴ 2 2− + Hence u + v = 1 is the functional relation between u and v. + + +

Maxima & Minima for functions of two Variables: Definition : Let f(x,y) be a function of two variables x and y. At x = a ; y = b , f(x ,y) is said to have maximum or minimum value , if f (a ,b) > f( a +h , b +k) or f (a ,b) < f( a +h , b +k) respectively where h and k are small values. Extremum : A function which have a maximum or minimum or both is called ‘extremum’ Extreme value :- The maximum value or minimum value or both of a function is Extreme value. Stationary points: - To get stationary points we solve the equations = 0 and

= 0 i.e the pairs (a1, b1), (a2, b2) ………….. are called

Stationary.

Working procedure:

1. Find and Equate each to zero. Solve these equations for x & y we get the pair

of values (a1,b1) (a2,b2) (a3 ,b3) ……………… 22ff 2 f 2. Find l = ,m  , n = x2  x  y y 2

2 3 i) If l n –m > 0 and l < 0 at (a1,b1) then f(x ,y) is maximum at (a1,b1) and maximum

value is f(a1,b1) 2 ii) If l n –m > 0 and l > 0 at (a1,b1) then f(x ,y) is minimum at (a1,b1) and minimum

value is f(a1,b1) . 2 iii) If l n –m < 0 and at (a1, b1) then f(x, y) is neither maximum nor minimum at (a1, b1).

In this case (a1, b1) is saddle point.

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2 iii) If l n –m = 0 and at (a1, b1) , no conclusion can be drawn about maximum or minimum and needs further investigation. Similarly we do this for other stationary points. Solved Problems: 1. Locate the stationary points & examine their nature of the following functions. u =x4 + y4 -2x2 +4xy -2y2, (x > 0, y > 0) Sol: Given u(x ,y) = x4 + y4 -2x2 +4xy -2y2 u For maxima & minima = 0, u = 0 x y

= 4x3 -4x + 4y = 0  x3 – x + y = 0 ------> (1)

= 4y3 +4x - 4y = 0  y3 + x – y = 0 ------> (2)

Adding (1) & (2), x3 + y3 = 0  x = – y ------> (3)

(1)  x3 – 2x  x = 0, 2, 2

Hence (3)  y = 0, - 2, 2

 2u 2u  2u l = = 12x2 – 4, m = = ( ) = 4 & n = = 12y2 – 4 x2  yx y 2 ln – m2 = (12x2 – 4 )( 12y2 – 4 ) -16 At ( , ), ln – m2 = (24 – 4)(24 -4) -16 = (20) (20) – 16 > 0 and l=20>0 The function has minimum value at ( , ) At (0,0) , ln – m2 = (0– 4)(0 -4) -16 = 0 (0,0) is not a extreme value. 2. Investigate the maxima & minima, if any, of the function f(x) = x3y2 (1-x-y).

Sol: Given f(x) = x3y2 (1-x-y) = x3y2- x4y2 – x3y3

= 3x2y2 – 4x3y2 -3x2y3 = 2x3y – 2x4y -3x3y2

For maxima & minima = 0 and = 0

 3x2y2 – 4x3y2 -3x2y3 = 0 => x2y2(3 – 4x -3y) = 0 ------> (1)  2x3y – 2x4y -3x3y2 = 0 => x3y(2 – 2x -3y) = 0 ------> (2) From (1) & (2) 4x + 3y – 3 = 0 2x + 3y - 2 = 0

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2x = 1 => x = ½ 4 ( ½) + 3y – 3 = 0 => 3y = 3 -2 , y = (1/3) 2 f l = = 6xy2-12x2y2 -6xy3 x2  2 f    = 6(1/2)(1/3)2 -12 (1/2)2(1/3)2 -6(1/2)(1/3)3 = 1/3 1/3 -1/9 = -1/9  2  (1/2,1/3) –  x 

2  f   f  2 3 2 2 m = =   = 6x y -8 x y – 9x y  yx x  y 

2   f  2 3 2 3   (1/2 ,1/3) = 6(1/2) (1/3) -8 (1/2) (1/3) -9(1/2) (1/3) = =  xy  2 f n = = 2x3 -2x4 -6x3y y2

 2 f    = 2(1/2)3 -2(1/2)4 -6(1/2)3(1/3) = - - = -  2  (1/2,1/3)  y  1 ln- m2 =(-1/9)(-1/8) –(-1/12)2 = - = = > 0 and l =  0 9 The function has a maximum value at (1/2 , 1/3)  1 1   1 1  1 1  1  1 1  1 Maximum value is f  ,     1        2 3 8 9 2 3 72 2 3 432        3. Find three positive numbers whose sum is 100 and whose product is maximum. Sol: Let x ,y ,z be three +ve numbers. Then x + y + z = 100  z = 100 – x – y Let f (x,y) = xyz =xy(100 – x – y) =100xy –x2y-xy2 For maxima or minima = 0 and = 0

=100y –2xy-y2 = 0 => y(100- 2x –y) = 0 ------> (1)

= 100x –x2 -2xy = 0 => x(100 –x -2y) = 0 ------> (2)

100 -2x –y = 0 200 -2x -4y =0 ------100 + 3y = 0 => 3y =100 => y =100/3 100 – x –(200/3) = 0 => x = 100/3

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2 f = =- 2y x2  2 f    (100/3 , 100/3 ) = - 200/3  2   x 

2 f   f  = =   = 100 -2x -2y  yx x  y   2 f    (100/3 , 100/3 ) = 100 –(200/3) –(200/3) = -(100/3)  xy  2 f = = -2x y2 2 f    (100/3 , 100/3 ) = - 200/3  2   y  ln -m2 = (-200/3) (-200/3) - (-100/3)2 = (100)2 /3 The function has a maximum value at (100/3 , 100/3) 100 100 100 i.e. at x = 100/3, y = 100/3 z = 100    3 3 3 The required numbers are x = 100/3, y = 100/3, z = 100/3 4. Find the maxima & minima of the function f(x) = 2(x2 –y2) –x4 +y4 Sol: Given f(x) = 2(x2 –y2) –x4 +y4 = 2x2 –2y2 –x4 +y4 For maxima & minima = 0 and = 0

= 4x - 4x3 = 0 => 4x(1-x2) = 0 => x = 0 , x = ± 1

= -4y + 4y3 = 0 => -4y (1-y2) = 0 =>y = 0, y = ± 1

 2 f  l =   = 4-12x2  2   x 

 2 f    f m =  =  = 0  xy  xy

 2 f  n =   = -4 +12y2  2   y  we have ln – m2 = (4-12x2)( -4 +12y2 ) – 0 = -16 +48x2 +48y2 -144x2y2 = 48x2 +48y2 -144x2y2 -16 i) At ( 0 , ± 1 ) ln – m2 = 0 + 48 - 0 -16 =32 > 0 l = 4-0 = 4 > 0

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f has minimum value at ( 0 , ± 1 ) f (x ,y ) = 2(x2 –y2) –x4 +y4 f ( 0 , ± 1 ) = 0 – 2 – 0 + 1 = -1 The minimum value is ‘-1 ‘. ii) At ( ± 1 ,0 ) ln – m2 = 48 + 0 - 0 -16 =32 > 0 l = 4-12 = - 8 < 0 f has maximum value at ( ± 1 ,0 ) f (x ,y ) = 2(x2 –y2) –x4 +y4 f ( ± 1 , 0 ) =2 -0 -1 + 0 = 1 The maximum value is ‘1 ‘. iii) At (0,0) , (± 1 , ± 1) ln – m2 < 0 l = 4 -12x2 (0 , 0) & (± 1 , ± 1) are saddle points. f has no max & min values at (0 , 0) , (± 1 , ± 1).

Lagrange’s method of undetermined multipliers Suppose it is required to find the extremum for the function f(x , y , z) subject to condition ( x , y , z) = 0 ------(1) Step 1 : Form lagrangean function F(x , y , z) = f(x , y , z) + ( x , y , z) where is called Lagrange’s constant, which is determined by the following conditions. Step 2: Obtain the equations F = 0 => + = 0 ------(2) x F = 0 => + = 0 ------(3) y F = 0 => + = 0 ------(4) z Step 3: Solving the equations (1) (2) (3) & (4) we get the stationary point (x, y, z). Step 4 : Substitute the value of x , y , z so obtained in equation (1) we get the extremum.

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Solved Problems: 1. Find the minimum value of x2 +y2 +z2, given x + y + z =3a Sol: u = x2 +y2 +z2 = x + y + z - 3a = 0 Using Lagrange’s function F(x , y , z) = u(x , y , z) + ( x , y , z) For maxima or minima F = + = 2x + = 0 ------(1) x F = + = 2y + = 0 ------(2) y

F = + = 2z + = 0 ------(3) z (1) , (2) & (3) = -2x = -2y = -2z

= x + x + x - 3a = 0 = a = y =z = a Minimum value of u = a2 + a2 + a2 =3 a2 2. Find the minimum value of , given that xyz = Sol: Let u = ……+ (1)+ 퐚 And = xyzx - += y 0 + z …… (2) Consider∅ the lagrangean function F(x,y,z) = u( x,y,z ) + i.e , F(x,y,z) = + ) ……(3)λ ϕx, y, z Now = 0 x + y =+ 2x z + λyz xyz = 0 − a ……(4) ∂F ∂ ∂∅ ∂ = 0 ⇒ ∂ + λ ∂ = 2y + λxz = 0 ……(5) ∂F ∂ ∂∅ ∂ = 0 ⇒ ∂ + λ ∂ = 2z + λyx = 0 ……(6) ∂F ∂ ∂∅ From∂ (4) , (5)⇒ and∂ + (6) λ ∂,we have λ ……(7) λ From the first two members , we have= = = − …(8) From the last members , we have = ⇒ x = y ……(9) From (8) and (9) , we have = ⇒ y = z ……(10) x = y = z ⇒ x = y = z DEPARTMENT OF HUMANITIES & SCIENCES ©MRCET (EAMCET CODE: MLRD) 96

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on solving (2) and (10) , we get , = a

u = x = y = z u 3. ∴Find Minimum the maximum value of value a of+ a = = a if 2x + 3y + 4z = a Sol: Given u = .….. (1) Let = x 2xy +z 3y + 4z - a = 0 ……(2) Considerϕx, y,the z lagrangean function F(x,y,z) = u( x,y,z ) + i.e, , F(x,y,z) = + (2x + 3y + 4z – a) ……(3)λ ϕx, y, z for maxima or minimax y z =λ 0 , = 0 , = 0 ∂F ∂F ∂F Now = 0 2x +∂ 2 = 0∂ x ∂ ……(4) ∂F ∂ = 0 ⇒ 3 +3λ = 0 ⇒ = − λ …….(5) ∂F and ∂ = 0 ⇒ 4 x y z + λ =0 ⇒ x y z = − λ …….(6) ∂F From (4)∂ and (5)⇒ , wex yhavez x = λ y ⇒ x y z = − λ …….(7) From (5) and (6) , we have y = z …….(8) Hence from (7) and (8) , we get x = y = z …….(9) On solving (2) and (9) ,we get x = y = z = a

a a a a 4. Show that the rectangular∴ Maximum solid value of maximumof u = volume that = can be inscribed in a sphere is a cube. Sol: Let 2x ,2y, 2z are the length , breadth and height of rectangular solid Then its volume V = 8 xyz ……..(1) Let the sphere have a radius of ‘r’ so that ……(2) Consider the lagrangean function F(x,y,z)x = +u( yx,y,z+ z) += r i.e, F(x,y,z) = V + λ ϕx, y, z = 8xyz 휆+ x + y + z − r …….(3) For maxima or minima 휆 x = 0 + , y =+ 0 z , − r = 0 ∂F ∂F ∂F = 0 8yz +2 = 0∂ ∂ ∂ ……(4) ∂F ∂ = 0 ⇒ 8zx +2 λx = 0 ……(5) ∂F ∂ = 0 ⇒ 8xz +2 λy = 0 ……(6) ∂F From∂ (4)⇒ ,(5) and λz(6) we have 2 = - 8xyz = - 2 = - 2 x =휆 y = z 휆 휆 ⇒ DEPARTMENT OF HUMANITIES & SCIENCES ©MRCET (EAMCET CODE: MLRD) 97

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Thus for a maximum value x = y = z which shows that the rectangular solid is a cube.

Taylor’s series for a function of two variables: Consider a function f(x,y) defined in a region enclosing (a,b) and having successive partial derivatives ,then taylor’s series gives an expansion of f(x,y) in powers of (x-a) and (y-b) and partial derivatives of f at (a,b) and is expressed in ascending powers of (x-a) and (y-b) . f(x,y) = f(a,b) + (x-a) fa, b + y − bfa, b + ! [ x − a fa, b + x − a y − bfa, b + (y − b fa. b] + ! [ x − a fa, b + + …….x − a y − bNote:f Thea, b above+x−ay−b expansion is calledf a,the b expansion+ (y − b off f(x,y)a, bat ](a,b) or in the neighbouhood of (a,b) or in powers of (x-a) and (y-b) . Solved Problems: 1. Expand near (1, 흅 Sol: Let f(x,y) = f(1, = . Then 휋 e cos y ⇒f(1, = √ π fx, y = e cos y ⇒ f(1, =√ π − fx, y = −e sin y ⇒ f(1, =√ π − fx, y = −e cos y ⇒ f(1, = √ π − fx, y = −e sin y ⇒ f(1, = √ π − fx, y = −e cos y ⇒ √ ′ By substituting above values in taylor s series , we get = π − π − π e cos y √ [ + x − − y − + ! − x − y − − ! + ⋯ ] 2. Expand using taylor’s thorem. + − − 퐚 + Sol: Let f(x,y) = , a = 1 , b = -2 Now f(a,b) = f(1,-2)x y = + -10y − = 2xy

fa, b ⇒ f, − = − = 4 fa, b == 2y x + ⇒ f, = − - 4 fa, b = 2x ⇒ f, − = 4 fa, b ⇒ f, − DEPARTMENT OF HUMANITIES & SCIENCES ©MRCET (EAMCET CODE: MLRD) 98

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= 0 = 0

fa, b = 0 ⇒ f, − = 0 fa, b = 2 ⇒ f, − = 2 fa, b = 0 ⇒ f, − = 0 fa, b = 0 ⇒ f, − = 0 fAll othera, b partial⇒ derivatives f, − of higher order will vanish

′ By substituting above values in taylor s series , we get x y + y − = + [x − −+y+]++3( [x − − + x − y + + y + + [x − x − y + + x − y + + y= 10 + - 4(x - 1)] + 4( y + 2) - 2( + 2 (x - 1) (y + 2) + ( (y + 2) x − x −

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UNIT – III

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ORDINARY DIFFERENTIAL EQUATIONS

I.Differential equations of first order and first degree Definition: An equation involving variables and its differentials is called a Differential equation. They are of two types 1. Ordinary differential equations 2. Partial differential equations

Ordinary differential equation: An equation is said to be ordinary if one or more variables are differentiated w.r.to only one independent variable. dy d2 y dy Ex . (1) 7xy x2 (2) 3  2y  ex dx dx2 dx Partial Differential equation: A Differential equation is said to be partial if the derivatives in the equation have reference to two or more independent variables.

2 2 zz E. g: 1. 4z xy zz 2. x y 2z xy Order of a Differential equation: It is the order of the highest derivative occurring in the Differential equation. Differential equation is said to be of order ‘n’ if the derivative is the highest derivative in that equation. dy E. g : (1). (x2+1) . + 2xy = 4x2 dx Order of this Differential equation is 1. d2 y dy (2) x 2x  1   x  1  y  ex dx2 dx

Order of this Differential equation is 2.

2 d2 y dy (3) 2 5  2y  0 . dx dx Order=2 22uu (4) 0. Order is 2. xy22

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Degree of a Differential equation: Degree of a differential Equation is the highest degree of the highest derivative in the equation, after the equation is made free from radicals and fractions in its derivations.

2 dy dy E.g : 1 ) y = x . 1  on solving we get dx dx

2 22dy dy 1 x   2xy.  1  y   0 . Degree = 2 dx dx

2 32 d2 y dy 2) a. 2 1  on solving . we get dx dx

3 2 2 2 2 d y dy a .2  1 . Degree = 2 dx dx

dy The general form of first order, first degree differential equation is  f x, y  or dx [i.e Where are functions of and ]. There ′ is, no , general = method to solve + any = first order differential equation The equation which belong to one of the following types can be easily solved. In general the first order first degree differential equation can be classified as: (1) Exact equations (2) Non exact equations (reducible to exact equations). Exact Differential Equations Def: Let M(x, y) dx +N(x, y) dy =0 be a first order and first degree Differential Equation where M & N are real valued functions of x, y . Then the equation M dx + N dy =0 is said to be an exact Differential equation if  a function f . ff d f x, y   dx dy xy 22 Eg: d ( x y ) 2 xydx x dy

Condition for Exactness: If & are two real functions which have continuous partial derivatives then, the necessary , and sufficient condition for the Differential equation is to be exact if MN  + = yx Hence solution of the exact equation is

, + , = DEPARTMENT OF HUMANITIES & SCIENCES ©MRCET (EAMCET CODE: MLRD) 102

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Mdx Ndy c . (y is taken as constant) (terms free from x are taken).

Solved Problems : dy ycos x sin y y 1. Solve 0 . dxsin x x cos y x Sol : Given equation can be written as (y cos x sin yydx  )  (sin xx  cos yxdy  )  0 …(1) It is of the form .

Here + = M ycos x  sin y  y

Nsin x  x cos y  x M cosxy  cos  1 y

N cosxy  cos  1 x MN Clearly  yx Equation is exact.

The⟹ general solution is given by Mdx Ndy c (y constant) (terms independent of x) (y cos x  sin y  y ) dx  (0) dy  c ysin x  (sin y  y ) x  c.

xxx 2. Solve 1 eyy dx  e 1  dy  0    y xx yyx Sol : Here M 1  e & N  e 1  y x x x My  x   N y   1   x  y  1  e2  &  e     1   e   y y  x  y   y   y  xx Myy  x   N   x   e22  &  e   y y  x  y  MN  equation is exact yx General solution is

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Mdx Ndy c. (y constant) (terms free from x) x 1  ey dx  0 dy  c.   x e y xc   1 y x x  yey  c

3. Solve the D.E (x+y-1) dy-(x-y+2) dx=0

Sol : Here M  x  y  2  ; N x  y  1  MN 1; 1 yx MN Clearly  yx Thus the equation is exact. General solution is Mdx Ndy c . (y constant) (terms free from x) (x  y  2) dx  ( y  1) dy  c ⟹ xy22   xy 2 x   y  c 22 22 x  y 2 xy  4 x  2 y  c1 4. Solveeyy 1  .cos xdx  e sin xdx  0 .

Sol. M eyy  1  cos, N  e sin x MN eyy cos x; e cos x yy MN ey cos x yx Equation is exact. Gen Sol. is Mdx Ndy c .

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(y constant) (terms free from x) ey  1  cos x dx  0dy  c ey sin x c

1 5. Solve y 1  cos y dx  x  log x  xsin y  dy  0 . x

1 Sol : Here M y 1   cos y, N  x  log x  x sin y x M 1 N 1 1   sin y  1   sin y y x x x MN  so the equation is exact yx General sol Mdx Ndy c . (y constant) (terms free from x) y y   cos y dx  0.dy  c. x y x  log x   xcos y  c.

6. Solve (cosx-xcosy)dy – (siny+ysinx)dx =0

Sol : N cos x  x cos y&M   sin y  ysin x NM  sin x  cos y   cos y  sin x xy MN  the equation is exact. yx General sol Mdx Ndy c . (y constant) (terms free from x)  sin y  ysin x  .dx  0.dy  c  x sin y  ycos x  c

 ycosx - xsiny = c 7. Solver sin  cos   dr  r  sin  cos   d   0 Sol : Given equation isr sin  cos   dr  r  sin  cos   d   0......  1  This is of the form Md Ndr  0 Where M r sin  cos   ;N  r  sin  cos 

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MN We have sin  cos  ;  cos   sin  r  MN Clearly  r  The given equation is exact. The general solution is given by Md  Ndr c (r constant) (terms independent of ) r(sin  cos  ) d   rdr  c 휃 r 2 rc(sin  cos )   2

2 The general solution is r2 r sin  cos   c1 .

REDUCTION OF NON-EXACT DIFFERENTIAL EQUATIONS TO EXACT FORM USING INTEGRATING FACTORS Definition: If the Differential Equation M(x,y) dx + N (x,y ) dy = 0 be not an exact differential equation it can be made exact by multiplying with a suitable function u (x,y) 0 called an Integrating factor(I.F). Note: There may exits several integrating factors.

Some methods to find an I.F to a non-exact Differential Equation Mdx+N dy =0 Case -1: Integrating factor by inspection/ (Grouping of terms).

Some useful exact differentials 1 . d (xy) = xdy +y dx x ydx xdy 2. d   2 y y x ydx xdy 3. d   2 y x

xy22 4. d  = x dx + y dy 2

x xdy ydx 5. d log = y xy

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x ydx xdy 6. d log = y xy

1 x ydx xdy 7. d tan  = 22 y xy

1 x xdy ydx 8. d tan  = 22 y xy xdy ydx 9. dlog xy   = xy

22 2xdx ydy  10. dlog x y  =   xy22

ex yexx dx e dy 11. d  = 2 y y

Solved Problems y() xy exx dx e dy 1) Solve  0 . y2

()xy2  yexx dx e dy Sol : It can be written as  0 . y2 xy2 exx ydx e dy dx   0 yy22 yexx dx e dy xdx   0 y2 ex xdx  d   0 y xe2 x On integrating, we get c 2 y This is the required solution.

2) Solve the differential equation y x22  dx   xcot y  x  dy  0

Sol : Given equation can be written as ydx xdy  x22 dx  xcot y dy

Dividing with x 2 , we get ydx xdy xdy ydx 22dx cot y dy or    dx  cot y dy xx y i.e , d  dx  cot y dy x

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y Integrating, we get  x cos ecy22  cos y   xx  cos ecycx   x which is the required solution.

xdy ydx 3) Solve xdx ydy   0 xy22

xdy ydx Sol : Given equation is xdx ydy   0 xy22

22 x y 1  y  ddtan 0 on Integrating we get 2 x

22 x y1  y tan c . 2 x 4) Solve yxe33.xy ydx   xy   xe . xy  dy  0 .

Sol : Given equation is on regrouping We get yx3 exy dx y 2 dx  xydy  x 4 e xy dy  0 x3 exy  ydx xdy   y  xdx  ydx   0 Dividing by x3

xy y   xdy ydx  e ydx xdy    . 2   0 xx   

xy yy    d e    .d   0 xx    on Integrating

2 y ecxy 1 is required G.S. 2 x 5) Solve (1+xy) x dy + (1- yx ) y dx = 0 Sol: Given equation is (1+xy) xdy + (1-yx) y dx =0. ( xdy + ydx ) + xy ( xdy – ydx ) = 0.

2 2 xdy ydx   xdy ydx  Divided by x y 22      0 x y   xy  1 1 1 d   dy  dx  0. xy y x 1 On integrating we get  log y  log x  logc xy

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1  log x  log y  logc . xy 6) Solve : ydx xdy  a x22  y  dx Sol : Given equation is ydx ydx xdy  a x22  y  dx ydx xdy a dx xy22 

1 x d tan a dx y x On Integrating tan1  ax c where c is an arbitrary constant. y Method -2: If Mx,ydx  Nx,ydy   0 is a homogeneous differential equation and 1 Mx Ny 0 then is an integrating factor of M dx+ N dy =0. Mx Ny Solved Problems : 1 . Solve x2 ydx x 3  y 3  dy  0 Sol : Given equation is x2 ydx x 3  y 3  dy  0 ------(1)

Where M x2 y & N   x 3  y 3  M N Consider  x2 & 3x2 y x MN  yx

equation is not exact . But given equation (1) is homogeneous differential equation then Mx Ny  x x2 y   y  x 3  y 3    y 4  0. 11 I.F =  Mx Ny y4 1 Multiplying equation (1) by  y4

x2 y x 3 y 3 dx  dy  0------(2) yy44

x2 x 3 y 3  dx  dy  0 yy34

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This is of the form M11 dx N dy 0

x2 xy33 For M  & N  1 y3 1 y4

M 3x2 N 3x2 1 & 1  yy4 xy4 MN 11 equation (2) is an exact D.E. yx

General sol M dx N dy c 11

(y constant) (terms free from x in N1) x12 dx  dy  c. yy3

x3  log y  c 3y3 2. Solve yy2 2x 2  dx  x2y  2  x 2  dy  0 Sol : Given equation is yy2 2x 2  dx  x2y  2  x 2  dy  0 ------(1) It is the form Mdx + Ndy = 0 Where M yy2  2x 2  ,N  x2y  2  x 2  M N Consider 3y22 2x & 2y22 3x y x MN  equation is not exact . yx Since equation(1) is Homogeneous differential equation then 2 2   2 2  Consider Mx Ny  xyy  2x    yx2y    x   3xy y22  x   0 . 1 I.F. 3xy y22 x  1 Multiplying equation (1) by we get 3xy y22 x  y y2 2x 2  x 2y 2 x 2  dx  dy  0 3xy y2 x 2 3xy y 2 x 2    

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y(y2 - 2x 2 ) x(2y 2 - x 2 ) M = ;N = 113xy(y2 - x 2 ) 3xy(y 2 - x 2 )

MN112xy ==2 2 2 y 3(y - x ) x Now it is exact y2 x 2   x 2 y 2  x y 2  x 2  dx dy 0 3x y2 x 2  3y y 2 x 2  dx xdx ydy dy     0 . x y2 x 2 y 2 x 2 y

dx dy 2ydy 2xdx     0 xy 2 y2 x 2  2 y 2 x 2  On integrating we get 11 log x log y  log y2  x 2   log y 2  x 2   log c  xy  c 22 Method- 3: If the equation Mdx Ndy 0 is of the form 1 y.f (xy)dx x.g(xy)dy 0 & Mx Ny 0 then is an integrating factor of Mx Ny Mdx+ Ndy =0.

Problems : 1 . Solve xysin xy cosxy  ydx   xysin xy  cosxy  xdy  0. Sol : Given equation xysin xy cosxy  ydx   xysin xy  cosxy  xdy  0------(1). Equation (1) is of the form y. f xy  dx x.g  xy  dy 0. Where M xysin xy cosxy  y N xysin xy cosxy  x MN  yx

equation (1) is not exact Now consider Mx-Ny ⟹ Mx-Ny =2xycosxy ≠ 0 1 Integrating factor = 2xycos xy

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So equation (1) x I.F gives xysin xy  y xysin xy cos xy  x dx  dy  0 2xycos xy 2xycos xy

11 y tan xy  dx  y tan xy  dy  0 xy

M11 dx  N dy  0 MN 11= tanxy + xysec2 xy = yx Now the equation is exact.

 General solution is M dx N dy c. . 11

(y constant) (terms free from x in N1) 11 y tan xy  dx  dy  c. xy y.log secxy  log x   log y   logc y x log sec xy   log  log c. y x .sec xy c. y 2. Solve 1 xy  ydx   1  xy  xdy  0 Sol : Here M 1  xyy:N    1  xyx  MN =1+ 2xy; =1- 2xy yx Hence, the equation is not exact Also Mx - Ny= 20xy22 11 I.F   0 Mx Ny 2x22 y Multiply the given equation by I.F, we get 1 1 1 2 dx  dy  0 2x y 2x 2y MN-1 11== y 2x22 y x Equation is exact.

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On integrating, we get 1 1 1 dx  dy  c 2 2x y 2x 2y 1 1 1  log x  log y  c . 2xy 2 2

1x  log  c1 where = 2c. xy y Method 4: If there exists a continuous single variable function such that MN  yx f x  dx  f x  , then I.F. of Mdx Ndy 0 is e N Solved Problems : 1 . Solve 3xy 2ay22  dx   x  2axy  dy  0 Sol : Given equation is 3xy 2ay22  dx   x  2axy  dy  0 This is of the form Mdx+ Ndy = 0

M  3xy  2ay22 & N  x  2axy MN 3x  4ay &  2x  2ay yx MN  equation is not exact . yx MN  yx3x 4ay    2x  2ay  Now consider  N x x 2ay 

MN  yx1    fx  . Nx

1  dx exx is an Integrating factor of (1) Multiplying equation (1) with I.F we get 3xy 2ay22   x 2axy  xdx  xdy  0 11 3x2 y 2ay 2 x  dx   x 3  2ax 2 y  dy  0

It is the form M1dx + N1dy =0

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M 3x2 y  2ay 2 x, N  x3  2ax 2 y 11 MN 113x22  4axy,  3x  4axy yx MN 11 Equation is exact yx General sol M dx N dy c. 11

(y constant) (terms free from x in N1) 3x22 y  2ay x  dx  0dy  c

x3 y  ax 2 y 2  c .. 2. Solve ydx xdy  1  x22  dx  x sin ydy  0 Sol: Given equation is y 1  x22  dx   x sin y  x  dy  0.

M y1x  22 &N  xsinyx  MN 1,  2x sin y  1 yx MN the equation is not exact. yx MN  yx1 2x sin y  1  2x sin y  2  2 x sin y  1   2 So consider     N x22 sin y x x sin y  x x x sin y  1  x

1 2 dx f x  dx  2log x 1 I.F e  ex  e  x2 y 1  x22 x sin y  x Equation (1) x I.F gives dx  dy  0 xx22

It is the form of M1dx+ N1dy =0.

MN111 ==2 y x x  Equation is exact y1 Gen. sol. is thus 22  1 dx  sin ydy  0 xx y1   x  cos y  c . xx

2 x  y  1  x cos y  cx.

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3. Solve 2xy dy – (x2+y2+1) dx = 0

Sol. Given equation is 2xy dy – (x2+y2+1) dx = 0 ………(1) This is of the form M dx + N dy = 0, where N = 2xy, M = -x2 - y2 – 1 We have = -2y and = 2y, so that The given equation is not exact. ≠ We have = (-2y-2y) = = f(x) ∴ − I.F. = − = = − ∫ − ∫푥 − Multiplying (1) with , we get ∴ = = ….(2) This is of the form M1 dx + N1 dy = 0, where M1 = and N1 = − + + We have = and = − − − − − Since = , (2) is exact. General solution is given by ∴ M dx N dy c 11

(y constant) (terms free from x in N1)

(y⇒ constant)∫− − − + ∫ = -x + = c This is the general solution o (2) and hence of (1) ⇒ + MN  yx Method -5: For the equation Mdx + Ndy = 0 if  gy() (is a function of y alone) M then eg() y dy is an integrating factor of M dx + N dy =0.

Solved Problems : 1 . Solve (3x2y4+2xy)dx +(2x3y3-x2) dy = 0 Sol : Given equation (3x2y4+2xy)dx +(2x3y3-x2) dy =0 ------(1). Equation is of the form Mdx + Ndy =0. where M =3x2y4+2xy & N = 2x3y3-x2

MN 12xy2 3  2x;  6xy 2 3  2x yx

MN equation(1) is not exact. yx MN  yx 2 So consider g() y   g y  My

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1 2 dy g() y dy yy2log 1 I. F e  e  e  . y2

3x2 y 4 2xy   2x 3 y 3 x 2  Equation (1) x I.F 22 dx   dy 0 yy   

2 2 22x 3 x 3xy  dx  2xy 2 dy0  yy 

It is the form M11 dx N dy 0 MNx 11= 6x2 y - 2 = y y2 x

 Equation is exact

General sol. is M dx N dy c 11

(y constant) (terms free from x in N1)

22 2x 3x y  dx  0dy  c . y 3x3 y 2 2x 2    c. 3 2y

x2 x32 y   c. y 2 . Solve (xy3+y) dx + 2(x2y2+x+y4) dy =0 Sol : Here M xyy;N2xy3  2 2  xy  4  MN 3xy22  1;  4xy  2 yx We see equation is not exact. MN  yx MN  yx Also  gy  M xy2 1 1    g(y) y xy2  1  y

1 g() y dy  dy Thus I.. F e  ey  y

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MN 11= 4xy3 + 2y = where M = xy4 + y 2 ; N = 2x 2 y 3 + 2xy + 2y 5 yx11 Gen Sol: xy4 y 2  dx   2y 5  dy  c x2 y 4 2y 6 y2 x   c. 26 3. Solve y4 2y  dx   xy 3  2y 4  4x  dy  0 Sol: The given equation is not exact. MN  33 yx yy4    4  2  3 Also g(). y    g y  M y4  2 y y  1 3 dy g y  dy y 1 I. F e  e  y3 2 4x HereM =y+ ;N =x+2y- 11yy23 MN2 11=1- = y y3 x Equation is exact. 2 Gen sol is y dx 2. ydy  c ∴ 2 y

2 2 y2 x  y  c. y

APPLICATIONS OF DIFFERENTIAL EQUATIONS OF FIRST ORDER FIRST DEGREE

ORTHOGONAL TRAJECTORIES (O.T) Def: A curve which cuts every member of a given family of curves at a right angle is an orthogonal trajectory of the given family. Orthogonal trajectories in Cartesian co-ordinates: Working rule: To find the family of O.T in Cartesian form : Let f(x,y,c) =0 ……(1) be the given equation of family of curves in Cartesian form. 1) Differentiate (1) with respect to ‘x’ and obtain F(x, y, y’) = 0 ------(2) of the given family of curves.

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dy dx 2) Replace by in (2) dx dy Then the Differential Equation of family of O.T is dx F(x, y, ) =0 ------(3). dy 3) Solve equation (3) to get the equation of family of O.T’s of equation(1).

Solved Problems 1 . Find the O.T’s of family of semi-cubical parabolas ay2=x3 where a is a parameter. Sol : The given family of semi-cubical parabola is ay2=x3 ------(1) dy Differentiating with respect to ‘x’ a 2y = 3 x 2 ------(2) dx 3 ⟹x dy 2 2y 3x Eliminating ‘a’ from (1) and (2) y2 dx ⟹ 2x3 dy  3x2 y dx

⟹ 3 dy dx 2x  dx  2 Replace by     3x dx dy y  dy  ⟹ -2 dx  x = y 3 dy -2  xdx - ydy = c 3 -x22 y  - = c 32 xy22  + =1 3c 2c

2. Find the O.T of the family of circles x2+y2+2gx+c =0, Where g is the parameter Sol : x2+y2+2gx+c =0. ------(1) represents a system of co- axial circles with g as parameter. dy Differentiating with respect to ‘x‘ 2x+ 2y + 2g =0------(2) dx Substituting equation from (2) in (1) ⟹ dy x2+y2 -(2x+ 2y )x +c =0. dx ⟹

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dy y2-x2-2xy + c=0 dx ⟹dy dx Replace by  dx dy

2 2 dx y -x -2xy + c=0 dy ⟹ dx y2-x2+2xy + c=0 dy This⟹ can be written as dx 1 -(c + y2 ) 2x  x2 = dy y y This is a Bernoulli’s equation in x dx du So put x2=u2x = dy dy

du 1 -(c + y2 )  - u = dy y y which is a linear equation in ‘u’

1 dy  y log y 1  I.F = ee y General solution is u(I.F) =  Q(y).I.Fdy + k 1 -(c + y2 ) 1  x2 =  dy k y y y -2 =-c - y + k y xc2  = - y + k yy

2 2 2 3. Find orthogonal trajectories of the family of curves x3 + y 3 = a 3

2 2 2 Sol : The equation of the given family of curves is x3 + y 3 = a 3 …(1) -1 2-1 2 -1 dy dy y 3 Differentiating (1) w.r.t x, we get x33 + y = 0 = …(2) 3 3 dx dx -1 x 3 This is the differential equation of the given family of curves.

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-1 -1 dy dx dx y33 dx y Changing to - in (2), we get - = - = …(3) dx dy dy-1 dy -1 xx33 11 Separating the variables in (3), we get x33 dx = y dy 11 Integrating, we get x33 dx = y dy +c 44 44 3333 33  x = y + c or x - y = c1 44 This is the orthogonal equation. 4.Find the O.T’s of the family of parabolas through origin and foci on y –axis.

Sol : The equation of the family of parabolas through the origin and foci on y-axis is

x2=4ay where ‘a’ is parameter. dy Differentiating with respective ‘x’ 2x =4a dx dy x  = dx 2a dx x O.T -= dy 2a dx⟹ dy  -= x 2a y On integrating, we get -logx = + c 2a

 y + 2alogx = c1 is the equation of required O.T.

5.Find the O.T of the one parameter family of curves ex + e -y =c. Sol : Given equation is ex + e -y =c. dy Differentiating with respective ‘x’  ex + e -y (- ) = 0 dx dx Its O.T  ex + e -y ( ) = 0 e-x dx +e y dy = 0 dy ⟹ On integrating, we get, -e-x + e y = c which is the required O.T

ORTHOGONAL TRAJECTORIES IN POLAR FORM Working Rule: To find the O.T of a given family of curves in polar-co ordinates. Let f(r, θ, c)=0-----(1) be the given family of curves in polar form.

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dr 1.) Differentiating (1) with respect to obtain F [r, θ, ] =0 by eliminating the dθ parameter c. dr dθ 2.) Replace by  r2 then the Differential Equation of family of O.T is dθ dr dθ F [ r, θ,- r2 ] =0 dr 3.) Solve the above equation to get the equation of O.T of (1)

Self-orthogonal A family of curves is said to be self orthogonal when the differential equation of the family of O.T is same as that of the original family. Solved Problems 1. Find the orthogonal trajectories of the family of curves rnn cosθ = a .

Sol : Given family of curves is rnn cosθ = a Taking logarithms on both sides, nlogr + logcosθ = nloga Differentiating w.r.t θ, we get n dr 1 + (-sinθ) = 0 rdθ cosθ n dr - tanθ = 0 …(2) rdθ is the differential equation of family of curves.

dr 2 dθ To get the differential equation of orthogonal trajectories, replace with -r dθ dr

nd2 θ Now, we have -r - tanθ = 0 r dr dθ  nr + tanθ = 0 …(3) dr This is differential equation of orthogonal trajectories. Separating variables, we get dr ncotθdθ + = 0 r Integrating, we get n nlogsinθ +logr = log logsin θ +logr = logc or log(rsinnθ) = logc n  c = rsin θ which is the required O.T.

2. Find the Orthogonal trajectories of the family of cardioids r = a(1- cosθ) where ‘a’ is the parameter. Sol : Given r = a(1- cosθ) …(1)

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Differentiating the given equation w.r.t ‘θ’, we get dr 1 dr = asinθ a = (2) dθ sinθ dθ 1- cosθ dr Eliminating ‘a’ from (1) and (2) we get r= sinθ dθ θθ    2rsin  cos   dr rsinθ 22  ==    dθ 1- cosθ 2 θ 2sin  2 dr θ  = rcot  …(3) dθ2 dr dθ The differential equation of required O.T is obtained by replacing with -r 2 in (3) dθ dr

2 dθ θ  dr  θ  -r = rcot  Þ = -tan   dθ dr 2  r  2  On integrating, we get logr = 2log | cos(θ / 2) | +log(2c) = log(2ccos2 (θ / 2)) or r = 2ccos2 (θ / 2) r = c(1+cosθ) This is the equation of family of orthogonal trajectories.

3. Prove that the system of parabolas y2 = 4a(x + a) is self orthogonal. Sol : Given parabola is y2 = 4a(x + a) 2  y' = 4ax + 4a …(1) yy Differentiating (1) with respect to x,we get 2yy = 4a a = 1 …(2) 1 2 yy y22 y Substituting (2) in (1), y2 = 411 x + 4 24 2 2 2 or y = 2xyy11 + y y ….. (3) Equation (3) is the differential equation of the given system of parabolas. Replacing with , we get equation of the orthogonal trajectories as − 2 2 2-1  2  -1  2 2xy y y = 2xy  + y    y = + 2 y1   y 1  y 1 y 1 y2 y 2 = -2xyy + y 2 or y2 = 2xyy + y 2 y 2 11 11…(4) which is differential equation of the orthogonal trajectories of the given family.

Equations (3) and (4) are same, hence the given system is self orthogonal. xy22 4. Show that the system of confocal conics + =1 where  is a a+22λ b + λ parameter, is self orthogonal.

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xy22 Sol : Given equation of family of confocal conics is + =1 …(1) a+22λ b + λ Differentiating (1) w.r.t x, we get 2x 2y dy 22+ = 0. For convenience, we write a+λ b + λ dx xy = + p = 0 x(b22 + λ) + py(a + λ) = 0 a+22λ b + λ -(b22 x + a yp)  λ= x + yp -(a22 - b )yp (a22 - b )x 2 a+2 λ= and b+λ= …(2) x + yp x + yp Eliminating  from (1) and (2), we get

x(x + yp) y(x + yp) x + yp y 2 2- 2 2 =1  2 2  x - =1 a-b (a-b)p a-b p …(3) y 22  (x+yp) x- =a -b p This is the differential equation of the family of curves (1). We get the differential equation of the family of orthogonal trajectories by replacing dy dx 1 1 p= with - = - = - . dx dydy p  dx Hence the differential equation of orthogonal trajectories is

y 22 x- (x+py)=a -b …(4) p which is same as (3). Thus we see that the differential equation of the family of orthogonal trajectories is same as that of the original family. Hence the given family of curves is orthogonal to itself. Hence it is a self orthogonal family of curves. NEWTON’S LAW OF COOLING STATEMENT: The rate of change of the temperature of a body is proportional to the difference of the temperature of the body and that of the surrounding medium.

Let be the temperature of the body at time ‘t’ and θ0 be the temperature of its surrounding medium(usually air). By the Newton’s law of cooling , we have dθ dθ α(θ -θ ) = -k(θ -θ ) k is +ve constant dt00 dt

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dθ   = -k dt (θ -θ0 )

 log (θ -θ0 ) = -kt +c.

If initially θ= θ1 is the temperature of the body at time t=0 then

c = log(θ10 -θ )

log (θ -θ0 ) = -kt+ log(θ10 -θ )

θ -θ  log0 = -kt. θ10 -θ

θ -θ -kt  0 = e θ10 -θ

-kt θ=θ0 +(θ 1 -θ)=e 0 which gives the temperature of the body at time‘t’.

Solved Problems

1. A pot of boiling water 1000 C is removed from the fire and allowed to cool at 300 C room temperature. 2 mins later, the temperature of the water in the pot is 900 C .What will be the temperature of water after 5 mins?

Sol : We have θ -30 = ce-kt …(1) When t = 0,θ = 100 from (1), we get c = 70

0 -kt θ -30 = 70e …(2) When t = 2,θ = 90

From (2), 90 -30 = 70e-2k  60 = 70e-2k 6  -2k = log  7 = 0.1542

 k = 0.0771 When t = 5,θ -30 = 70e-5k

0  θ = 77.46

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2. A body is originally at 800C and cools down to 600 C in 20 min . If the

temperature of the air is 400 C find the temperature of body after 40 min. Sol : By Newton’s law of cooling we have dθ = -k (θ -θ ) where θ is the temperature of the air. dt 0 0 dθ  = -k dt log( θ -θ0 ) = -kt + logc (θ -θ0 )

0 Here θ0 = 40 C  log(θ - 40) = -kt + log c θ - 40  log = -kt c

θ - 40 -kt  =e c  θ = 40 + ce-kt ------(1) When t=0,θ = 800 C

 80 = 40 +c c = 40 ------(2). When t=20,θ = 600 C  60 = 40+ ce-20k ------(3). Solving (2) & (3)  ce-20k =20

2k  40e =20  k = -20log2

1 - log2 40 When t = 400 C then equation (1) is θ = 40 + 40e 20 = 40 +40 e-2log2 1 = 40 +40X 4

 θ = 500 C

3. An object whose temperature is 750C cools in an atmosphere of constant temperature C, at the rate of k , being the excess temperature of the body over

0 that of the temperature. If after훉 10min, the temperature of the object falls to 65 C , find its temperature after 20 min. Also find the time required to cool down to 550C. Sol : We will take one minute as unit of time.

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dθ It is given that = -kt dt  θ = ce-kt ------(1). Initially when t=0  θ = 750 - 25 0 = 50 0

0  c= 50 Hence c = 50  θ = 50e-kt ------(2) When t= 10 min  θ = 650 - 25 0 = 40 0 40= 50 e-10k 4  e=-10k ------(3). 5

The value of θ when t=20  θ = ce-kt θ = 50e-k θ = 50(e-10k ) 2

2 4   50 5

When t=20 θ=320 C . Hence the temperature after 20min = 320 C + 25 0 C = 57 0 C When the temperature of the object = 550C θ = 550 C - 25 0 C = 30 0 C

Let t, be the corresponding time from equation (2) 30 = 50e-kt ------(4)

1 10 (-k)10 44 -k  From equation (3) e = i.e e =  55

t1 10 4t1 4 3 From equation (4), we get 30 = 50 log = log 5 10 5 5

3 log 5  t1 = 10 = 22.9min 4 log 5

4. A body kept in air with temperature 250C cools from 1400C to 800C in 20 min. Find when the body cools down in 350C.

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dθ dθ Sol : By Newton’s law of cooling = -k(θ -θ0 ) = -kdt dt θ -θ0

 log( . Here, 0 θ -θ0 ) = kt +c θ0 = 25 C  log(θ - 25) = kt + c ------(1).

When t=0, θ = 1400 C log (115) =c c =log (115). kt = - log (θ - 25) + log 115------(2)

When t=20,θ = 800 C  log (80-25)= -20k + log 115  20 k =log (115) - log(55) ------(3) kt log115-log(θ - 25) Divide equation (2) by (3), we get = 20k log115-log55 t log115-log(θ - 25) = 20 log115-log55 ⟹ t log115- log(10) When θ = 350 C  = 20 log115- log55 t log(11.5)  = =3.31 20 28 log 11  temperature = 20  3.31 = 66.2 The temp will be 350 C after 66.2 min.

5. The temperature of the body drops from 1000 C to 750 C in 10 min. When the surrounding air is at 200 C temperature. What will be its temp after half an hour. When will the temperature be 250 C .

dθ Sol : = -k(θ -θ ) dt 0 log(θ - 20) = -kt + logc

when t=0, θ = 1000 c=80 11 when t=10,θ = 75⟹0 e=-10k . 16 ⟹ 1331 0 when t =30min θ = 20 +80 = 46 C 4096 ⟹

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log5-log80 when θ = 250 C  t =10 = 74.86min (log11-log18)

6. If the air is maintained at 150 C and the temperature of the body drops from 7000 C to 40 C in 10 minutes. What will be its temperature after 30 minutes?

dθ Sol : If θ be the temperature of the body at time t, then = -k(θ -15) , where k is dt constant dθ Integrating, = -k dt + logc θ -15 i.e log(θ -15) = -kt + logc i.e , θ -15 = ce-kt …(1) When tC0, 700 and when tC10, 400 70  15 ce0  c  55 40 15 ce10k and 25 5  =e-10k or e=-10k …(2) 55 11 Then (1) becomes θ -15 = 55e-kt When t=30 min, θ = 15 + 55e-30k 3 -10k 3 5 θ = 15 + 55(e ) = 15 + 55 using (2) 11 625 2441 = 15 + = = 20.160 C . 121 121 7. In a pot of boiling water 1000 C is removed from the fire and allowed to cool at 300 C room temperature. Two minutes later, the temperature of the water in the pot is 900 C . What will be the temperature of the water after 5 minutes?

Sol : We have θ -300 C = ce -kt …(1) when t = 0,θ = 100 c = 70 θ -300 C = 70e -kt …(2) when, t = 2,θ = 900 from (2), 60 = 70e-2k 6  -2k = log = -0.1542 7  k = 0.0771 when t = 5,θ -300 = 70e -5k  θ = 77.460 8. The temperature of a cup of coffee is 920 C when freshly poured, the room temperature being 240 C . In one min it was cooled to 800 C . How long a period must elapse, before the temperature of the cup becomes 650 C . Sol : By Newton’s Law of cooling,

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dθ = -k(θ -θ ) ;k>0 dt 0 0 θ0 = 24 C  log(θ - 24) = -kt + log c------(1). When t=0; θ = 92 c =68 68 When t =1; θ = 800 C  e=-k 56 56  k = log . 68 65x41 When θ = 650 C , t = = 0.576min 682 LAW OF NATURAL GROWTH OR DECAY Statement : Let x(t) or x be the amount of a substance at time ‘ t’ and let the substance be getting converted chemically . A law of chemical conversion states that the rate of change of amount x(t) of a chemically changed substance is proportional to the amount of the substance available at that time dx αx dt Note: a) In case of Natural growth we take dx = kx (k > 0) dt dx b) In case of Natural decay, we take = -kx (k>0) dt where k is a constant of proportionality RATE OF DECAY OF RADIO ACTIVE MATERIALS Statement : The disintegration at any instant is proportional to the amount of material present in it.

du If u is the amount of the material at any time ‘t’ , then = - ku , where k is any dt constant (k >0). i.e Law of Natural Decay is applied.

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Solved Problems

1. The number N of bacteria in a culture grew at a rate proportional to N . The value of N was initially 100 and increased to 332 in one hour. What was the value 1 of N after 1 hrs . 2

dN Sol : The differential equation to be solved is = kN dt dN  = kdt N dN  = kdt N logN = kt + log c N = c e-kt ------(1). When t= 0sec, N =100 100 = cc =100 When t =3600sec, N =332332 =100 e3600k 332  e3600k = 100 3 Now when t = hours = 5400 sec then N=100 e5400k 2 3 3600k 2 N =100 e

3 332 2 N=100 = 605. 100 N = 605. 2. A bacterial culture, growing exponentially, increases from 100 to 400 gms in 10 hrs. How much was present after 3 hrs, from the initial instant?

Sol : Let N be the weight of bacteria culture at any . Then N = ce-kt …(1) By data, when t = 0, N = 100g > 100 = c Substituting in (1), we get N = 100e-kt …(2) When t = 100, N = 400g from (2), 400 = 100e-10k  4 = e-10k  -10k = log4 11  k = - log22 = - log(2) 10 5 …(3) When t = 3, N =100e-3k

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1 3 -3 - log2 = 100e5 = 100e(log2)5 31 = 100X(2)55 = 100X8 = 100X1.414 = 141.4gms

3.If a radioactive Carbon-14 has a half life of 5750 years, what will remain of one gram after 3000years? Sol : Let mass of radioactive Carbon-14 at any time be denoted by . dx Then it is known that = -kt where k is a constant dt  x = Ae-kt where A is also a constant. It is known that at t=0, we have 1gm of Carbon-14 1 = Ae0 A = 1

x = e-kt However when t=5750 years, we have 1/2gm of Carbon-14. 11 = e-k(5750) k = log2 2 5750 Suppose t=3000years, we have to find x. 3000 - log2 x = e-kt = e -3000k = e 5750

3000 - gms  x = (2) 5750

4. If 30% of a radioactive substance disappears in 10 days, how long will it take for 90% of it to disappear?

Sol : The differential equation of the diffusing radioactive material is, dm = -km …(1) dt Separating the variables and integrating, we get m = ce-kt …(2)

When t = 0, let m = m1

 m1 = c …(3) 70m By data, when t = 10,m = 100 70m  1 = ce-10k = m e -10k 100 1

-10k 7 1 7 e = k = - log 10 10 10 1 10 k = log  10 7  …(4)

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Required time at is 10m 1 1 = ce-kt = m e -kt = e -kt 1001 10 1  t = log(10) k 10log(10) 64.5 days. log10 log7

II .LINEAR DIFFERENTIAL EQUATIONS OF SECOND AND HIGHER ORDER dn y d n-1 y d n-2 y Definition: An equation of the form + P (x) + P (x) +…+ P (x)y = Q(x) dxn1 dx n-1 2 dx n-2 n

where P1 (x),P 2 (x),P 3 (x)…P n (x) and Q(x) (functions of x) are continuous is called a linear differential equation of order n. LINEAR DIFFERENTIAL EQUATIONS WITH CONSTANT COEFFICIENTS dn y d n-1 y d n-2 y Def: An equation of the form +P +P +…+Py=Q(x) where dxn1 dx n-1 2 dx n-2 n

P,P,PP1 2 3 n , are real constants and Q(x) is a continuous function of x is called an linear differential equation of order ‘ n’ with constant coefficients. Note: d d2 dn 1. Operator D = ; D2 = ; …………………… Dn = dx dx2 dxn

dy dy2 dyn D y = ; D2y= ;…………………… Dn y= dx dx2 dxn 1 2. Operator Q = Qdx i e D-1Q is called the integral of Q. D 

To find the general solution of f(D).y = 0 :

n n-1 n-2 Here f(D)=D +PD1 +PD 2 +…+P n is a polynomial in D. Now consider the auxiliary equation: f(m) = 0

n n-1 n-2 i.e f(m)=m +Pm1 +Pm 2 +…+P n  0

where P1 ,P 2 ,P 3 …P n are real constants.

Let the roots of f(m) =0 be m1 ,m 2 ,m 3 …m n .

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Depending on the nature of the roots we write the complementary function as follows: Consider the following table S.No Roots of A.E Complementary function(C.F) f(m) =0

m x m x m x 1. m1, m2, ..mn are 1 2 n y=cec 1 +ce 2 +…ce n real and distinct.

2. m1, m2, ..mn and

two roots are m1n xmx3 m x yc =(c+c)e 1 2 +ce 3 +…ce n

equal i.e., m1,

m2 are equal and real (i.e repeated twice) &the rest are real and different.

2 m x m x m x 3. m1, m2, ..mn are 1 4 n yc =(c+cx+cx)e 1 2 3 +ce 4 +…ce n real and three roots are equal i.e., m1, m2 , m3 are equal and real (i.e repeated thrice) &the rest are real and different. 4. Two roots of αx mx3 mxn yc = e (c 1 cosβx + c 2 sinβx) + c 3 e +…c n e A.E are complex say +i -i and rest are real and distinct. 5. αx mx5 mxn If ±i are yc = e (c 1 +c 2 x)cosβx +(c 3 +c 4 x)sinβx  +c 5 e +…c n e repeated twice & rest are real and distinct 6. If ±i are αx 2 2 yc = e (c 1 + c 2 x + c 3 x )cosβx + (c 4 + c 5 x + c 6 x )sinβx repeated thrice mx7 mxn +c7n e +…c e & rest are real and distinct 7. If roots of A.E. y = eαx  c cosh mx3 mxn c 1βx + c 2 sinh βx + c 3 e +...... + c n e

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irrational say    and rest are real and distinct.

Solved Problems dy3 dy 1. Solve - 3 + 2y = 0 dx3 dx Sol : Given equation is of the form f(D).y = 0 Where f(D) = (D3 -3D + 2)y = 0 Now consider the auxiliary equation f (m) = 0 f(m) = (m3 -3m + 2)y = 0  (m-1) (m-1) (m+2) = 0  m =1, 1, -2

Since m1 and m2 are equal and m3 is -2

x -2x We have yc =(c 1 +c)e 2 +ce 3 2. Solve (D4-2D3-3D2+4D+4)y = 0 Sol : Given f(D) = (D4 -2 D3 - 3 D2 + 4D +4) y = 0 …(1) Auxiliary equation is f(m)=0  m4 - 2m 3 -3m 2 + 4m + 4 = 0 …(2) By inspection m+1 is its factor. (m +1)(m32 -3m + 4) = 0 …(3) By inspection m+1 is factor of (m32 -3m + 4) . (3) is m +1  m +1  m2 - 4m + 4  = 0

 (m +1)22 (m - 2) = 0  m = -1,-1,2,2 Hence general solution of (1) is -x 2x y = (c1 + c 2 x)e + (c 3 + c 4 x)e

3.Solve (D4 +8D2 + 16) y = 0 Sol : Given f(D) = (D4 +8D2 + 16) y = 0 Auxiliary equation f(m) = (m4 +8 m2 + 16) = 0

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 (m2 + 4)2 = 0  (m+2i)2 (m+2i)2 = 0  m= 2i ,2i , -2i , -2i Here roots are complex and repeated Hence general solution is

yc = [(c1+c2x)cos x + (c3+c4x) sin x)]

4.Solve y11+6y1+9y = 0 ; y(0) = -4 , y1(0) = 14 Sol : Given equation is y11+6y1+9y = 0 Auxiliary equation f(D) y = 0  (D2 +6D +9) y = 0 A.equation f(m) = 0  (m2 +6m +9) = 0  m = -3 ,-3 -3x yc = (c1+c2x)e ------> (1) 1 -3x -3x Differentiate of (1) w.r.to x  y =(c1+c2x)(-3e ) + c2(e )

Given y1 (0) =14  c1 = -4 & c2 =2 Hence we get y =(-4 + 2x) (e-3x ) 5.Solve 4y111 + 4y11 +y1 = 0 Sol : Given equation is 4y111 + 4y11 +y1 = 0 That is (4D3+4D2+D)y=0 Auxiliary equation f(m) = 0 4m3 +4m2 + m = 0 m(4m2 +4m + 1) = 0

m (2m +1)2 = 0 m = 0 , -1/2 ,-1/2 -x/2 y =c1 + (c2 + c3x) e 6.Solve (D2 - 3D +4) y = 0 Sol : Given equation (D2 - 3D +4) y = 0 A.E. f(m) = 0 m2-3m + 4 = 0 3 ± 9 -16 3 ± i 7 m = = 22 3 7 α ± iβ   i 2 2 3 77 y = e2 (c cos x + c sin x) 1222

To Find General solution of f(D) y = Q(x)

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It is given by y = yc + yp i.e. y = C.F+P.I Where the P.I consists of no arbitrary constants and P.I of f (D) y = Q(x) 1 Is evaluated as P.I = Q x  f(D) Depending on the type of function of Q(x) , P.I is evaluated .

1 1. Find (x2 ) D 1x3 Sol : (x22 ) = x dx = D3

1 2. Find Particular value of (x) D +1 1 Sol : (x) = e-x xe x dx (By definition) D +1 

= e-x (xe x -e x )

= x-1

General methods of finding Particular integral :

1 P.I of f(D)y=Q(x), when is expressed as partial fractions. f(D) Q.Solve (D22 +a )y = secax

Sol : Given equation is …(1)

Let f(D) = D22 + a

The AE is f(m) = 0 i.e m22 + a = 0 …(2)

The roots are m= -ai , -ai

yc = c 1 cosax +c 2 sinax 1 1 1 1 y = secax = - secax p 22 …(3) D + a 2ai D - ai D + ai

1 cosax -isinax secax = eiax secaxdx = e iax dx D - ai cosax

i = eiax (1-itanax)dx = e iax x + logcosax …(4) a

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1i Similarly we get secax = e-iax x - logcosax …(5) D + ai a

From (3),(4) and (5), we get

1iax i  -iax  i  yp = e x + logcosax  -e  x - logcosax  2ai a   a  x(eiax - e -iax ) 1 (e iax + e -iax ) = +2 (logcosax) 2ai a 2 x1 = sinax + cosaxlog(cosax) aa2 The general solution of (1) is x1 y = y + y = c cosax + c sinax + sinax + cosaxlog(cosax) c p 1 2 aa2 RULES FOR FINDING P.I IN SOME SPECIAL CASES;

Type 1. P.I of f (D)y=Q(x) where Q(x) =eax , where ‘a’ is constant. 1 1 1 Case1.P.I = .Q(x) = eax = e ax f(D) f(D) f(a) provided f(a) ≠ 0 i.e In f(D), put D=a and Particular integral will be calculated.

k Case 2: If f(a)=0 then the above method fails. Then if f D  = D-a   D  (i.e ‘a’ is repeated root k times). 1 1 Then P.I = eax . x k provided (a) 0 (a) k! Type 2. P.I of f(D)y = Q(x) where Q(x) = sinax or Q(x) = cosax where ‘a’ is constant 1 then P.I= Q(x) . f(D) sinax Case 1: In f(D) put D2 = -a 2 f(-a 2 ) 0 then P.I = f(-a)2

Case 2: If f(-a2)=0 then D2 + a2 is a factor of (D2 ) and hence it is a factor of f(D). Then

let f(D)=(D2 + a2) f(D) = (D2 + a 2 ) (D 2 ). sinax sinax 1 sinax 1 -xcosax Then = = = f(D) (D2 + a 2 ) (D 2 ) (-a 2 ) D 2 + a 2  -a 2  2a cosax cosax 1 cosax 1 xsinax = = = 2 2 2 2 2 2 2 f(D) (D + a ) (D ) (-a ) D + a -a  2a

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Type 3.P.I for f(D)y=Q(x) where Q(x)=xk where k is a positive integer .f(D) can be expressed as f(D) = [1 (D)] 11 Express  [1  (D)]1 f(D) [1 (D)] 1 Hence P.I = Q(x) [1 (D)]

=[1 (D)]1k x Type 4.P.I of f(D)y=Q(x) when Q(x)=eax V where ‘a’ is a constant and V is function of x. where V =sinax or cosax or xk 1 Then P.I = Q(x) f(D) 1 = eVax f(D)

ax 1 1 = eV& V is evaluated depending on V. f(D + a) f(D + a) Type 5. P.I of f(D)y=Q(x) when Q(x)=xV where V is a function of x. 1 Then P.I = Q(x) f(D) 1 = V f(D)

11 = x - f'(D) V f(D) f(D) Type 6. P.I. of f(D)y = Q(x) where Q(x)= xm v where v is a function of x. 11 When P.I. = ×Q(x) = xm v, where v = cosax or sinax f(D) f(D) 11 i. P.I. = xm sinax = I.P.of x m e iax f(D) f(D) 11 ii. P.I. = xm cosax = R.P.of x m e iax f(D) f(D)

Formulae

1. = (1 – D)-1 = 1 + D + D2 + D3 + ------

2. = (1 + D)-1 = 1 - D + D2 - D3 + ------

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3. = (1 – D)-2 = 1 + 2D + 3D2 + 4D3 + ------

4. = (1 + D)-2 = 1 - 2D + 3D2 - 4D3 + ------

5. = (1 – D)-3 = 1 + 3D + 6D2 + 10D3 + ------

6. = (1 + D)-3 = 1 - 3D + 6D2 - 10D3 + ------

Solved Problems 1. Solve (4D2 - 4D+1)y = 100 1 -1 Sol : A.E is 4m22 - 4m +1 = 0 (2m -1) = 0 m = , 22

x 2 C.F = (c12 + c x)e

100 100e0x 100 Now P.I = = = 100 { since 100e0x  100} 4D2 - 4D +1 (2D-1) 2 (0-1) 2

x 2 Hence the general solution is y = C.F+ P.F = (c12 + c x)e 100 2. Solve the differential equation (D2 + 4)y = sinh2x + 7 .

Sol : Auxillary equation is m2 + 4 = 0 m2 = -4 m = ±2i

C.F is yc = c 1 cos2x +c 2 sin2x …(1) To find P.I : 1 y = (sinh2x + 7) p D2 + 4 2x -2x 1 e + e 0 =2  + 7e D + 4 2 1 e2x 1 e -2x e 0 = . + + 7 2 D2 + 4 2 D 2 + 4 (D 2 + 4) e2x e -2x 7 = + + 2(4 + 4) 2(4 + 4) (0 + 4) e2x + e -2x 7 1 7 = + = sinh2x + …(2) 16 4 8 4

y = ycp + y 17 = c cos2x + c sin2x + sinh2x + 1284

3. Solve (D+ 2)(D-1)2 y = e -2x + 2sinhx

Sol : The given equation is (D + 2)(D-1)2 y = e -2x + 2sinhx …(1)

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This is of the form f(D)y = e-2x + 2sinhx A.E is f(m) = 0 (m + 2)(m-1)2 = 0 m = -2,1,1 The roots are real and one root is repeated twice. -2x x C.F is yc = c 1 e + (c 2 + c 3 x)e . e-2x + 2sinhx e -2x + e x -e -x P.I= = =y+y+y 22p1 p 2 p 3 (D + 2)(D-1) (D + 2)(D-1) e-2x Now y= p1 (D + 2)(D-1)2 Hence f(-2) = 0. Let f(D) = (D-1)2 . Then (2) 0 and m 1 e-2x x xe -2x y = = p1 99 ex and y= . Here f(1)=0 p2 (D + 2)(D -1)2 ex x 2 x 2 e x == (3)2! 6 e-x and y= p3 (D + 2)(D-1)2 ee-x -x Putting D=-1, we get y = = p3 (1)(-2)2 4 The general solution is y = y + y + y + y c p1 p 2 p 3 xe-2x x 2 e x e -x i.e y=ce-2x +(c+cx)e+ x + - 1 c 3 9 6 4

4. Solve the differential equation (D2 + D+1)y = sin2x .

Sol : A.E is m2 + m +1 = 0 -1± 1-4 -1± 3i  m = = 22 -x x 3 x 3 y = e2  c cos + c sin …(1) c 122 2 

To find P.I : sin2x sin2x y = = p D2 +D+1 -4+D+1 sin2x (D + 3)sin2x (D + 3)sin2x = =2 = D -3 D -9 -4 -9 Dsin2x + 3sin2x 2cos2x + 3sin2x = = -13 -13 -x x 3 x 3 1 y = y + y = e2 c cos + c sin - (2cos2x + 3sin2x) c p 1 2 2 2 13

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5. Solve (D22 - 4)y = 2cos x

Sol : Given equation is (D22 - 4)y = 2cos x …(1) Let f(D) = D2 - 4 A.E is f(m) = 0 i.e m2 - 4 = 0 The roots are m=2,-2. The roots are real and different. 2x -2x C.F = yc = c 1 e + c 2 e

112 P.I = yp =22 (2cos x) = (1+ cos2x) D - 4 D - 4 e0x cos2x = + = P.I + P.I D22 - 4 D - 4 12 e0x e10x P.I = y = [Put D=0] = = - 1p1 2 D - 4 -4 4 cos2x cos2x P.I = y = = [Put D22 = -2 = -4 ] 2p2 D2 - 4 -8 The general solution of (1) is y = y + y + y c p12 p 1 cos2x y = c e2x + c e -2x - - 1248 6. Solve (D2 +1)y = sinxsin2x

Sol : Given D.E is (D2 +1)y = sinxsin2x A.E is m2 +1 = 0 m = ±i The roots are complex conjugate numbers.

C.F is yc = c 1 cosx + c 2 sinx w.k.t 2sinAsinB=cos(A-B)–cos(A+B) sinxsin2x 1 cosx -cos3x P.I = = = P.I + P.I (D22 +1) 2 (D +1) 12 1 cosx Now P.I = 1 2 D2 +1 Put D2 = -1 we get D2 +1 = 0 1 xsinx xsinx cosax x P.I = = Case of failure : = sinax 1 2 2 2 4 D + a 2a 1 cos3x and P.I = - 2 2 D2 +1 Put D2 = -9, we get 1 cos3x cos3x P.I = - = 2 2 -9 +1 16 General solution is xsinx cos3x y = y + y + y = c cosx + c sinx + + c p12 p 1 2 4 16 7. Solve the differential equation (D32 - 3D -10D+ 24)y = x + 3 .

Sol : The given D.E is (D32 -3D -10D + 24)y = x +3 A.E is m32 -3m -10m + 24 = 0 m=2 is a root.

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The other two roots are given by m2 - m - 2 = 0  (m - 2)(m +1) = 0 m=2 (or) m = -1 One root is real and repeated, other root is real. ⟹ 2x -x C.F is yc = e (c 1 + c 2 x) + c 3 e

x + 3 1 x3 + 3 y = = p (D32 -3D -10D + 24) 24 D32 -3D -10D 1+ 24 -1 1 1+D32 -3D -10D = (x + 3) 24 24 1 D32 -3D -10D = 1- (x + 3) 24 24 1 10 24x +82 = x + 3+ = 24 24 576

General solution is y = ycp + y

2x -x 24x +82  y=e (c1 +cx)+ce 2 3 + 576 8. Solve the differential equation (D2 - 4D+ 4)y = e 2x + x 2 +sin3x .

Sol : The A.E is (m2 - 4m + 4) = 0 (m-2)2 =0 m=2,2 2x yc = (c 1 + c 2 x)e …(1) 1 To find y : y = (e2x + x 2 + sin3x) p p D2 - 4D + 4 e2x x 2 sin3x = + + (D - 2)2 (D - 2) 2 D 2 - 4D + 4 22 x2x x sin3x = e +2 + 2!D -9 - 4D + 4 4 1- 2 2 -2 x2x 1 D 2 (4D -5)sin3x = e + 1- x - 2 4 2 (5 + 4D) 22 x2x 1 2D 3D 2 (4D -5)sin3x = e + 1+ + x - 2 2 4 2 4 16D - 25 x22 x x 3 (12cos3x -5sin3x) = e2x + + + - 2 4 2 8 -144 - 25 x22 x x 3 (12cos3x -5sin3x) = e2x + + + + …(2) 2 4 2 8 169 x22 x x 3 (12cos3x -5sin3x) y=y+y=(c+cx)e2x += e 2x + + + + c p 1 2 2 4 2 8 169 9. Solve the differential equation (D2 + 4)y = xsinx .

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Sol : Auxiliary equation is m2 +4-0 m 2 =(2i) 2 m = ±2i. The roots are complex and conjugate.

Hence Complementary Function, yc = c 1 cos2x + c 2 sin2x 1 Particular integral, y = xsinx p D2 + 4 1 = I.P of xeix D2 + 4 11 = I.P ofeix x = I.P of e ix x (D + i)22 + 4 D + 2Di + 3 -1 eix D 2 + 2Di = I.P of 1+ x 33 eix D 2 + 2Di = I.P of 1- +... x 33 ix e22 = I.P of 1- Di x D (x) = 0,etc 33 12 = I.P of (cosx + isinx) x -i 33 12 = - cosx + xsinx 33 Hence the general solution is 12 y = yc + y p = c 1 cos2x + c 2 sin2x + xsinx - cosx 33 where c1 and c2 are constants. 1 Other Method (using type 5): yp =2 xsinx D + 4

2D sinx = x- D22 + 4 D + 4 xsinx 2(Dsinx) = - 3 3(D2 + 4) xsinx 2cosx = - 39 Hence the general solution is 12 y = yc + y p = c 1 cos2x + c 2 sin2x + xsinx - cosx 33

10. Solve the Differential equation(D2+5D+6)y=ex

Sol : Given equation is (D2+5D+6)y=ex Here Q(x) =e x Auxiliary equation is f(m)=m2+5m+6=0 m2+3m+2m+6=0 m(m+3)+2(m+3)=0

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m=-2 or m=-3 The roots are real and distinct

-2x -3x C.F=yc= c12 e + c e 1 Particular Integral=yp= Q(x) f(D) 1 1 = ex = ex D2 + 5D + 6 (D + 2)(D + 3) Put D = 1 in f(D) 1 P.I= ex (3)(4)

1 x Particular Integral = yp = e 12

General solution is y = ycp + y

ex y = c e-2x + c e -3x + 1212 11. Solve y''- 4y'+ 3y = 4e3x ,y(0) = -1,y'(0) = 3

Sol : Given equation is y''- 4y'+3y = 4e3x

d2 y dy i.e - 4 +3y = 4e3x it can be expressed as dx2 dx

D2 y - 4Dy +3y = 4e 3x

(D2 - 4D +3)y = 4e 3x Here Q(x) = 4e3x; f(D)= D2-4D+3 Auxiliary equation is f(m)=m2-4m+3 = 0 m2- 3m - m+3=0 m(m-3) -1(m-3)=0 m=3 or 1

The roots are real and⟹ distinct. 3x x C.F= yc = c 1 e + c 2 e 1 P.I= y = Q(x) p f(D) 1 = 4e3x D2 - 4D + 3 1 = 4e3x (D -1)(D -3) Put D=3

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4e3x 4 e 3x x' y = = = 2 e3x = 2xe 3x p 3-1  D -3  2 D -3  1!

General solution is y = yc+yp

3x x 3x y=ce12 +ce +2xe …(3) Equation (3) differentiating with respect to ‘x’

3x x 3x 3x y' = 3c12 e + c e + 2e + 6xe …(4) By data, y(0) = -1 , y1(0)=3

From (3), -1=c1+c2 …(5)

From (4), 3=3c1+c2+2

3c1+c2=1 …(6)

Solving (5) and (6) we get c1=1 and c2 = -2 y=-2ex + (1+2x) e3x 12. Solve y''+ 4y'+ 4y = 4cosx + 3sinx,y(0) = 0,y'(0) = 0

Sol : Given differential equation in operator for (D2 + 4D + 4)y = 4cosx +3sinx A.E is m2+4m+4 = 0 (m+2)2=0 then m=-2, -2

-2x C.F is yc = (c 1 + c 2 x)e 4cosx + 3sinx P.I is= y= put D2 = -1 p (D2 + 4D + 4) 4cosx + 3sinx (4D -3)(4cosx + 3sinx) y = = p (4D + 3) (4D -3)(4D + 3)

(4D -3)(4cosx + 3sinx) = 16D2 -9 (4D -3)(4cosx + 3sinx) yp = -16 -9 -16sinx +12cosx -12cosx -9sinx -25sinx = = = sinx -25 -25

General equation is y=yc+yp

-2x y = (c12 + c x)e + sinx …(1)

By given data y(0) = 0, c1 = 0 and

-2x -2x Differentiating (1) w.r.t ‘x’, y' = (c1 + c 2 x)(-2)e + e (c 2 ) + cosx …(2) given y'(0) = 0

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Substitute in (2)  -2c1 + c2+1=0 c2 = -1 Required solution is y = -xe-2x +sinx 13. Solve (D2+9)y = cos3x

Sol : Given equation is (D2+9)y = cos3x A.E is m2+9 = 0  m = ±3i

yc = C.F = c 1 cos3x + c 2 sin3x cos3x cos3x y = P.I = = p D2 + 9 D 2 + 3 2 xx = sin3x = sin3x 2(3) 6

General equation is y=yc+yp x y = c cos3x + c sin3x + sin3x 126 14. Solve y'''+ 2y''- y'- 2y = 1- 4x3

Sol : Given equation can be written as (D3 + 2D 2 - D- 2)y =1- 4x 3

A.E is m32 + 2m - m - 2 = 0 (m2 -1)(m + 2) = 0

m2 =1or m = -2 m = 1, -1, -2

x -x -2x C.F= c1 e + c 2 e + c 3 e

1 3 -1 P.I = 32 (1- 4x ) = 32 (D + 2D - D - 2) (D + 2D - D) 3 2 1- (1- 4x ) 2

32 -1 -1 (D + 2D - D) 3 = 1- (1- 4x ) 22

3 2 3 2 2 3 2 3 -1 (D + 2D - D) (D + 2D - D) (D + 2D - D) 3 = 1+ + + +…(1-4x) 2 2 4 8 -1 1 1 1 = 1+ D3 + 2D 2 - D  + D 2 - 4D 3  + -D 3   1- 4x 3  2 2 4 8

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-1 5 5 1 = 1- D3 + D 2 - D (1- 4x 3 ) 2 8 4 2 -1 5 5 1 = (1- 4x32 ) - (-24) + (-24x) - (-12x ) 2 8 4 2 -1 = -4x32 + 6x -30x +16 2  =[2x32 -3x +15x -8] The general solution is y= C.F + P.I

x -x -2x 3 2 y=ce1 +ce 2 +ce 3 +[2x -3x +15x-8]

15. Solve (D3 - 7D 2 +14D- 8)y = e x cos2x

Sol : Given equation is (D3 -7D 2 +14D-8)y = e x cos2x

A.E is (m32 -7m +14m -8) = 0 (m-1)(m-2)(m-4)=0 Then m=1,2,4

x 2x 4x C.F= c1 e + c 2 e + c 3 e

ex cos2x P.I= (D32 -7D +14D-8) 1 = ex cos2x (D +1)32 -7(D +1) +14(D +1) -8

11ax ax P.I = e v = e v f(D) f D + a  1 = ex cos2x (D32 - 4D +3D) 1 = ex cos2x (D32 - 4D +3D) 1 = ex cos2x (Replacing D2 with -22) (-4D +3D +16) 1 = ex cos2x (16 - D) 16 + D = ex cos2x (16 - D)(16 + D)

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16 + D = ex cos2x 256 - D2 16 + D = ex cos2x 256 -(-4)2 ex = (16cos2x - 2sin2x) 260 2ex = 8cos2x -sin2x  260 ex = 8cos2x -sin2x  130 General solution is y = yc + yp ex y = c ex + c e 2x + c e 4x + 8cos2x -sin2x  1 2 3 130

16. Solve (D2 - 4D+ 4)y = x 2 sinx +e 2x + 3

Sol : Given (D2 - 4D + 4)y = x 2 sinx + e 2x +3 A.E is (m2 - 4m + 4) = 0 (m - 2)2 =0 then m=2,2 2x C.F= (c12 + c x)e xsinx+e+32 2x 1 1 1 P.I= = (x2 sinx) + e 2x + (3) (D- 2)2 (D- 2) 2 (D- 2) 2 (D- 2) 2 11 Now (x22 sinx) = (x ) (I.P of eix ) (D - 2)22 (D - 2) 1 = I.P of (x2 )e ix (D - 2)2 1 = I.P of (eix ) (x 2 ) (D +i - 2)2 1 I.P of (eix ) (x 2 ) (D +i - 2)2 On simplification, we get 11 (x2 sinx) = (220x + 244)cosx + (40x +33)sinx  (D +i - 2)2 625 1x2 and e2x = e 2x , (D- 2)2 2 13 (3) = (D - 2)2 4 1 x2 3 P.I= (220x + 244)cosx + (40x +33)sinx  + e2x + 625 2 4 y=yc+yp 1 x2 3 y = (c + c x)e2x + (220x + 244)cosx + (40x +33)sinx  + e2x + 12 625 2 4

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Linear equations of second order with variable coefficients d2 y dy An equation of the form + P x  + Q(x)y = R (x), where P(x), Q(x), R(x) are real dx2 dx valued functions of ‘x’is called linear equation of second order with variable coefficients.

Variation of Parameters : This method is applied when P,Q in above equation are either functions of ‘x’ or real constants but R is a function of ‘x’.

Working Rule :

1. Find C.F. Let C.F= yc = c12 u(x) + c u(x) vRdx uRdx 2. Take P.I= y =Au+Bv where A= - and B = p uv'- vu' uv'- vu'

3. Write the G.S. of the given equation y = ycp + y dy2 1. Apply the method of variation of parameters to solve + y = cosecx dx2

Sol : Given equation in the operator form is (D2 +1)y = cosecx …(1) A.E is (m2 +1) = 0 m = ±i The roots are complex conjugate numbers.

C.F is yc = c 1 cosx + c 2 sinx Let yp= A cosx+ B sinx be P.I. of (1) dv du u - v = cos22 x + sin x = 1 dx dx A and B are given by vRdx sinxcosecx A= - =- dx=-dx=-x uv'- vu'  1  uRdx B= = cosx.cosecxdx = cotxdx = log(sinx) uv11 - vu  

yp = -xcosx + sinx.log(sinx)

General solution is y= yc + yp.

y = c12 cosx +c sinx - xcosx +sinx.log(sinx)

2. Solve (D2x - 2D+ 2)y = e tanx by method of variation of parameters.

Sol : A.E is m2 - 2m + 2 = 0 2 ± 4 -8 2 ± i2 m = = = 1± i 22

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x x x We have yc = e (c 1 cosx + c 2 sinx) = c 1 e cosx + c 2 e sinx

= c12 (u) + c (u) where u = exx cosx,v = e sinx du dv =eex (-sinx) + x cosx, =e x cosx +e x sinx dx dx dv du u - v =ex cosx(e x cosx + e x sinx) - e x sinx( e x cosx - e x sinx) dx dx =e2x (cos22 x + cosxsinx -sinxcosx +sin x) = e2x Using variation of parameters, vRex tanx A = - = - (ex sinx)dx dv du 2x u - v e dx dx sin22 x (1-cos x) = - tanxsinxdx =  dx =  dx cosx cosx = (secx -cosx)dx = log(secx + tanx) -sinx uR B = dx  dv du u - v dx dx exx cosx.e tanx = dx = sinxdx = -cosx e2x General solution is given by x x x x i.e y = c12 e cosx + c e sinx +[log(secx + tanx) -sinx]e cosx -e cosxsinx = + + or y = c ex cosx + c e x sinx +[log(secx + tanx) - 2sinx]ex cosx 12 3. Solve the differential equation (D2 + 4) y = sec2x by the method of variation of parameters. Sol. Given equation is (D2 + 4) y = sec2x …..(1) A.E is m2 +4 = 0 m = The roots are complex conjugate numbers. ∴ yc = C.F = c1 cos2x⇒ + c2 sin2x± 푖 Let yp = P.I = A cos2x + B sin2x ∴Here u = cos2x, v = sin2x and R = sec2x. = -2 sin2x and = 2 cos2x u (cos 2x) (2 cos2x) (sin2x) (-2sin2x) ∴ – = 2 cos22x + 2 sin22x = 2(cos22x + sin22x) = 2 A∴ and B− are given= by : A = - dx = - 푖 || A =∫ 푥− 푥 ∫ = − ∫ = || B = dx = ⇒ yp ∫= P. 푥 I −= 푥 ∫ (cos2x) + (sin2x) = ∫ = The general solution|| is given by : ∴ y = yc + yp = C.F. + P.I ∴ i.e., y = c1 cos2x + c2 sin2x + (cos2x) + (sin2x) || DEPARTMENT OF HUMANITIES & SCIENCES ©MRCET (EAMCET CODE: MLRD) 150

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UNIT-IV

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PARTIAL DIFFERENTIAL EQUATIONS Definition: A Differential equation involves a dependent variable and its derivatives with respect to two or more independent variables is called partial differential equation.

Ex:

Linear & + non linear = p.d.e: + If the partial derivatives of the dependent variable occur in first degree only and separately, Such a P.D.E is said to the linear P.D.E, otherwise it is said as non –linear P.D.E Homogeneous & non homogeneous p.d.e: A P.D.E is said to the Homogeneous if each term of the equation contains either the dependent variable or one of its derivatives. Otherwise it is said to be Non - Homogeneous Formation & partial differential equations: Partial Differential equations can be formed by two methods 1.By the elimination of arbitrary constants 2.By the elimination of arbitrary functions

1.By elimination of arbitrary constants Let the given function be where a and b are arbitrary constants. ,,,, = ………. To eliminate a and b, differentiating (1) partially w.r.t. ‘ ’ ‘ ’ f  f  z  f  f .  0   .p  0 ……………….(2) and x  z  x  x  z f  f  z  f  f .  0   .q  0 ……………….(3) y  z  y  y  z Now eliminate the constants a and b from (1), (2) and (3). We get a partial differential equation of the first order of the form.  x, y , z , p , q   0 Note : 1. If the number of arbitrary constants is equal to the number of variables, a partial differential equation of first order can be obtained. 2.If the number of arbitrary constants is greater than the number of variables, a partial differential equation of order higher than one can be obtained. Solved Problems

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1. Form the partial differential equation by eliminating the arbitrary constants from (i) Sol: we have = + + Differentiating (1) = partially + w.r.t. + ‘ .’ ...... ‘ ’, we get z z a  p  a ……… (2) and b  q  b ………..(3) x y Putting the values of a and b from equation (2) and (3) in (1), we get

This is the required partial = differential + + equation 2. Form the partial differential equation by eliminating the arbitrary constants a and b a from (a) z ax  by  a22  b (b) z ax  by   b b Sol. (a) we have z ax  by  a22  b ...... (1) Differentiating (1) partially w.r.t. ‘x’ and ‘y’, we get z z a  p  a ……… (2) and b  q  b ………..(3) x y Putting the values of a and b from equation (2) and (3) in (1), we get z px  qy  p22  q

This is the required partial differential equations a (a) We have z ax  by   b ……………(1) b Differentiating (1) partially w.r.t. ‘x’ and ‘y’, we get z z a  p  a ..... (2) and b  q  b ………… (3) x y Putting the values of a and b from equation (2) and (3) in (1), we get p z px  qy   q q This is the required partial differential equation. 3. Form the partial differential equation by eliminating the arbitrary constants from x a 22   y  b   z22  r (OR ) Find the differential equation of all spheres of fixed radius having their centre on the xy-plane. Sol. The equation of sphere of radius r having their centers on xy-plane is

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22 x a    y  b   z22  r ……….(1)

Differentiating (1) partially w.r.t. ‘x’ and ‘y’, we get. z 2x a  2 z .  0 x a  zp 0 or x  a zp (2) x z And 2yb   2 z .  0 oryb    zq  0 oryb    zq  (3) 2y Putting the values of (x-a) and (y-b) from (2) and (3) in (1), we get

2222 zp     zq   z  r is the required partial differential equation. 4. Form the partial differential equation by eliminating the arbitrary constants a and b from

Sol:-The given = equation + + Differentiating (1) w.r.t.,= x+ +−−−−−−−−−− P = = 1 Differentiating . +(1) −−−−−−−−−− w.r.t., x q = = 1 from (2 . + − − − − − − − − −

= + Substituting in (1) we get = +

This is = the . required partial differential equations 5. Form the partial differential by eliminating the arbitrary constants from logaz 1   x  ay  b

Sol. We have logaz 1   x  ay  b………….(1) Differentiating (1) partially ‘ ’ ‘ ’ we get 11z .a . 1 or ap  1. or .ap .  az  1…………….(2) , az11   x az   1 z and a.1 a  aq  az   a ………….(3) az1  y q (3) (2), gives  a  ap  q …………..(4) p Putting (4) in (2), we get

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q q z 11 orpq  qz  porpq    q2 p is the required partial differential equation.

11 6.Form the differential equation by eliminating a and b from 2z x  a 22  y  a   b

11 Sol: We have 2z x  a 22  y  a   b …………..(1)

Differentiating (1) partially w.r.t. ‘ ’ ‘ ’, we have, z 11 2 2pp    4 x 2 x a x a 1 or xa 4p 1 or xa   (2) 16 p2 z 1 1 1 And 22or q  or y  a  yq22y a y a 4 1 yq    ………… (3) 16q2 Adding (2) and (3), we get 1 1 1 xy   22 16 pq or16 x y  p2 q 2  p 2  q 2 is the required partial differential equation. 7.Form the partial differential equation by eliminating the arbitrary constants a and b from z ax33 by

Sol. We have z ax33  by  (1)

Differentiating (1) partially w.r.t. ‘ ’ ‘ ’, we get zp 33ax22 or p  ax  a   (2) xx3 2 zq And 33by22 or q  by  b   (3) yy3 2 Putting the values of ‘a’ and ‘b’ from (2) and (3) in (1), we get pq z x y 33

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3z px qy

8.Form the partial differential equation by eliminating the arbitrary constants a and b from Sol:-The = given equation+ + Differentiating (1) w.r.t., = + +−−−−−−−−−− P = = +−−−−−−−−−− Differentiating (1) w.r.t ,we get ∴ + = q = = 2y . , +−−−−−−−−− ∴ + = Substituting in (1) we get implies that =

− = is the required partial differential equation.

9.Form the partial differential equation by eliminating the arbitrary constants from

Sol:Given − + − = 휶 ……..(1) Differentiating (1) w.r.t., − + − = 휶 Differentiating (1) w.r.t., − = 휶 Substituting (2),(3) in (1),we get − = 휶

The required 휶 Partial+ differential 휶 equation= is휶 ∴ + = 퐚 휶 Formation of the partial differential equation by the elimination of arbitrary functions:

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Derive a p.diff.eqn by the elimination of the arbitrary function from  uv,0   where u, v are functions of x, y and z.  uv,0   …. (1) Differentaite partially equation (1) w.r.to.  u  u  z     v  v  z  .     .   0 , u  x  z  x   v   x  z  x   u  u     v  v  i.e., pp       0 ……(2) u  x  z   v   x  z   44     vv   and qq       0……(3) 4  y  z   v   y  z    Eliminating and from (2) and (3) u v

u  u   v  v    u  u   v  v p   q     q   p y  z   y  z    y  z   x  z

uvuv  uvuv  uvuv i.e.  pq     yzzy  zxxz  xyyx is the P.D.E after the elimination of  from  uv,   0. Written in a simpler form u,,, v   u v   u v  pq y,,, z   z x    x y  Above equation is generally written as pP+qQ=R where u  v  u  v  u  v  u  v u  v  u  v P , Q   and R  y  z  z  y  z  x  x  z x  y  y  x Solved Problems x2 y 2 z 2 1.Form the partial differential equation by eliminating a,b,c from   1 a2 b 2 c 2 x2 y 2 z 2 Sol. Given   1 ….. (1) a2 b 2 c 2 Differentiating (1) partially w.r.t. ‘x’ and ‘y’. 22x z x z .p  0 or  . p  0  (2) a2 c 2 a 2 c 2 22y z y z And .q  0 or  . q  0  (3) b2 c 2 b 2 c 2 Since it is not possible to eliminate a, b,c from eqn (1), (2) and (3). We require one more relation.

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Differentiating (2), partially w.r.t. ‘x’, we get

2 1 1p  z 1 1  z 1 2 2 z.  p .  0 or2  2 . z . 2  2 . p a c x  x a c  x c

11 p2  .0zr    (4) a2 c 2 c 2 Multiplying (4) by ‘x’ and then subtracting (2) from it, we get

2 xz xp z 1 2 2.r 2  2 . p  0 or 2  xzr  xp  zp   0 c c c c

2 pz  xp  xzr is the required partial differential equation. 2. Form a partial differential equation by eliminating the arbitrary function Sol:-Given 흋 + , − = This can휑 be written+ , as − = f( ------(1) Now we have to eliminate − f from= (1)+ Differentiating (1) w.r.t., x

′ −′ = + ------(2) Differentiating (2) w.r.t., y − = + ′ ------(3) Dividing (2) by (3) − = + p y – q x = is the required partial − differential equation. 3. Form a partial differential equation by eliminating the arbitrary function from z f x22 y  Sol. We have z f x22  y   (1)

22 Put u x y , we have z f u  (2) Differentiating (2) partially w.r.t. ‘x’ and ‘y’,

zu11 f u . f u  .2 x xx p  f1  u 2 x  (3) Similarly we get (u) (4) DEPARTMENT = − OF HUMANITIES & SCIENCES ©MRCET (EAMCET CODE: MLRD) 158

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3   4  , gives = −

is the required∴ partial + differential= equation. 4. Form the partial differential equation by eliminating the arbitrary functions from xyz f x2  y 2  z 2  Sol. We have xyz f x2  y 2  z 2   (1) Differentiating (1) partially w.r.t. x and y

1 2 2 2 z yzxypfx.   y  z  . 2 x  2 z . x (or) yz xyp  f1 x 2  y 2  z 2 . 2 x  2 zp   (2) And xzxyq.  fx1 2  y 2  z 2  . 2 y  2 zq .   (3)

2   3  , gives yz xyp22 x zp  xz xyq22 y zq

yz xyp  y  zq    xz  xyq  x  zp 

2 2 2 2 2 2 y z z yq  xy p  xyzpq  x z  x zp  x yq  xyzpq

2 2 2 2 2 2 xyzpyz   xq  x  yz       is the required partial differential equation. 5. Form the partial differential equation by eliminating the arbitrary functions From

Sol: Given equations = + + ------(1) Differentiating (1) partially = w.r.t. + ‘x’ + y(xp+z) = ′ ------(2)

Differentiating (1) + partially + w.r.t. + ‘x’ x(y q + z ) = ′ ------(3)

Dividing (2) by (3) + + + + + + = +(

Yxp + z + q = xyq + z +

− + − = −

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This is the required partial differential equation.

6. Form the partial differential equation by eliminating the arbitrary function z from xy yz zx f  xy z Sol. We have xy yz  zx  f   (1) xy Differentiating (1) partially w.r.t. ‘x’ and ‘y’, we get x y. p z 1 z   y y.. p  z  x p  f  2  (2) xy xy 

z x y  q z And 1  (3) x z  yq  xq  f  2  xy xy  Dividing (2) by (3), we get

x y  p  y  z x  y  p  z  x y  q  x  z x  y  q  z

is the required partial differential equation.

7. Form the partial differential equation by eliminating the arbitrary function

y from z f x  e. g  x 

y Sol. We havez f x   e. g  x   (1)

Differentiating (1) partially w.r.t. ‘x’ and y, we get

z 1yy 1 1 1 fx   egxorp..    fx    egx    (2) x

yyz And q e.. g x  or  e g x   (3) y

Differentiating (3), partially w.r.t. ‘y’, we get

2 zzy e. g x  [using (3)] yy2

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2 zz    0 yy2 tq   0 Which is the required P.D.E. 8.Form a partial differential equation by eliminating the arbitrary function

Sol. We have = +

22 Put u x y , we have z f = u  (2)+ … . . Differentiating (2) partially w.r.t. ‘x’ and ‘y’,

zu11 f u . f u  .2 x xx p  f1  u 2 x  (3) zu And f11 u . f u  .2 y yy q  f1  u 2 y  (4)

pxf1  u .2 x 3   4  , gives  q f1  u 2 y y

py  qx  0

This is the required partial differential equation. 9.Form a partial differential equation by eliminating the arbitrary function Sol:Given function can be written as 흋 + + , + + =

+ + = + + …………………… Differentiating (1) partially w.r.t. ‘x’ and ‘y’,we get

+ = + + + … . implies + = + + + … .

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+ + + = + Which is the required complete solution of given Partial differential equation.

SOLUTION OF PARTIAL DIFFERENTIAL EQUATIONS : Complete integral: A solution in which the number of arbitrary constants is equal to the number of independent variables is called complete integral or complete solution of the given equation. Particular integral : A solution obtained by giving particular values to the arbitrary constants in the complete integral is called a particular integral or particular solution.

Singular integral: Let f x, y , z , p , q   0 (1) be the partial differential equation. Let  x, y , z , a , b   0 (2) Be the complete integral of (1). Where a and b are arbitrary constants.   Now find 0 (3) 0 (4) a b Eliminate a and b between the equations(2), (3) & (4) When it exists is called the singular integral of (1). General integral : In the complete integral (2). Assume that one of the constant is a function of the other i.e. b=f(a) Then (2), becomes  x, y , z , a , f a    0 (5)  Differentiating (5) partially w.r.t. ‘a’, we get .0fa1     (6) af Eliminate ‘a’ between (5) and (6), when it exists is called the general integral or general solution of (1).

LINEAR PARTIAL DIFFERENTIAL EQUATIONS OF THE FIRST ORDER: A differential equation involving partial derivatives p and q only and no higher order derivatives is called a first order equation. If p and q occur in the first degree, it is called a linear partial differential equation of first order; otherwise it is called a non-linear partial differential equation of the first order. For example: px qy2 z is a linear p.d.e of first order and pq221 is non-linear

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Lagrange’s linear equation: A linear partial differential equation of order one involving a dependent variable z and two independent variables x and y of the form Pp Qq R Where P, Q, R are functions of x, y, z is called Lagrange’s linear equation. Lagrange’s auxiliary equations or Lagrange’s subsidiary equations dx dy dz The equations are called Lagrange’s auxiliary equations. PQR WORKING RULE TO SOLVE LAGRANGE’S LINEAR EQUATION dx dy dz Step 1: Write down the auxiliary equations  + = PQR Step 2 : Solve the auxiliary equations by the method of grouping or the method of multipliers or both to get two independent solutions u=a and v=b where a,b are arbitrary constants Step 3: Then Q(u,v) =0 or u=f(v) is the general solution of the equation dx dy dz To solve  …… (1) + = Pxyz(,,)(,,)(,,) Qxyz Rxyz (i) Method of grouping : In some problems, it is possible that two of the equations dx =dy or P Q

dy dz dx dz  or = are directly solvable to get solutions u(x,y) = constant or v(y,z)=constant QR P R or w(x,z)=constant. These give the complete solutions of (1) dx dy Sometimes one of them, say  may give rise to solution u(x,y)=c1 PQ

dy dz From this we may express y, as a function of x. Using this in  and integrating we get QR

2. Method of multipliers: This is base on the following elementary result. , = . ℎ 푖 = , = 푖 ℎ 푖 aaaa l a l a ....  l a If 12 3 ......  n then each ratio is equal to 1 1 2 2 nn b b b b l b l b ....  l b 1 2 3 n 1 1 2 2 nn dx dy dz Consider  PQR If possible identity multipliers l, m, n, not necessarily constant, so that each ratio ldx mdy ndz  lP mQ nR Where Then

Integrating this + we get + = + + = Similarly we get another, solution , = . ndependent of the earlier one. We have the complete solution of ,(1) constituted , = 푖

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Linear Partial P.D.E: Solved Problems 1. Solve

Sol. The given equations + can be =written as tanxp tan yq  tan z  (1) Comparing with , we have Ptan x , Q  tan y , R  tan z dx dy dz The auxiliary equations + are =  tanx tan y tan z dx dy Taking the first two members, we have  tanxy tan

Integrating logsinx log sin y l og c1 sinxx sin or log log c or  c  (2) sinyy11 sin dy dt Taking the last two members, we have  tanyz tan

Integrating,

푖 = 푖 + sinyy sin or log log c or  c  (3) sinzz22 sin From (2) and (3). The general solution of (1) is

 cc12,0  

sinxy sin ie. .  , 0 sinyz sin is the required Complete Solution. 2. Find the general solution of y2 zp x 2 zq y 2 x

Sol. We have y2 zp x 2 zq  y 2 x  (1) Comparing

2 2 2 P y z,, Q 푖ℎ x z R  + y x = , ℎ dx dy dz  The auxiliary equations are 2 2 2 y z x z y x Taking the first two members, we have dx dy dx dy   or x22 dx  y dy y2 z x 2 z y 2 x 2

x3 x 3 y 3 Integrating, y3  c or   c  (2) 33 11 3 3

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Taking the first and last two members, we have dx dz or xdx zdz y22 z y x

x2 z 2 x 2 z 2 Integrating  c or   c2  (3) 2 22 2 2 From (2) and (3) The general solution of (1) is

 c12, c   0 i . e .

x3 y 3 x 2 z 2  ,0   3 3 2 2  is the required Complete Solution.

3. Solve p x q y z Sol. The given equation can be written as xp yq  z  (1) Comparing with Pp+Qq=R, we have P x,, Q  y R  z

dx dy dz  The auxiliary equations are  x y z dx dy From the first two members, we have  xy

Integrating, 2x 2 y  c11 or 2 x  2 y  c or x y  a  (2) dy dz From the last two members, we have  yz

Integrating, 2y 2 z  c22 or 2 y  2 z  c

or y z  b  (3) From (2) and (3). The general solution of (1) is  ab,0   ie. .,

 x y,0 y  z   is the required Complete Solution. 4. Solve xyzpyzxq       zxy    Sol. We have xyzpyzxq       zxy     (1) Comparing with , we have

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PxyzQyzxRzxy  ,,         dx dy dz  The auxiliary equations are  x y z  y z  x  z x  y  Using l=1,m=1, n=1 as multipliers, we get dx dy dz dx dy dz  x y  z   y  z  x   z  x  y   0 x y z y z  x z x  y 0        dx  dy  dz  0 Integrating, x y  z  a  (2) 1 1 1 Again using lmn,,   as multipliers, we get x y z

111 dx dy dz x y z Each fraction  =k(say) 0 111 dx  dy  dz  0 x y z Integrating, logx+logy+logz=logb. or xyz=b……. (3) From (2) and (3). The general solution of (1) is  ab,0 ie. .,    x y  z,0 xyz   is the required Complete Solution. 5. Solve xyzpyzxqzxy2   2     2    Sol. Given xyzpyzxqzxy2   2     2     (1) Comparing with we have

2 2 2 PxyzQyzxRzxy  ,,   +  = ,    dx dy dz  The auxiliary equations are  x2 y z  y 2 z  x  z 2  x  y  1 1 1 Using l2,, m  2 n  2 as multipliers, we get x y z 111 dx dy dz x2 y 2 z 2 Each fraction  =k(say) 0 111 dx  dy  dz  0 x2 y 2 z 2

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1 1 1 1 1 1 Integrating ,     a or   c1  (2) x y z x y z 1 1 1 Again using lmn,,   as multipliers, we get x y z 111 dx dy dz x y z Each fraction  =k(say) 0 111 dx  dy  dz  0 x y z

Integrating logx log y  log z  log c2

or xyz c2 (3) From (2) and (3), The general solution of (1) is .

1 1 1   ,0xyz  x y z is the required Complete Solution.

6. Solve mz ny  p   nx  lz  q  ly  mx Sol. Given eqn is mz ny  p   nx  lz  q  ly  mx  (1) Comparing with , we have

+ =  The auxiliary equations are = − , = − , = −

dx dy dz  mz ny nx  lz ly  mx Using l=x, m=y, n=z as multipliers, we get xdx ydy zdz Each fraction  0 xdx  ydy  zdz  0

x2 y 2 z 2 Integrating,   a or x2  y 2  z 2  c  (2) 2 2 2 1 Again using as multipliers, we get ldx mdy nz Each fraction , , =k(say) 0 ldx+mdy+ndz=0

Integrating, lx my  nz  c2  (3)

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From (2) and (3), the general solution of (1) is

 x2 y 2  z 2 ,0 lx  my  nz   is the required Complete Solution. 7. Solve xp yq  y22  x

Sol. Here P x,, Q  y R  y22  x dx dy dz  The auxiliary eqn’s are 22 x y y x dx dy From the first two members,  xy

Integrating, logx log y  log c11 or xy  c  (1) Using l=x, m=y, n=1 as multipliers, we get xdx ydy dz Each fraction  0 xdx  ydy  dz  0 11 Integrating, x2 y 2  z  c or x 2  y 2 2 z  c  (2) 22 2 From (1) and (2), The general solution is

 xy, x22 y  2 z  0   is the required Complete Solution.

8. Find the integral surface of xy2 zpyx    2  zq    x 2  yz 2  Which contains the straight line x+y=0, z=1 Sol. Given that xy2 zpyx    2  zq    x 2  yz 2  ……..(1) Comparing with we have

2 2 2 2 PxyzQ  ,,   yxzR+  =  ,   xyz   The auxiliary equations are dx dy dz  x y2 z   y x 2  z   x 2  y 2  z 1 1 1 Using lmn,,   as multipliers, we get x y z

111 dx dy dz Each fraction = x y z 0

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111 dx  dy  dz  0 x y z Integrating, logx log y  log z  log a or xyz a (2) Again using l=x, m=y, n=-1 as multipliers, we get xdx ydy dz  Each fraction  =k(say) 0 xdx  ydy  dz  0

xy22 Integrating,  z  cor x22  y 2 z  b  (3) 22 Given that z=1, using this (2) and (3), we get xy=a and x22 y 2  b

2 Now b+2a= x22 y 2  2 xy  x  y   2=0-2 xy   0  =-2 2ab   2  0 Hence the required surface is

22 x y 2 z  2 xyz  2  0 is the required Complete Solution. 9.Solve

Sol:Given + = is a Lagrange’s linear equation

The Auxillary + equations = are

By Consider first group, we get = =

∫ = ∫ …..(1) log = log + log By = Consider second group, we get

∫ = ∫ log = log + log DEPARTMENT OF HUMANITIES & SCIENCES ©MRCET (EAMCET CODE: MLRD) 169

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…..(2) = ∴ , = is the required solution.

10. Solve Sol: The auxiliary − equations − are + − − = −

= = − − − − − Taking 1,-1-1 multipliers,we get

= −− ( − −− + +−+) ( − −)

Integrating,we get − − = ………(1)

Taking as multipliers,we get − − = , −,

− = − − − + + − − = − − − ) 퐥 − = 퐥 − = … … … … … … …

− ∴ 흋 − − , = 11. Solve Sol: The auxiliary equations− − are + = +

= = − − + + DEPARTMENT OF HUMANITIES & SCIENCES ©MRCET (EAMCET CODE: MLRD) 170

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Taking , multipliers, we get

, , = ++ ( − − − + + ) ( − )

+ + = + + = …………………. Taking , multipliers, we get - − , Integrating, + we +get = …….(2) From (1),(2), =

∴ 흋 , + + =

12.Solve Sol:Comparing with Pp+Qq=R, we have − = −

The auxiliary equations are   From the= first− two= − members, we have

Type equation here. Integrating,we get = − …..(2) + = From the last two members, we have

= − − − = − ………..(3) − = From (2) and (3). The general solution of (1) is − =

DEPARTMENT OF HUMANITIES & SCIENCES ©MRCET (EAMCET CODE: MLRD) 171

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ie. ., =0 12. Solve∅ − , +

Sol:Comparing + with + Pp+Qq=R, + =we have +

The auxiliary equations are   Taking + 1,1,1 = and+ 1,-1,0= + and 0,1,-1 as multipliers , we have ++ − − ++ = − = − From the last two members, we have

− − Integrating,we get = − −

− log = log − …..(1) − − = From the first two members, we have

+ + − = + + − 푰, log + + = log − + log ………..(2) From (2) and (1). The general solution of given pde is + + − = ie. ., =0 − 13. Solve∅ − , + + − Sol:Comparing − with = − we have

+ = , The auxiliary equations are   From the= −first= two− members, we have

DEPARTMENT OF HUMANITIES & SCIENCES ©MRCET (EAMCET CODE: MLRD) 172

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Integrating,we get = − …..(1) Taking 1,1,0 as multipliers,we get + = = + − −= + +−=− + Integrating,we get +

+ = …. From (2) and (1). The general solution is

ie. ., =0 + 14. Solve ∅ , + Sol: The auxiliary − equations + are − = −

Taking( − 1,-1,0) = ( and− )0,-1,-1= − as multipliers,we get

and also − − ( −− −) (− ++ −) = − − ∴ ( −− −) (− ++ −) = − − solving−++ it,we get− ++

= …..(1) − − multipliers,we get

푖 , , = ,, ++ ++ + + − + + −=−− ++ ++ ++ + + −−− + + −−−

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Integrating,we get

+ + + + = + …..(2)∴ + + = + + + + + =

− ∴ 푖 푖 푖 휑 + + , = −

NON-LINEAR PARTIAL DIFFERENTIAL EQUATIONS OF FIRST ORDER A partial differential equation which involves first order partial derivatives with degree higher than one and the products called a non-linear partial differential equations. 푖 CHARPIT’S METHOD In this method give D.E of the form to find another relation of the form

which is compatible,,,, with = the we solve for and substitute 휑,,,, these values in= the relation ℎ , , , , = , Which on integration gives the required = solution + . of

Charpit’s equation : ,,,, =

= = = = Solved Problems − − − − [ + ] [ + ] 1.Solve

Sol: Given + = The auxillary equations = + are−

= = = = Here − − − − [ + ] [ + ] Substituting = ; them = in ; above = ,we ; get = ; =

By considering first and last groups,= = = = − − +

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Integrating,we get

log = log + log ∴ = … .. Integrating,we get = log = log + log Substitute (1),(2) in given pde ,we get ∴ = … . .

+ But we have = + ; =

= +

+ = + +

Integrating it we get required solution as

2.Solve = + + Sol: Given = = − = = = = Here − − − − [ + ] [ + ] Substituting = −; them in above= −; ,we get = − ; = − ; =

Considering = = = = − + − + and Taking= −+ = −+

, 푖푖 푖 , , ℎ 푖 ℎ = + + Solving−++ , −++

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Integrating,we get =

log = log + log Substituting in given pde,we get =

; = ∴But =we √have = √

= +

= √ + √ Integrating,we get, ∫ = ∫ √ + ∫ √ Required solution is

∴ √ √ = Now let us start solving some standard forms of first order partial differential equations by using Charpit’s method

STANDARD FORM I: Equation of the form f(p,q)=0 i.e., equations containing p and q only. Given partial differential equation is

The auxillary equations are , = … …

= = = = Here − − − − [ + ] [ + ] Substituting = above; = and ; considering = last group, we get

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Put .

But we = have , , 흋

= = +

= + ……(2)

+ 흋 Integrating (2),we get required complete solution of (1) is

z ax   a  y  c

Which contains two arbitrary constants a and c. PROCEDURE: Given partial differential equation is

STEP1:Put , = … … .then we can obtain ‘p’ value. = 푖 , ℎ 푖 .

: = , 푖 = + STEP3:Integrating it ,we get required complete solution of (1) . . + Solved Problems 1.Solve where k is a constant. Sol. Given that = , Since (1) is of the= form … . Put p=a in (1),we get , = But we have = =

+ Integrating,we = + get , k z ax  y  c a which contains two arbitrary constants a and c. 2.Solve Sol : Given +that = 1) + = … … … … . . DEPARTMENT OF HUMANITIES & SCIENCES ©MRCET (EAMCET CODE: MLRD) 177

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Since (1) is of the form

Put p=a in (1),then we get, = But we have = [ ± √ − ] =

+ Integrating,we = + get[ , ± √ − ] a z ax  n  n2 4 = y ∫ c + [ ± √ − ] ∫ 2  This is the complete integral of (1), which contains. two arbitrary constants a and c. 3.Find the complete integral of Sol. Given that + = Since (1) is of the+ form = … … … . Put , = But we = have 푖 , = √ − = …….(2) Put the values of p,q in (2), we get +

z ax  m22  a y  c Which is the complete integral of (1)

STANDARD FORM II : Equation of the form (i.e., not containing x and y)

, , = PROCEDURE Given partial differential equation is

STEP1:Put , , = … … .then = 푖 , ℎ 푖 , : = , 푖 = + STEP3:Integrating it ,we get required complete solution of (1) . . + Solved Problems : Solve the following partial differential equations 1. 2. 3. = + + = = + DEPARTMENT OF HUMANITIES & SCIENCES ©MRCET (EAMCET CODE: MLRD) 178

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Sol. 1. We have

Since (1) is of the form=+……. Put , , = = 푖 , ℎ = √+

Putting the values =∴ = √ , we get + = ,푖 + √ √+ + Integrating ,we get

∫ = ∫ + √ √ + This is the required∴ √ solution= √+ of (1) + 2. Given that p2 z 2 q 2  p 2 q  (1) Since (1) is of the form

Put , , = + = 푖 , ℎ =

+ ∴ = Putting the values of p and q in = ,we get

= + Integrating,we + get + = ∫ + ∫ + which is the required complete solution− of (1) ∴ a tan = + +

3. Given that …..(1)

Since (1) is of the form = + Put , , = + = 푖 , ℎ = + Putting the values = ∴ = , we get = , 푖 + + + DEPARTMENT OF HUMANITIES & SCIENCES ©MRCET (EAMCET CODE: MLRD) 179

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Integrating ,we get

+ ∫ = ∫ +

∴ + = + + This is the required solution of (1)

STANDARD FORM III : Equation of the form f x,, p f y q i.e. Equations not involving z and the terms 12    containing x and p can be separated from those containing y and q. We assume that these two functions should be equal to a constant say k.

f12 x,, p   f  y q   k Solve for p and q from the resulting equations

f12 xp,,   kandf  yq   k Solve for p and q, we obtain

p Fxk12,,  andq F  yk  Since z is a function of x and y zz dz dx dy [By total differentiation] xy =

dz  F 12 xkdx,,+   F  ykdy  Integrating on both sides z  F x,, k dx  F y k dy  c 12    Which is the complete solution of given equation Solved Problems: 1. Solve p22 q  x  y

Sol :. Given that p22 q  x  y ………..(1) Separating the given equation can be written as

22 p x   q  y , Let p22 x   q  y  k (constant)

p22  x  k and  q  y  k

p22  k  x and q  y  k

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p  k  x and q  y  k

zz Since dz dx  dy  pdx  qdy xy

dz  k  x dx  y  k dy Integrating on both sides

11 z  k  x22 dx   y  k dy  c    

2233 z  k  x 22  y  k   c 33

Which is the complet solution of (1) 2. Solve xp yq  y22  x

Sol: Given that xp yq  y22  x  (1) Separating and from and . The given equation can be written as. xp x22  yq  y Let xp x22  yq  y  k (arbitrary constant)

xp  x22  k and yq  y  k

k x22 k y p  and q  xy zz We have dz dx  dy  pdx  qdy xy

kk dz   x dx   y dy xy Integrating on both sides

kk z   x dx    y dy  c xy xy22 klog x   k log y   c 22

1 22 z  klog xy   x  y   c 2 Which is the complete integral of (1)

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22 pq    3. Solve xy      1 22    Sol: Separating the given equation can be written as.

22 pq    , xy  1     22   

22 pq    2 Let x  1    y   k (arbitrary constant) 22   

22 pq 22    x   k and1    y   k 22   

2 p q22 q  xkand   y 11  kor  y   k 2 2 2

p 2 k  x and q  2 1  k2  y    zz We have dz dx  dy  pdx  qdy xy

dz 2( k  x ) dx  2 1  k2  y dy  Integrating on both sides

z2 ( k  x ) dx  2 1  k2  y dy  c 

22 xy2 z2( kx  )  2 1  k y   c 22

z 2 kx  x2  2 1  k 2 y  y 2  c

This is the complete solution of (1) 4.Solve Sol:Let − = + Then − = and + = − = + = ∴ = + = + But we have zz dz dx  dy  pdx  qdy xy

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Integrating,we get

+c is the required = +complete + solution. +

5. Solve Sol: Let− = − - Then − = = = + = √ + But zz dz dx  dy  pdx  qdy xy Integrating,we get

+C is the required = + complete + solution. + 6.Solve Sol:Let = + Then we get = + =

Solving,we get + − = = and −±√ + But = = zz dz dx  dy  pdx  qdy xy Integrating,we get

)+k +C − is the required = − complete+ [ √ solution. + + √

STANDARD FORM IV:

An equation analogous to the =clairaut’s + equation + , it is complete solution is which is obtained by writing The differential equation =which +satisfies + some , specified conditions known as the boundary conditions.The differential equation together with these boundary conditions, constitute a boundary value problem

MATHEMATICS - I

Solved Problems: 1. Solve Sol : The given PDE is form IV = + + Therefore complete solution is given by

2. Find = the+ solution + of p q  z  px  qy  1 Sol. The given equation can be written as 1 z px  qy  pq 1 z  px  qy   (1) pq Hence the complete solution of (1) is given by

1 z ax  by  ab

3. Solve pqz p2 qx  p 2   q 2  py  q 2  Sol. The given equation can be written as

22 22pq    pqz p q x    q p  y   qp   

pq22    z  p x    q  y   qp   

pq33 z  px  qy    (1) qp Since it is in the form z px  qy  f p, q  Hence the complete solution of (1) is given by ab33 z ax  by   ba

4. Solve Sol. We have Since (1) is = of the+ form + + Hence the complete = solution+ + of (1)+ is given… … …by … … . = + + , . For singular solution, differentiating (2) partially w.r.t. a and b, we get = + + + … … .

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Implies that = , = , 4) Eliminating a, b between (2), (3) and (4), we get = + … . = + + … … … … = − − − − + is the singular solution ∴ = EQUATIONS REDUCIBLE TO STANDARD FORMS:

EQUATIONS OF THE FORM The above form of the equation of the type can be transformed to an equation of the form f(p,q)=0 By substitutions given below. , = Case (i):- when Put ≠ ≠ − − − = = ℎ = = = − ℎ = − Now the given = equation − reduces to = = = − ℎ = → = − Case(ii):- when Put [ − , − ] = ℎ푖ℎ 푖 ℎ , = = , = p= = = = ℎ = = 푖푖 = ℎ = now the given equation reduces to the푖푖 form = ℎ = EQUATIONS OF THE FORM : , = This can be reduced to an equation of the form by the substitutions given for the equation as above. , , = ,, = Solved , Problems: , = xy22 1. Solve the partial differential equation z pq Sol. Given equation can be written as 11 xp2 1 yq 2  1  zorxp  2    yq  2   z  (1) This is of the form f xmn p, y q , z   0 with

13m 12  1n 1 2 3 Put X x  x  x and Y y  y = y−, = −. z  z  X z Then p .  P .3 x2 where P  x  X  x X x2 p3 P z  z  Y z and q .  Q .3 y2 where Q  y  Y  y Y

MATHEMATICS - I

y2 q3 Q Now equation (1), becomes. 33P 11  Q   z  (2) Since (2) is of the form f P, Q ,z   0 Put + = 푖 , ℎ = + ∴ = Putting the values of Pand Q in = ,we get = + Integrating,we get + + = ∫ + ∫ + ∫ = + + + 2a 1 3 3 32z  x  ay   c1 a

,taking + Which is the required solution of (1) = pq 2. Solve the partial differential equation z xy22 Sol. The given equation can be written as px22 qy  z  (1) Since (1) is of the form f xmn p, y q , z   0 With

13m 13n Put X x x , and Y y y = −, = − z  z  X z Now p .  P .3 x2 where P  x  X  x X x2 p3 P z  z  Y z and q .3  Q y2 where Q  y  Y  y Y y2 q3 Q Equation (1) becomes, 33P Q  z  (2) Since (2) is of the form f P, Q ,z   0 Put = 푖 , ℎ = + ∴ = Putting the values of Pand Q in = + ,we get

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= Integrating,we get + + = ∫ + ∫ + ∫ log = + + 1 + log z  x33  ay   c 31 a  This is the complete solution of (1) 3. Solve q22 y z z px  Sol. Given equation can be written as q2 y 2 z 2  zpxor xp  z   qy 2  z 2  (1)

mn Since (1) is of the form f x p, y q , z   0 with Put Xx log and Yy log = = z  z  X 1 z Now pP ..  where P  x  X  x x X xp P z  z  Y 1 z and qQ ..  where Q  y  Y  y y Y qy Q 22  Equation (1), becomes, Pz Q  z  (2) Since (2) is of the form f P, Q , z   0 Put = 푖 , ℎ = [−a ± √a + ] ∴ = [−a ± √a + ] Putting the values of Pand Q in = ,we get = + Integrating,we get [−a ± √a + ] + = ∫ [−a ± √a + ] ∫ + ∫ log = [−a ± √a + ] + + ∴ log = [−a ± √a + ] + + is the complete integeral of (1) 4. Solve the partial differential equation p2 x 4 y 2 zq2 z 2 Sol. Given that p2 x 4 y 2 zq2 z 2 Then given equation can be written as

MATHEMATICS - I

2 px2   qy 2  z 2 z 2  (1) Since (1) is of the form f xmn p, y q , z   0 with m=2 and n=2 1 1 Put X x1m  x 1  2  x  1  and Yy1 x x z  z  X  1 z Now PP ..,  2 where P  x  X  x x X x2 p   P z  z  Y  1 z and qQ ..,  where Q  2 y  Y  y y Y y2 q   Q Now equation (1) becomes, P2 Qz 2 z 2 or P 2  Qz  2 z 2  (2) Since (2) is of the form Put , , = = 푖 , ℎ = [ ± √a + ] Putting the values of Pand Q in ∴ = = [ ± √a ,we+ get] = + Integrating,we get [ ± √a + ] + = ∫ [ ± √a + ] ∫ + ∫ log = [ ± √a + ] + + ∴ log = [ ± √a + ] + + Which is the complete integral of (1). 5. Solve x2 p 2 xpq z 2 Sol. The given equation can be written as xp 2   xp  q  z 2  (1) Since (1) is of the form f xmn p. y q , z   0 with m=1 and n=0

Put Xx log z  z  X 1 Now PP .,  where x  X  x x z P  X xp P Equation (1) becomes, P22 Pq  z  (2) Since (2) is of the form Put , , = = 푖 ,

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, P=a But = we√+ have √+ = Substuting P,q ,we get + = √+ +

Integrating on both sides

∫ / = ∫ + √ + =

√be the +complete log integral + of (1)+ 6.Solve z p22 x q y Sol. Given that z p22 x q y The given equation can be written as

1122 22    px   qy   zorpx22    qy   z  (1)    

This is of the form

1 1 1 1 11 Put X x1m  x2  x 2 and Y  y 2  y 2 z  z  X1 1  z Now p .,  P x2 where P  x  X  x2  X z  z  Y1 1  z and q .,  Q y2 whereQ  y  Y  x2  Y 11PQ px22  and qy  22 22 PQ    22 Then equation (1) becomes,     z i. e . P  Q  4 z  (2) 22    This if of the form f P, Q , z   0 Put

= 푖 , + = , P=a , But = we√ have+ √ + = Substuting P,Q ,we get +

MATHEMATICS - I

= √ + + = + Integrating on both sides √ √ +

∫ /√ = ∫ + ∫ = √ +

= √ + √ + + Which is the complete integral of (1) √ + √ (√ + √) + 7.Solve Sol: Given + = + = … … . . Since (1) is of the form f xmn p, y q , z  0 with   + = Put Xx log and Yy log = = z  z  X 1 z Now pP ..  where P  x  X  x x X xp P z  z  Y 1 z and qQ ..  where Q  y  Y  y y Y qy Q  Equation (1), becomes

……(2) + = But we have = 푖 , = ; = = √ + √ + Substuting P,Q ,we get + = √ + + = + Integrating on both sides √ +

∫ / = ∫ + ∫ = √ +

= √ + log + + is the Complete solution of (1) √ + log log + log + DEPARTMENT OF HUMANITIES & SCIENCES ©MRCET (EAMCET CODE: MLRD) 190

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8. Solve Sol: Given + = + = … … . . Since (1) is of the form + with = Put Xx log and Yy log , = = = z  z  X 1 z Now pP ..  where P  x  X  x x X xp P z  z  Y 1 z and qQ ..  where Q  y  Y  y y Y qy Q  Equation (1), becomes

……2 + = But we have √ = = 푖 , = − Substuting P,Q ,we get + = − Integrating on both+ √ sides

∫ = ∫ + √ − ∫ z =

z= ( + √ − ) + (is logthe Complete + √ − solution log ) of + (1) EQUATIONS OF THE FORM Use the following substitution to reduce the above form to an equation of the form f(P,Q)=0 , = :

+ 푖 ≠ − EQUATIONS OF THE FORM = { : An equation of the above form can be reduced to logan equation , of 푖 the form = − f(P,Q)=0 by the substitutions given for the equation, = , as above Solved Problems : , = 1. Solve z2 p 2 q 2   x 2  y 2 Sol. Given that z2 p 2 q 2   x 2  y 2

MATHEMATICS - I

The given equation can be written as zp2222 zq  x 2  yorzp 2 222  x  y 222  zq

Or zp 22 x22  y   zq   (1) Since (1) is the of the form f x,,. pznn  g  y qz  with n=1

put Z  zn1  z 1 1  z 2 Zz z Then 2z .  P  2 zp where P  xx x P pz 2 Zz z Q and 2z .  Q  2 zq where Q  qz yy y 2 PQ22 Equation (1) becomes, xy22   44 i. e ., P2 4 x 2  4 y 2  Q 2  (2)

This is of the form f12 x,, P  f  y Q  Let P24 x 2  4 y 2  Q 2  4 k 2 (say) P2 4 x 2  4 k 2 and 4 y 2  Q 2  4 k 2 P2 4 x 2  4 k 2 and Q 2  4 y 2  4 k 2

P 22 x2  k 2 and Q  y 2  k 2

ZZ We have dZ. dx dy xy Pdx Qdy [By total differentiation] dZ 22 x2  k 2 dx  y 2  k 2 dy Integrating on both sides Z22  x2  k 2 dx   y 2  k 2 dy

2 2 x2 2 k 1  x x2 2 k 1  y 2xk   sinh 2y  k  cosh   c 22k 22k

2 2 2 1xy  2 2 2 1   x x  k  ksinh   x y  k  k cosh    c kk   

2 2 2 2 2 2 1xy  1   orzxxkyykk    sinh   cosh    c kk    x x22 k orzxxkyykk2 2  2  2  2  2 log  c 22 y y k

This is the complete solution of (1)

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2. Solve the partial differential equation. p2 z 2sin 2 x q 2 z 2 cos 2 y 1 Sol. Given that p2 z 2sin 2 x q 2 z 2 cos 2 y 1 The given equation can be written as 22 2 2 2 2 2 2 pz sin x  qz  cos y  1 or  pz  sin x  1   qz  cos y  (1)

nn Since (1) is of the form f x,, pz  g  y qz  with n=1.

Put Z zn12 z Z z P Now 2z .  P  2 zp or pz  where xx 2 Z z Q = ; = and 2z .  Q  2 zqor qz  yy 2

22 PQ 22   Then equation (1) becomes,  sinxy 1   cos 22    PQ22 i. e . sin22 x 1  cos y  (2) 44

This is of the form f12 x,, p  f  y q  PQ22 Let sin2x 1  cos 2 y  k 2 (constant) 44 PQ22 sin2x  k 2 and 1  cos 2 y  k 2 44 P2sin 2 x  4 k 2 and Q 2 cos 2 y  4 1  k 2 

2kk 2 1 2 P  and Q  sinxy cos ZZ We have dZ dx dy [By total differential] xy dZ  Pdx  Qdy

2kk 2 1 2 dZ dx dy sinxy cos Integrating on both sides

= ∫ csc + √ − ∫ sec 2k log cos ecx  cot x   2 1  k2 log sec y  tan y   c

z22 2 k log cos ecx  cot x   2 1  k log sec y  tan y   c

This is the required complete solution of (1) 3.Solve Sol:Given …….(1) + + + = + + + = DEPARTMENT OF HUMANITIES & SCIENCES ©MRCET (EAMCET CODE: MLRD) 193

MATHEMATICS - I

Put = 푖 푖 ℎ , , , = = + Differentiating = partially w.r.t ‘x’,we get But = implies = 푖푖 ℎ = Substitute in (1),we get = = = ; 푖푖 = =1 Separating the given equation can be written as. + + +

, + = − + = + = − + =

+ = Implies that

)

= √ − − ZZ = −dZ  dx dy We have xy

dz 2( k  x ) dx  2 1  k2  y dy  Integrating on both sides z2 ( k  x ) dx  2 1  k2  y dy  c  xy22 z2( kx  )  2 1  k2 y   c 22  

z 2 kx  x2  2 1  k 2 y  y 2  c

This is the complete solution of (1). 4. Solve Sol: Given ( − ) = −

…..2 ( − ) =−………… − = −

Put = 푖 푖 ℎ , , , = = + Differentiating partially w.r.t ‘x’,we get = =

MATHEMATICS - I

푖푖 ℎ = But =Pimplies Substitute in (2),we get = = ; 푖푖 = Separating the given equation− can= be written − as. , Solving, we get − = − + = and = √ +ZZ = √ + We have dZ dx dy [By total differential] xy dZ  Pdx  Qdy

Integrating on both sides = [√ + + √ + ]

= [∫ √ + + ∫ √ +

This is the required complete solution = of +(1) + + +

Methods Of Separation Of Variables: This method is used to reduce one partial differential equation to two or more ordinary differential equations,each one involving one of the independent variables.This will be done by separating these variables from the beginning. This method is explained through following examples.

1. Solve by the method of separation of variables where U(x,0)=6 − Solution:- Given equation is ------(1) = + Let U(x,t)= X(x) T(t) =XT ------(2) = + be a solution of (1) Differentiating (2) partially w.r.t x ant t

′ ′ Put these values in equation (1), we =have , = Dividing by XT ′ ′ = + ------(3) ′ ′ = + DEPARTMENT OF HUMANITIES & SCIENCES ©MRCET (EAMCET CODE: MLRD) 195

MATHEMATICS - I

Since L.H.S is a function of ‘x’ and the R.H.S is a function of ‘t’ where x and t are independent variables, the two sides of (3) can be equal to each other for all values of ‘x’ and ‘t’ if and only if both sides are equal to a constant. =k------(4) where k is a constant ′ ′

Therefore = +

Now from (4) = k----(5) and =k------(6) ′ ′ kx Now consider (5) = k X  kX 0 +  X  C e ′ 1 k 1 t k 1 2 Now consider (6) =k T  T 0  T  C2 e    (8) ′  2 Substituting the values of+ X and T in (2) we get k1 t kx 2 U(,) x t X C12 e C e k1 t kx 2 U( x,)() t XAee w here A=C12 C kx Put t=0 in the above equation ,we have U(x,0) = A e -----(9) but given that U(x,0)=6 ------(10) from (9) and (10) we have− A ekx =6 A=6 and k=-3 the solution of the given− equation becomes U( x , t ) X  6 e3x e2  t  6 e  (3 x  2 t )

2. Solve the equation by the method of separation of variables Sol: Given equation is ------(1) = + Let U(x,y)= X(x) Y(y) =X Y ------(2) = + be a solution of (1) Differentiating (2) partially w.r.t x ant y

′ ′ ′′ Put these values in equation (1),= we have , = = Dividing by XY on both sides we have′′ ′+2 ′′ = ′′ + ------(3) ′′ = ′′ Since L.H.S is a function of ‘x’ and the R.H.S is a function of ‘y’ where x and y are − = independent variables, the two sides of (3) can be equal to each other for all values of ‘x’ and ‘y’ if and only if both sides are equal to a constant. ------(4) ′′ ′′ Now from (4) − = = DEPARTMENT OF HUMANITIES & SCIENCES ©MRCET (EAMCET CODE: MLRD) 196

MATHEMATICS - I

′′ And −=−−−−−− ′′

=−−−−−− From (5) Which is second′′ order differential equation′′ − = − + = Auxiliary equation is m= Solution of the given equation (5) is − + = → ±√ + √+ Now consider equation (6) = ′ Integrating on both sides we get′ logy = ky + log = → =  Y ky log ky  Y  C3 e    (8) C3 Substituting the values of X and Y in (2) we have U C e(2k ) x C e  (2  k ) x C eky 1 2 3 U Ae(2k ) x Be  (2  k ) x eky 

Where A C1 C 3 and B C 1 C 2

MATHEMATICS - I

UNIT V

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LAPLACE TRANSFORMS INTRODUCTION Laplace Transformations were introduced by Pierre Simmon Marquis De Laplace (1749-1827), a French Mathematician known as a Newton of French. Laplace Transformations is a powerful technique, it replaces operations of calculus by operations of algebra. An Ordinary (or) Partial Differential Equation together with Initial conditions is reduced to a problem of solving an Algebraic Equation by this method.

USES  Particular Solution is obtained without first determining the general solution.  Non-Homogeneous Equations are solved without obtaining the complementary integral.  L.T is applicable not only to continuous functions but also to piecewise continuous functions, complicated periodic functions, step functions and impulse functions.

APPLICATIONS:  L.T is very useful in obtaining solution of linear differential equations, both ordinary and partial, solution of system of simultaneous differential equations, solution of integral equations, solution of linear difference equations and in the evaluation of definite integrals. DEFINITION: Let f (t) be a function of‘t’ defined for all positive values of t. Then Laplace transforms of f (t) is denoted by L {f (t)} is defined by  L f t    est f t  dt  f s   (1) 0 provided that the integral exists. Here the parameter‘s’ is a real (or) complex number. The relation (1) can also be written as f t   L1  f s  

In such a case the function f(t) is called the inverse Laplace transform of fs  .The symbol ‘L’ which transform f(t) into fs  is called the Laplace transform operator. The symbol ‘L-1’ which transforms (s) to f (t) can be called the inverse Laplace transform operator. ̅

MATHEMATICS - I

Conditions for Laplace Transforms Exponential order: A function f (t) is said to be of exponential order ‘a’ If lt est f t   a t finite quantity. Ex: (i). The function t2 is of exponential order

3 (ii). The function et is not of exponential order (which is not finite quantity) Piece – wise Continuous function: A function f (t) is said to be piece-wise continuous over the closed interval [a,b] if it is defined on that interval and is such that the interval can be divided into a finite number of sub intervals, in each of which f (t) is continuous and has both right and left hand limits at every end point of the subinterval. Sufficient conditions for the existence of the Laplace transform of a function: The function f (t) must satisfy the following conditions for the existence of the L.T. (i).The function f(t) must be piece-wise continuous (or sectionally continuous) in any limited interval 0 a  t  b . (ii).The function f (t) is of exponential order. Laplace Transforms of standard functions: 1 1. Prove that L1   s Proof: By definition

  est e  e0 L1 est .1 dt     0 1 if s  0     s 0 s0  s  s

Le101     s   1 2. Prove that Lt   2 s Proof: By definition

  st st st ee L t   e. tdt  t .   1. dt  ss 0  0

 st st ee 1 t. 2  2 s s  s 0

n n! 3. Prove that Lt   n1 where n is a +ve integer s

st  st  ee Proof: By definition L tn e st.... t n dt  t n  n t n 1 dt   0  ss0 0

MATHEMATICS - I

n  00   est t n 1 dt 0 s n  Ltn1  s n 1 Similarly L tnn12   L t  s n  2 L tnn23   L t  s By repeatedly applying this, we get n n1 n 2 2 1 L tn   ...... L t n n  s s s s s n! n ! 1 n ! nL1.   n  n1 s s s s Note: Ltn  can also be expressed in terms of Gamma function.

n n! n 1  i. e .,L t  nn11   n  1   n !  ss

 Def: If n>0 then Gamma function is defined by n  exn x1 dx 0

 We have L tn e st. t n dt   0 Putting x=st on R.H.S, we get

n x st  x 1 L tnx t e .. dx    0 n 1 ss dx dt s

1  When t0, x 0  e xn. x dx n1 0  s When t , x  

n 1 L t  n1 .1  n   s If' n 'is a +veinteger then n  1   n ! n n! Lt  n1 s Note: The following are some important properties of the Gamma function. 1. n 1   n .   n  if n  0 2. nn 1!   if n is a +ve integer

3. 1  1, 1     2 

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Note: Value of  n in terms of factorial   2   1   1   1! 3   2   2   2! 4  3  3  3!     and so on. In general nn 1!    provided ‘n’ is a +ve integer. Taking n=0, it defined 0! = 11   1 4. Prove that Leat   sa Proof: By definition,  L eat e st. e at dt es a  t dt   00

 es a  t   sa   0

ee 0 1   if s  a s a s  a s  a 1 Similarly L eat   if s   a sa a 5. Prove that Lsinh at   sa22 at at 1 Proof: ee at at Lsinh at   L L e  L  e  2 2

1 1 1  1 s a  s  a  2 a a      22s a s  a   s2  a 2  2 sa22 s 2  a 2   s 6. Prove that Lcosh at   sa22 eeat  at Proof: Lcosh at   L 2 

1at at 1 1 1 L e   L  e    22s a s a

12s a  s  a s s    2 s2 a 22 sa22 s 2 a 2   a 7. Prove that Lsin at   sa22

 Proof: By definition, Lsin at   est sin atdt 0

MATHEMATICS - I

 est  ssinat  a cos at  sa22 0 eax eax sin bxdx a sin bx b cos bx  22  ab a  22 sa s 8. Prove that Lcos at   sa22 1 Proof: We know that Leat   sa Replace ‘a’ by ‘ia’ we get 1 s ia Leiat   s ia s  ia  s  ia  s ia i. e .,L cosat i sin at  sa22 Equating the real and imaginary parts on both sides, we have sa Lcos at   and L sin at  s2 a 2 s 2 a 2 Solved Problems : 1. Find the Laplace transforms of (t22 1)

Sol: Here f(t) (t2  1) 2 tt 4  2 2  1

2 2 4 2 4 2 LLLLL{(t+ 1) } = {t + 2t + 1} = {t } + 2 {t } + {1} 4! 2! 1 4! 2! 1  2.   2.  s4 1 s 3 s s53 s s

24 4 1 1 24 53   5 (24  4ss  ) s s s s eat 1 2. Find the Laplace transform of L  a

eat 1 11at at Sol: L = L e11   L e   L  a aa 1 1 1 1     a s a s s() s a 3. Find the Laplace transform of Sin2tcost 11 Sol: W.K.T sin 2t cos t [2sin 2 t cos t ]  [sin 3 t  sin t ] 22

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11 L{sin 2 t cos t }  L [sin 3 t  sin t ]  L sin 3 t   L  sin t  22

1 3 1 2(s2  3)    2 2 2 2 2s 9 s  1 ( s  1)( s  9) 4. Find the Laplace transform of Cosh22t 1 Sol: w.k.t cosh2 2tt 1 cosh 4  2 1 Lcosh 2 2t  L )1(  L{cosh t}4  2 11s s2 8   2 2 2ss 16 ss( 16) 5. Find the Laplace transform of Cos33t

Sol: Since cos9t=cos3(3t)

1 cos9t=4cos3 3t-3cos3t (or) cos3 3t= cos9tt 3cos3  4 13 L{cos3 3 t }=+ L {cos9 t } L {cos3 t } 44 13ss = . . 4ss22 81 4 9

∴ 2 s 1 3 ss 63  =  4ss22 81 9 ss229  81 

6. Find the Laplace transforms of sintt cos 2

2 Sol: Since sint cos t  sin22 t  cos t  2sin t cos t 1 sin 2t   L{(sin t+ cos t )2 } = L {1 + sin 2 t }

=+L{1} L {sin 2 t }

1 2ss2  2 4    ss2  4 ss2  4  7. Find the Laplace transforms of cost cos2t cos3t 1 Sol: cost cos 2 t cos3 t .cos t 2.cos 2 t .cos3 t  2 11 =cost [cos5 t + cos t ] = [cos t cos5 t + cos2 t ] 22 11 2cost cos5 t  2cos2 t  cos6 t  cos 4 t    1  cos 2 t  44

MATHEMATICS - I

1 = 1[  cos2t  cos4t  cos t]6 4 1 L{cos t cos 2 t cos3 t }  L {1  cos 2 t  cos 4 t  cos6 t } 4 1 = [L }1{  L{cos t}2  L{cos t}4  L{cos 6t}] 4

1 1 s s s  =    4 s s2  4 s2 16 s2  36 8. Find L.T. of Sin2t

2 1 cos 2t Sol: L{sin t }  L  2 1 1 1 s [L{1} Lt {cos 2 }]   2 2 2ss 4 9. Find L

1 (√) 1 1  2 2 Sol: L t L t 1 where n is not an integer    1 s 2 11  22  33n 1.   n   n 

ss222 10. Find , where  a constant is

Sol: 푳 { + 휶} {푖 + 훼} = {푖훼 + 푖훼}

= 훼 {푖} + 푖훼 { } = 훼 + + 푖훼 + Properties of Laplace transform: Linearity Property: Theorem1: The Laplace transform operator is a Linear operator. ie. . ( iLcft ).    c .L ft   ( iiLft ).    gt     Lft    Lgt  Where ‘c’ is constant Proof: (i) By definition

 L cf t    est cf t  dt  c e st f  t  dt  cL f  t   00 (ii) By definition

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 L f t  g  t    est  f t   g  t   dt 0  est f t  dt  e st g  t  dt  L f  t    L g t   00 Similarly the inverse transforms of the sum of two or more functions of ‘s’ is the sum of the inverse transforms of the separate functions. Thus, Lfsgs1      Lfs  1    Lgs  1     ftgt    

Corollary: Lcft1  cgt 2     cLft 1    cLgt 2   , where c1, c2 are constants Theorem2: If a, b, c be any constants and f, g, h any functions of t, then Laft{ ()+ bgt () - cht ()} = aLft .{()} + bLgt .{()} - cLht {()} Proof: By the definition

 L{() af t bg () t  ch ()} t  est {() af t  bg () t  ch ()} t dt  0

   ae. st ftdt  be  st gtdt  ce  st htdt         0 0 0 a.{()} L f t  bL {()} g t  cL {()} h t

Change of Scale Property:

- 1  s If L{ f ( t )}= f ( s ) then L{ f ( at )} . f  aa Proof: By the definition we have

 L{ f ( at )}  est f ( at ) dt 0 du Put at u  dt  a when t  then u  and t = 0 then u = 0

 su  s  du 11.u L{ f ( at )} ea f ( u ) ..ea f u du f s     a  0 a aa0 Solved Problems : 1. Find

Sol: 푳{ } {sinh } = − =(Change of scale property) DEPARTMENT ∴ {sinh OF HUMANITIES} = ⁄ & SCIENCES ©MRCET (EAMCET CODE: MLRD) 206

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1 1 3 2 2 39s 1 s  3  2. Find

Sol: 푳{ } {cos } = + = (Change of scale property) {cos } = ( ⁄) ⁄ + First shifting {property:cos } = ( ⁄) + =

If Lft{ ( )}= fs ( ) then L {eat ft ( )} = fsa ( - ) Proof: By the definition

 L{eat f ( t )} e st e at f ( t ) dt  0

  es a  t f t dt    0

 eut f t dt whereu  s  a    0

= fu()f s a 

Note: Using the above property, we have L{e- at f ( t )}=+ f (s a) Applications of this property, we obtain the following results nn!! 1. L{eat t n } L (t n ) nn11 ()s a s b b 2. L{eat sin bt } L (sinbt) 2 2 2 2 ()s a  b s  b s a s 3. L{eat cos bt } L (cosbt) 2 2 2 2 ()s a  b s  b bb 4. L{eat sinh bt } L (sinhbt) 2 2 2 2 ()s a  b s  b s a s 5. L{eat cosh bt } L (coshbt) 2 2 2 2 ()s a  b s  b

Solved Problems : 1. Find the Laplace Transforms of te33 t 3! Sol: Since L{t3 }= s4

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Now applying first shifting theorem, we get 3! Le{t33- t }= (s + 3)4

2. Find the L.T. of ett cos 2 Sol: Since L Now applying{cos first} = shifting + theorem, we get

− + + { cos } = = 3. Find L.T of + + + + Sol: - L [ ] = L [ + = ] = {[ +] + [ ]} = −+ {[]} →− − = −+ − + − − −+ Second translation (or) second Shifting theorem:

If −,> − Proof:{ By} = the definition = { < ℎ {} =

∞ − − ∞ − {} = ∫ = ∫ + ∫ ∞ − ∞ − ∞ − Let t-a = u so that= dt∫ = du And. also + ∫ u = 0 when −t = a and= u∫ →  when −t → 

∞ −+ − ∞ − − ∞ − ∴ { } = ∫ = ∫ = ∫ − − Another =Form of{ second} = shifting theorem: If − where{ H (t)} = = and > H(t) ℎ is called{ −Heaviside − unit} step= function. , > Proof: By the { definition , <

∞ − Put t-a=u{ so that −− dt= du and also} = ∫when t=0, u=- −a when −t →  , u→ → 

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0  es()() u  a FuHudu    e  s u  a FuHudu     a 0

∞ −+ −+ = ∫ − . + ∫ [Since By the definition. of H (t)]

∞ −+ − ∞ −  = sa st =  e∫  e F t  dt by property of∫ Definite Integrals 0

− − Note: =H t a { isalsodenoted} = byu  t a  Solved Problems

cos t if t  33  1. Find the L.T. of g (t) when gt     0 if t    3 Sol. Let f t   cos t

∴ {} = {} = + =

휋 휋 휋 ( − ⁄) = ( − ⁄), 푖 > ⁄ Now= { applying second shifting theorem, then휋 we get , 푖 < ⁄

−휋 −휋 . 2. Find{ the} = L.T. of (ii)+ = + ) (ii) Sol: (i). Comparing the given function with f(t-a)− u(t-a), we have a=2 and f(t)=t3 − − −

! Now∴ { applying} = second{ } = shifting = theorem,= then we get

− − {(ii). − − } = = − −− − − −− { − } = { . − } = { − } − Now applying =second thenshifting theorem = + then, we get

−+ − − − { − } = . + = +

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Multiplication by‘t’: Theorem: If − 푳{} =  푳{} = Proof: By the definition f s   est f  t  dt 0

dd f s    est f t  dt ds ds 0 By Leibnitz’s rule for differentiating under the integral sign, d   f s    est f t  dt ds0  s

  test f t  dt 0

∞ − Thus = − ∫ {} = − {} − {} = 푛 푛 Note: Leibnitz’s ∴Rule { } = − =  If f x,,  and f x  be continuous functions of x and  then 

Where훼 {∫ a, ,b are 훼 constants} = ∫ 훼 independent, 훼 of  Solved Problems: 1. Find L.T of tcosat Sol: Since { } = + { } == − [ + ] − + −. − 2. Find t2sin at + = + Sol: Since {푖 } = + { . 푖 } = − +

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22 d2 as 23a s a  = ds 2 223 2 2 s a   s a  3. Find LT. of tet sin3 t

Sol: Since {푖 } = + Now using the shifting property, we get − ∴ {푖 } = [ + ] = + − + + 4. Find{ 푖 } = + + = ++ Sol: Since 푳{ } 2t{푖 } = +33 L etsin 3  2  2 s 29  ss4 13

−− { 푖 } = −3 2ss[ 4  −+] 6=− 2  [ −+ ] =22 s224 s  13   s  4 s  13 

2 5. Find the L.T. of 1 tet  Sol: Since

− 2 − − LLLL +1 tet =  1+   2 te+  t  t22 e  t       2 1dd 1 2  1   2  1      1  2   s ds s12  ds  s 

1 2d  1     22 sss12  ds   1 2 2    s ss12 23  6. Find the L.T of t3e-3t (already we have solved by another method)

Sol: − − { } = − { } −!− = = − + = + ! 7. Find +

MATHEMATICS - I

= − 1 [{ aasin } + { sin }]  2222 2 s a   a s  a   a

f t t 12 , t  1 8. Find the L.T of the function     0 0 t  1 Sol: By the definition

∞ − − ∞ − 1  {} =∫ est01 dt  e = st ∫ t  2 dt + ∫ 01

− ∞ − ∞ − ∞ = ∫ 2  − = [ − − ] − ∫ − − 01 est  t   dt 1 s

 st st 2 ee t 1   dt s s1 s 1

st  2 1 2e  2  0 est dt   e st 1 23  1 s s s s1 s

22ss 330 ee   ss 9. Find the L.T of f (t) defined as ft( ) 3 , t >2  0 , 0

∞ − 2  { }  = ∫est f t  dt e st f  t  dt 02

− ∞ −  33   =0∫  est.3 dt  + ∫  e  st   0  e 2 s  2 ss2 3  e2s s 10. Find

Sol: 푳{ + } { + } = {cos cos − sin sin } = cos . {cos } − sin { sin } = cos . + − sin . + DEPARTMENT OF HUMANITIES & SCIENCES ©MRCET (EAMCET CODE: MLRD) 212

MATHEMATICS - I

− + + {. + } = [cos. 22− sin . ] s22 a.0 a .2 s  s a.1 s .2 s     = cosbb .22 sin 2 2   2 2  s a    s a  

1 2 s22  acos b  2 as sin b 2   sa22   11. Find L.T of L [t Sol: - We know that ] L[sint] = L[tsint] = (-1) + L[sint] = - ( = - − = + + By First Shifting Theorem + L [t = = − − Division by‘t’: 푖] = [ + ]→− − + −+

Theorem: If ∞  Proof: 푳 We{ have} = f s   e 푳st { f  t  dt} = ∫ 0 Now integrating both sides w.r.t s from s to , we have

   st fsds()()  e ftdtds    00s 

  f t  est dsdt (Change the order of integration) 0 s

  f t   est ds dt ( t is independent of‘s’) 0 s 

st   e  f t  dt 0  t s

= ∞ − Solved Problems:∫ { } 1. Find Sol: Since 푳 { } Division{푖 by‘t’,} = we have+ =

MATHEMATICS - I

− ∞ − −  Tan11 s  cot s = [2 ] = ∞ − sin at 2. Find the L.T of t Sol: Since Division{sin by t, we} = have + =

ia ∞ ∞  1 1 +  1  1 { } =a∫.  Tan ss = ∫  Tan   Tan a aas

 Tan11ss  cot 2 aa 

3. Evaluate − 퐚 Sol: Since 푳 { } { − cos a } = {} − { } = − + −c a ∞ +  { } = ∫ −1 22 logs  log s  a  2 s

 11 s2 22 2logs  log s  a    log22 22s sa s

  2 1 1 1 s l og2  log1  log 22 22a sa 1 2  s s

1 1 s2   s 2  2 s 2 a 2  l og2 2   log  2 2   log 2 2 s a   s a  s

s2 1 Note: = log (Putting a=1 in the above problem) − s 푳 { } 4. Find − − − Sol: 푳 { } = − − − ∞ + +  { } ∫ −  sa   logs  a  log s  b  log     s  sb s

MATHEMATICS - I

a 1 s sa lt  log log s  b 1 sb s sb log1  log(s  a )  log( s  b )  log  sa

1 cost 5. Find L  t 2

1 costt   1 1 cos  Sol: LL  .  ..... 1  t2   t t 

 1 costs   1   1  Now L  ds log s  log s2  1 s 2   t   s s 12   s

 1s2  1  s 2  1 s 2 1 log  log  log 2s2 1  2  s 2 1  2 s 2  s   2 1 costs 1 1 L log ds 22s ts2  22 1ss 1  2 log .s . sds 2s 2 3 21s s  s s

2 1 1 s  1  ds lt s.log 1 2  s log 2  2 2 21s s ss s    2 1 1 1 1s  1 1 lt s2  4  6 ....  s log 2  2 Tan s 2s s 2 s 3 s s s

2 3 4 11   1  x x x 0s log 1 2  2  Tan s   log 1  x  x ..... 2s   2   2 3 4

1 11 cotss  log 1  2 2 s  6.Find L.T of − − − Sol: W.K.T L [ , L [ − − L = ] = + ] = + ∞ [ ] ∫ ̅

− − ∞ − ∴ L [ ] = ∫ − = + + ∞ = [loglog + − log + ] + ∞ + DEPARTMENT OF HUMANITIES & SCIENCES ©MRCET (EAMCET CODE: MLRD) 215

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= log + ∞ =log (1)-log+ ( ) + =0- log ( ) =+ log ( ) + + Laplace transforms of Derivatives:+ +

1 If ft  be continuous and {} = ℎ { } = − Proof: By the definition

∞ −   { } =∫est f t     s  e st f  t  dt 0 0 (Integrating by parts)

  est f t  s e st f  t  dt 0 0

st lt f ( t ) f (0) = t e Since f (t) is exponential order+. {}

 ltst f ( t ) t e =0

∴ { } = − + {} The Laplace Transform of the second derivative f11(t) is similarly obtained. = −

1 ∴ { } =s .. s{ f s  } f−  0   f  0 

21 s f s   sf 00   f  

∴ { } = . { } − = [ {} − − ] − Proceeding similarly, we have = {} − − −

− − − { } = {} − − … … Note 1: Note 2: Now − { } = 푖 = = , = … = We have| | . . − − | | ==M. || . if s>a −− DEPARTMENT OF HUMANITIES & SCIENCES→ ©MRCET → ∞ (EAMCET CODE: MLRD) 216

MATHEMATICS - I

− Solved Problems:∴ →∞ = > Using the theorem on transforms of derivatives, find the Laplace Transform of the following functions. (i). eat (ii). cosat (iii). t sin at (i). Let f t  eat Then f1  t   a. e at and f  0   1

{ } = . { } − 푖. . , { } = . { } − 푖. . , { } − . { } = − 푖. . , − { } = − (ii). ∴ { } = − = ℎ = −푖 = − 1 ∴ Now{ f0} = cos0 {  1 and} − f .  − 0    a sin0  0

ℎ {− cos } = {cos } − . − ⟹ − {cos } − {cos } = − (iii). Let f t ⟹ tsin − at then+ f1  t{ cos  sin at} =  at− cos ⇒ at { cos } = + f11  t   acos at  a cos at  at sin at   2 a cos at  a2 t sin at Also f = and = { } = { } − − 푖. . , { cos − sin } = {tsin } − − 푖. . , {cos } − { sin } − {tsin } = − Laplace푖. . , − Transform + {t ofsin Integrals:} = + ⇒ {t sin } = +

If t Proof: Let g{ t } = f  x  dxℎ {∫ } = 0

Then Taking Laplace = Transform[∫ ] on =both sides =

MATHEMATICS - I

But { } = { } − = {} − [푖 = ] ∴ { } = {} ⟹ {} = {} ⇒ { } = {} = ∫ ∴ {∫ } = Solved Problems:

t 1. Find the L.T of sin atdt 0 Sol: L Using{sin the} =theorem + =of Laplace transform of the integral, we have

{∫ } = ∴ {∫ sin } = tsin +t 2. Find the L.T of dt 0 t

Sol: i { sin } = + → = 푖 i ∞ ∞  ∴ { }  =∫Tan {1 ssin } Tan =  1 ∫  Tan+  1 s   Tan  1 s  cot  1 s ( or ) Tan  1 1 s 2 s 

i − − 푖. . , { } = ( ⁄) i − − 3. Find∴ { ∫L.T of } = ( ⁄) − Sol: L [ ]∫ − 푖 We know∫ that L {sint} = L { = + = ̅ = 푖 ∞ ∞ } =( ∫ ̅ ∫ + − ∞ = − − 휋 − − ∞ − = − = 푖 − ∴ L { } =

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Hence L { 푖 − By First Shifting∫ Theorem} = L [ ] = − − 푖 ∫ ̅ + = →+ − 푖 − ∴ L [ ∫ ] = + + Laplace transform of Periodic functions: If f (t) is a periodic function with period ‘a’. i.e, f t a  f  t  then

1 a L f t   est f t  dt   sa 0 1 e Eg: sin x is a periodic function with period 2 i. e ., sin x sin 2  x   sin  4  x  ...... Solved Problems:

1. A function f (t) is periodic in (0,2b) and is defined as ft  1 if 0  t  b  1 if b  t  2 b Find its Laplace Transform.

1 2b Sol: L f t   est f t  dt   2bs 0 1 e 1 bb2 est f t  dt e st f  t  dt 2bs 0 b 1 e  1 bb2 est dt e st dt 2bs 0 b 1 e 

bb2 1 eest   st  2bs     1e  s    s  0 b 1  esb 1  e 2 bs  e  sb 2bs     se1   1 L f t   12  esb  e 2 bs se1 2bs    2. Find the L.T of the function f t  sin t if 0  t   22   0if  t  where f t  has period   

MATHEMATICS - I

2 Sol: Since f (t) is a periodic function with period 

1 a L f t   est f t  dt   sa 0 1 e 1 2 L f t    est f t  dt   s2 0 1 e 

2 1 st st esin t dt e .0 dt 2s 0  1 e  

 1 est  ssin t  cos  t    2s 22  s  1 e 0

at b e eat sin bt a sin bt b cos bt  a 22 ab

11s e  . 2 s 22  s    1 e  Laplace Transform of Some special functions: 1. The Unit step function or Heaviside’s Unit functions:

It is defined as < Laplace Transform of unit − step = function: { > To prove that − Proof: Unit step{ function − } is= defined as < Then − = { > ∞ − a  { − } =∫eutst  adt −   eut st   adt  0 a

a  est.0 dt e st .1 dt 0 a

st  as  ee1 est dt   . e   e as  a   sa s s

− Laplace Transforms∴ { − of } Dirac= Delta Function:

The Dirac delta function or Unit impulse function ∈ = { ⁄∈ ∈ 2. Prove that hence show that −∈ >∈ − 푳{∈} = ∈ 푳{휹} =

MATHEMATICS - I

Proof: By the definition

∈ ⁄ ∈ And Hence = { ∈ >∈ ∞ − {∈} = ∫ ∈ ∈ − ∞ − = ∫ ∈ + ∫∈ ∈ ∈ − ∞ − ∈ = ∫ ∈ + ∫ . − ∈ −∈ −∈ − = ∈ [ − ] = − ⁄∈ [ − ] = ∈ −∈ − ∈ Now ∴ { } = ∈ −∈ − {훿} =∈→ { L-Hospital∈} = ∈→ rule.∈

∴Properties {훿} = of Dirac 푖 Delta Function: 1. ∞ 2. ∫ 훿 = where G(t) is some continuous function. ∞ 3. ∫ 훿 = where G(t) is some continuous function. ∞  훿11 − = 4. ∫G()() t  t a    G a 0 Solved Problems: 1. Prove that Sol: By Translation theorem − 푳{휹 − } = L − {훿 − } = {훿} − 2. Evaluate = [sin {훿 } = ] ∞ Sol: By using property (3) then we흅 get ∫ 휹( − ⁄)

∞ ∫Here 훿 − = G a = 휋 G⁄ , cos2= cos  1    3  3 2

∞  휋 휋 −휋 ∴ cos 41훿(t − ) = cos = 3. Evaluate∫ e  t 2⁄  dt ⁄ ⁄ 0 Sol: By the 4th Property then we get

MATHEMATICS - I

  11t a  G  t  dt   G() a 0 G t   e4t and a 2 G14 t  4. e t G1 a   G 1 2    4. e 8

 e4t 1 t 2  dt   G 1  a   4. e 8 0

Inverse Laplace Transforms: If fs  is the Laplace transforms of a function of f (t) i.e. then f (t) is called the inverse Laplace transform of fs  and is written as {} = − is called the inverse L.T operator. = {} − ∴ Table of Laplace Transforms and Inverse Laplace Transforms

S.No. − 1. {} = {} = − 2. {} = ⁄ { ⁄} = − ⁄ 3. { } = { − } = − − − ⁄ = 4. { } = { + } + ! − 5. { } = + 푖 + 푖 { +} = ! − − − ! − 6. { } = { ⁄ } = , = ,, … − ! − 7. {sin } = { } = . sin + + − 8. {cos } = { } = cos + + − 9. {sinh } = { } = sinh − − − {cosh } = { } = cosh 10. − − − 11. { sin } = { } = . sin − + − + − − − { cos } = { } = cos − + − + DEPARTMENT OF HUMANITIES & SCIENCES ©MRCET (EAMCET CODE: MLRD) 222

MATHEMATICS - I

12.

− 13. { sinh } = { } = . sinh − − − − − − − 14. { cosh } = { } = cosh − − − − − − − 15. { sin } = { } = . sin + + + + − + − + − 16. { cos } = { } = cos + + + + − − 17. { } = − { − } = {} − − − − − { } = + { + } = {}

Solved Problems :

ss2 34 1. Find the Inverse Laplace Transform of s3

Sol: − −+ − { } = 1{1⁄ − .  1⁄ 1+ ⁄  1} 4  LLL  3 23 s ss    t 2 1  3t  4.  1  3 t  2 t 2 2!

2. Find the Inverse Laplace Transform of + Sol: −+

− + − + − −+ { −+} = {− +} = {− + }

− − − = {− +4 } +. {− + } e22ttcos3 t e sin 3 t 3 25s  3. Find the Inverse Laplace Transform of s2  4

Sol: − − − { −} = { − − −}

− − − 1 − =2.cosh { 2tt} 5. − sinh 2{ } 2

4. Find − + 푳 {+}

MATHEMATICS - I

Sol: − ++ − { + } = {+ + } − − − 5. Find = {+ } + {} = + − − 푳 { +} Sol: − − − − 푳 { +} = { +} − { +}

− − =3 { 5 +( 8⁄ 2 ) } − 5 { +( ⁄ ) } .costt . sin 4 2 4 5 2 3 5 4 5 costt sin 4 2 5 2 s 6. Find the Inverse Laplace Transform of sa 2

Sol:

− − +− − − − {+ } = {+ } = { }

− − = { − }

− − − =e at 1[ at {} − . { }]

7. Find − + 푳 { −−} 37s A B Sol: Let  s2 2 s  3 s  1 s  3

− + + = +

= , = ⇒ = 3s  7 1 4  = −, − =  ⇒ = − s2 2 s  3 s  1 s  3

− + − − − − { }  =  eett { 4. 3 + } = − { } + { } − − + − + − 8. Find − Sol: 푳 {+ ( +)} + + + = + + + + + DEPARTMENT OF HUMANITIES & SCIENCES ©MRCET (EAMCET CODE: MLRD) 224

MATHEMATICS - I

Equating Co-efficient of s3, + + + + + A+C=0……..(1) + + = Equating Co-efficient of s2, A+B+2C+D=0…….(2) Equating Co-efficient of s, A+C+2D=1…….(3) 1 put s 1,2 B   1  B   2 1 Substituting (1) in (3) 21DD   2 Substituting the values of B and D in (2) i.e. − + + = 11 ⇒ + = , + = ⇒ = , = s  22  s1 22 s2  1   s  1  s2 1

− − − { + + } = [ { +} − {+ }]

− − 1 =[sinsin −t tet { }] 2 

9. Find − 2 Sol: Since s4푳4 a{ 4 + s 2}  2 a 2   2 as 2

s As  B Cs D Let = + + − + s44 a 4 s 2  2 as  2 a 2 s 2  2 as  2 a 2

11 Solving + we get− ACBD0,+  0, +  + ,  + + = 44aa

− − − { + } = { ++ } + { −+ } 11  1 1 1 1 a.L.22 .L 22 4 (s a ) a 4 a ()s a a 1 1 1 1 . .eat sin at . e at sin at 44a a a a

1 at at 11 2 sin at e e  22.sinat .2sinh at sin at sinh at 4a 42aa

MATHEMATICS - I

10. Find i. ii. − −+ − ( −) Sol: 푳 { } 푳 { }

i. − −+ − − { } = { − + } = { − + } − − − = {} − { } + { } ii. = − + ! = − + − ( −) − ( −) − − + { } = { } = { }

− − − − = { 2− + 4} + { { 4} − { } + { }} 3 t 4 t  3  2 t  1 4 2 1  4     1  2t    t  6 t  6 2 2! 4!  2  6  4

11. Find − Sol: 푳 [ − ]

− − − − [ ] = [ ] = [ ] = [ + ] − − − + − + 1 eat  e at  cosh at 2 

1 4 12. Find L  (s  1)(s 2)

1 4 11  1  1 1 tt 2 Sol: L 4LL  4   4[e  e ] (s  1)(s 2) (s 1)(s  2) s  1 s  2

1 1 13. Find L 22 (ss 1) ( 4) 1 A B Cs D Sol: 2 2  2  2 (1)(s s  4) s  1(1) s  s  4 2 1 2 3 ABCD,,,    25 5 25 25

1 1 2 1 1  1  1  1  2  1 s  3  1  1  LLLL2 2  L    2    2    2  (s 1) ( s  4) 25  s  1  5  (s  1)  25  s  4  25  s  4 

2 t 111  1 t   1  2 3 1  e L  e L   cos 2 t  . sin 2 t 25 ss 5 2  25 25 2

MATHEMATICS - I

2 1 2 3 ett  e. t  cos 2 t  sin 2 t 25 5 25 50

2 1 ss2 14. Find L  ss( 3)(s 2) s2  s2 A B C Sol:    s s3  s  2  s s  3 s  2 Comparing with s2, s, constants, we get

ABC1,,  4  2 3 15 5

2 1 ss2 1 1 4 2 L L    ss( 3)(s 2) 3s 15( s 3) 5( s 2)

11  1 4   1  2  LLL      3s 15( s 3)   5( s 2)  1 4 2  ee32tt  3 15 5

2 1 ss24 15. Find L 2 (s  9)(s 5) ss2 24 A Bs C Sol: 2 =  2 (s  9)(s 5) ss59 Comparing with s2, s, constants, we get

ABC31,,  3  83 34 34 34

2 2 1 ss24 1 ss24 L 2  L 2 (s  9)(s 5) (s  9)(s 5)  31   3   83  LLL1   1   1 34(s 5)   34(ss22  9)   34(  9)        31 83 et5t 1 3cos3t  sin 3 3434  3  First Shifting Theorem: If Lfs11( )   ft ( ), thenL fsa (  )   eftat ( ) Proof: We have seen that L eat f()() t   f s a L11 fsa()()()    eftat  eL at  fs 

MATHEMATICS - I

Solved Problems : 1 1. Find 11 L2  L f( s 2)  (s  2) 16

1 1 21t 1 Sol: L 2  eL2 (s  2) 16 s 16 1et2t sin 4 et2t . sin 4 44

1 32s  2. Find L  ss2 4 20

13s 2  1 3 s  2   1  3( s  2)  4  Sol: LLL2  2   2 2  s4 s  20 ( s  2)  16   ( s  2)  4 

11s  21    34LL2 2   2 2  (ss 2)  4   (  2)  4 

2tt 1s  2 1 1  34e L2 2  e L  2 2  ss44    1 3e22tt cos 4 t 4 e sin 4 t 4

1 s  3 3. Find L  ss2 10 29

1s3  1 s  3   1  s  5  8  Sol: LLL2  2 2   2 2  s10 s  29 ( s  5)  2   ( s  5)  2 

5tt 1s  81  5   eL 22   e cos 2 t  8. sin 2 t  s  22    Second shifting theorem: f t  a  if t  a  If L11 fs( ) ftthenLe ( ),  { as fs ( )} G (t), where G(t )     0 if t a  f t  a  if t  a  Proof: We have seen that G(t )   0 if t a then L G t    eas .() f s

L1 eas f() s G t   

MATHEMATICS - I

Solved Problems :

 s 3s 1 1 e 1 e 1. Evaluate (i) L 2 (ii) L 2 s 1 (s  4)

 s  s 1 1 e 1 1 1 e Sol: (i) L 2 = L 2 + L 2 s 1 s 1 s 1

1 1 Since Lsin t f ( t ) , say s2 1

 s 1 e sin(t ) , if t By second Shifting theorem, we have L 2   s 1 0,if t   ∴  s 1 e or L =sin(t-π)H(t-π)= -sint. H(t-π) s2 1 

 s 1 1 e Hence L 2 =sint-sint. H (t-π) =sint [1- H (t-π)] s 1 Where H (t-π) is the Heaviside unit step function

111 4t 1  (ii) Since L22  e L   (ss 4) 

= e4t .() t f t , say

3s 4(t 3) 1 e et.( 3) ,3if t  By second Shifting theorem, we have L 2    (s  4) 0 ,3if t  ∴ 3s 1 e or L = et4(t 3).( 3) H(t-3) (s  4)2  Where H (t-3) is the Heaviside unit step function Change of scale property:

1 1 t If L f t    f s , Then L f as    f,0 a aa Proof: We have seen that L f t    f s 

1 t Then f as   L f,0 a  aa

1 1 t L f as    f,0 a  aa

MATHEMATICS - I

Solved Problems :

1 s 1 1 8s 1. If L 22= ttsin , find L 22 (s  1) 2 (4s  1)

1 s 1 Sol: We have L 22= ttsin , (s  1) 2 Writing as for s,

1 as 11tttt L 2 2 2 = . . sin = 2 .sin , by change of scale property. (as  1) 2 a a a 2aa Putting a=2, we get

1 2s tt 1 8s 1 t L 22= sin or L 22= sin (4s  1) 82 (4s  1) 22 Inverse Laplace Transform of derivatives:

n n n d Theorem: L1 f()() s f t , then L1 f( s ) ( 1)nn t f ( t ) where f()() s  f s     dsn 

n nnd Proof: We have seen that L t f( t )   ( 1)n f (s) ds

n L1  f( s )   (  1)nn t f ( t )

Solved Problems :

1 s 1 1. Find L log s 1

1 s 1 Sol: Let Llog f ( t ) s 1 s 1 Ltf ( )   log s 1 d s 1 L tf() t   log  ds  s 1 1 1 L tf( t)    s 11s

1 1 1 tf( t)  L  s 1 s 1

1  1  1 1 tf(. t) 1L  L  s 1 s 1 = − +

MATHEMATICS - I

2sinh t t f t 2sinh t  f  t   t

1 st1 2sinh L log  st1

t 1 1 s 1 e Note: L log = s t 2. Find L1cot 1 ( s )  Sol: Let L1cot  1 ( s )   f ( t ) L f( t )   cot1 ( s )

d 11 L ttfs( ) [cot1 ( )      22 ds 11 s s

1 1 tf() t L 2  sin t s 1 sin t ft   t 1 L11cot ( s )  sin t t

Inverse Laplace Transform of integrals:

 1 1 ft() Theorem: L f()() s   f t , then L f() s ds  s t

ft()  Proof: we have seen that L  f() s ds t s

 1 ft() L f() s ds s t Solved Problems :

1 s 1 1. Find L 22 (ss 2 2) s 1 Sol: Let fs() (ss22 2 2)

 s 1 Then L1 fs()= L1 ds    22 s (s  2s 2)

MATHEMATICS - I

1 s 1 = L 22 [(s  1) 1]

t 1 s = eL22, by First Shifting Theorem (s  1)

tttt 1 st esin t e sin t L2 2 2  sin at 22 (s a ) 2 a

Multiplication by power of’s’:

Theorem: L1  f()() s   f t , and f (0),then L11 s f()() s   f t Proof: we have seen that L f1( t )   s f ( s ) f (0) L f1()() t  s f s [f (0) 0] or L11 s f()() s   f t Note: L1  sn f( s )   f n ( t ),if f n (0)  0forn  1,2,3...... n  1 Solved Problems :

1 s 1 s 1. Find (i) L 2 (ii) L 2 (s  2) (s  3) 1 Sol: Let fs() Then (s  2)2

1 1 L11 f() s L = 21t = 2t ,    2 eL2 e.() t f t (s  2) s Clearly f (0) =0 s s 1 = 1 = 1 = 1 Thus L 2 L s. 2 L s.()fs  ft() (s  2) (s  2) d = ()te2t = t( 2ee22tt ) .1= e2t (1 2t) dt

1 s 3t Note: in the above problem put 2=3, then L 2 = e (1 3t) (s  3)

MATHEMATICS - I

Division by S:

t 1 1 fs  Theorem: If L f()() s f t , Then L  f u  du   s  0 t fs  Proof: We have seen that L f u  du  0 s

t 1 fs  L  f u  du s  0

tt 1 1 fs  Note: If L f()() s f t , then L  f u  du. du   s2  00 Solved Problems : 1 1. Find the inverse Laplace Transform of s2() s 2 a 2

1 1 1 Sol: Since L 22= sinat , we have ()sa a 1 t 1 L1 = sin atdt 22  s(sa ) 0 a

t 1 cosat 1 1 = = 2 (cosat 1) = 2 (1 cosat) aa0 a a 1 t 1 Then L1 = (1 cosat ) dtdt 2 2 2  2 s() s a 0 a

t 1 sin at 1 sin at = 2 t  = 2 t  aa0 aa

1 1 1 sin at L 2 2 2 = 2 t  s() s a aa

Convolution Definition: If f (t) and g (t) are two functions defined for t  0 then the convolution of f (t) and g (t) is

t defined as ft * gt   fugtudu     0 f t ** g  t  canalsobewrittenas  f g  t  Properties: The convolution operation * has the following properties

MATHEMATICS - I

1. Commutative i.e. f** g  t   g f  t 

2. Associative f**** g h    t    f g  h   t 

3. Distributive f* g h  t  f * g  t  f * h  t for t  0

Convolution Theorem: If ft() and g(t ) are functions defined for t  0 then LLLftgt *..     ft    gt   fsgs    i.e., The L.T of convolution of f(t) and g(t) is equal to the product of the L.T of f(t) and g(t)

 t Proof: WKT L  t   est fugtududt      00    t est f u  g  t u  du dt 00 The double integral is considered within the region enclosed by the line u=0 and u=t On changing the order of integration, we get

 L  t   efugtudtdust       0 u

 esu fu  es t u  gtudtdu  0 u  

  esu f u  e sv g  v  dv du put t u v 00    esu fugsdu     gs  esu fudu    gsfs .   00 

LLLftgt*.. ft gt fsgs               

Solved Problems :

1 s 1. Using the convolution theorem find L 2 2 2 ()sa

1 s 1 s 1 Sol: L 2 2 2 =L 2 2. 2 2 ()sa s a s a s 1 Let f s   and g s  s2 a 2 s 2 a 2 s So that LL11 f( s )  22  cos at  f ( t )  say sa

MATHEMATICS - I

s 1 11 L g( s )   L22  sin at  g ( t )  say s a a By convolution theorem, we have s t 1 L1 cos au . .sin a ( t u ) du 2 2 2  ()s a0 a

1 t sin(au at  au )  sin( au  at  au )  du 2a  0 1 t sinat  sin(2 au  at )  du 2a  0 t 11 sinat . u  .cos(2 au  at ) 22aa 0 1 1 1 tsin at  cos 2 at  at   cos  at  2a 2 a 2 a 1 1 1 tsin at  cos at  cos at 2a 2 a 2 a t  sin at 2a

2 1 s 2. Use convolution theorem to evaluate L 2 2 2 2 (s a )( s b )

2 11s s s Sol: LL2 2 2 2   2 2. 2 2  (s a )( s  b )  s  a s  b ss Let f s   and g s  s2 a 2 s 2 b 2 s So that 11 L f( s )   L22  cos at  f ( t )  say sa

11s Lgs( )   L22  cos btgt  ( )  say ()sb  By convolution theorem, we have

t 1 ss L .  cosau .cosb( t u ) du s2 a 2 s 2 b 2  0 1 t cos(au  bu  bt )  cos( au  bu  bt )  du 2  0

MATHEMATICS - I

t 1 sin(au bu  bt ) sin( au  bu  bt )  2 a b a b 0 1 sinat sin bt sin at  sin bt a sin at  b sin bt    2 a b a  b a22  b

1 1 3. Use convolution theorem to evaluate L 22 s(s  4)

1111  s  Sol: LL2 2   2. 2 2  s(s 4)   s ( s 4)  1 s Let f s  2 and g s  2 s s2  4 

1 So that 11 L g()() s   L2  t  g t  say s

1  1 s  t.sin 2 t1  s  ts in 2 t L f()() s   L2 2    f t  say  L 2 2 2   (s 4)  4  ( s a )  2 a

t 1 1 s u L . sin 2u (t u )d u s2 (s22 4)  4  0 tt1 t usin 2 udu u2 sin 2 udu 4 4 00 t u 1 t cos2uu  sin 2 4 240 t 11uu2  cos 2u  sin 2 u  cos 2 u 4 2 2 4 0 1 1 t sin 2 t  cos 2 t  16

1 1 4. Find L 2 (ss 2)( 1) 1 1 1 Sol: LL11 . 22 (s 2)( s  1) s  2 s  1 11 Let f s   and g s  2 ss21

1 11 2t So that L f()() s   L  e  f t  say s  2

MATHEMATICS - I

111 L g( s )   L2  sin t  g ( t )  say s 1 t 1 11 L . f( u ).g(t u )du (By Convolution theorem) ss212  0

t t e2u sin(t u)du (or) sinu.e2(tu ) du 0 0 t  e22tusin ue du 0 t e2u e2t  2sin u  cos u  212  0 1 e22tt e 2sin t  cos t 1  1   5   5 1 e2t 2sin t  cos t  5

1 1 5. Find L  (ss 1)( 2)

111 1 1 Sol: LL  .  (s 1)( s  2) s  1 s  2 11 Let f s   and g s  ss12

11 1 t So that L f()() s   L  e  f t  say s 1

1 11 2t L g()() s   L  e  g t  say s  2 By using convolution theorem, we have

t 11 u 2( t  u ) L e e du (s  1)(s 2)  0

tt3u t 2t 3 u 2 t  3 u 2 te 1 2 t t eedueedue    ee  33 000

1 1 6. Find L 2 2 2 s() s a

111 1 1 Sol: LL2 2 2   2. 2 2  s() s a s s a

MATHEMATICS - I

11 Let f s   and g s  s2 s 2 a 2 1 So that LL11 f()() s  2  t  f t  say s 11 LL11 g( s )  22  sinh at  g ( t )  say s a a By using convolution theorem, we have

t 1 11 L u. sinh a ( t u ) du s2() s 2 a 2  a  0 1 t usinh( at au ) du a  0 t 1 u sinat au  coshat  au   2 a a a 0 11t cosh(at at )  0  [0  sinh at ] 2 = a a a 11t sinh at 2 a a a 1  at  sinh at  a3 3. Using Convolution theorem, evaluate { − Sol: = 푳 + +} − − − {+= . +} = { + . + } {̅ . ̅ }------(1) − − ̅ =+ = {} ⇒ = {+} = ------(2) − By̅ Convolution + = { theorem} ⇒ we have = { + } =

− Where {̅ . ̅ } = ∗ ∗ = ∫ − − −− ∴ {+ . + } = = ∫ − = ∫ − = . + [ − 푖] − = [ − − 푖] − − DEPARTMENT OF HUMANITIES [ & SCIENCES −©MRCET 푖] −(EAMCET CODE: MLRD) 238

MATHEMATICS - I

Application of L.T to ordinary differential equations:

(Solutions of ordinary DE with constant coefficient): 1. Step1: Take the Laplace Transform on both the sides of the DE and then by using the formula

n n n1 n  1 1 n  2 2 n  1 L{f ( t )} s L {f(t)}  s f (0)  s f (0)  s f (0)  ...... f (0) and apply given initial conditions. This gives an algebraic equation. 2. Step2: replace f (0), f 1(0) , f 2 (0) ,……… f n1(0)with the given initial conditions.

Where f1 0   s f 0  – f  0 

f2(0) s 2 fs  – s f  0  f 1  0  , and so on 3. Step3: solve the algebraic equation to get derivatives in terms of s.

4. Step4: take the inverse Laplace transform on both sides this gives f as a function of t which gives the solution of the given DE

Solved Problems : 1. Solve y1112 y 11  y 1  2 y  0 using Laplace Transformation given that

yy(0)1 (0) 0 and y11(0) 6

Sol: Given that y1112 y 11  y 1  2 y  0 Taking the Laplace transform on both sides, we get LLLLy111( t )  2  y 11 ( t )    y 1   2 y   0 s3LL y() t   s 2 y (0)  sy 1 (0)  y 11 (0)2  s 2  y () t   sy (0)  y 1 (0)   sLL y( t )  y (0)  2 y ( t )   0   s3 2 s 2  s  2L  y ( t )   s 2 y (0)  sy 1 (0)  y 11 (0)  2 sy (0)  2 y 1 (0)  y (0)   0  0  6  2.0  2.0  0 s32 2 s  s  2L  y ( t )   6   66 Lyt()   s322 s  s  2 ( s  1)( s  1)( s  2) ABC    s1 s  1 s  2 A( s  1)( s  2)  B ( s  1)( s  2)  C ( s  1)( s  1)  6

2 2 2 A( s  32) s   B ( s  s  2)  C ( s  1)6 

MATHEMATICS - I

Comparing both sides s2,s,constants,we have ABCABABC   0,3   0,2  2   6

A B  C  0

2A 2B  C  6 ______36AB 30AB ______6AA 6   1 3ABBAB  0    3    3

ABCCAB 0   1 3 2

1 3 2 L yt()     s1 s  1 s  2

11   1  1   1  1  yt( )LLL   3.    2.   = et3 e t 2. e 2 t s1   s  1   s  2  Which is the required solution

2. Solve y113 y 1  2 y  4 t  e 3t using Laplace Transformation given that y0  1 and y1  0    1

Sol: Given that y113 y 1  2 y  4 t  e 3t Taking the Laplace transform on both sides, we get

11 1 3t L{LLLLy( t )  3  y ( t )   2 y ( t )   4  t    e  41 s21L{L{L{y( t )  sy(0) y (0) 3s  y ( t )  yt (0) 2  y ( )    ss2  3 41 (ss2  3 2)L{yt()    s  4 ss2  3 4s  12 s4  s 2  3 s 3  4ss 3  12 2 (ss2  3 2)L{y(t)  ss2 (3 )

s47 s 3  13 s 2  4 s 12  L{yt( )   s22( s 3)( s  3s 2)

s47 s 3  13 s 2  4 s  12 L{y()t  2 s( s 3)( s  1)(s  2) s437 s  13 s2  4 s  12 As B C D E      s2 ( s 3)( s 1)( s  2) s2 s 3 s  1 s  2

MATHEMATICS - I

(AsBs )( 1)( s 2)( s 3) Css (2 )(  1)( s 2) Dss ( 2 )(  2)( s 3) Ess ( 2 )(  1)( s 3)  s2 ( s 3)( s  1)( s  2) s47 s 3 13 s 2  4 s 12 ( As B )( s 3  6 s 2 11 s 6) 2 2 2 2 2 2 Css()( 3 s  2)  Dss ()(  5 s  6)  Ess .(  4 s  3) Comparing both sides s4,s3,we have ACDE   1...... (1) 6ABCDE   3  5  4   7...... (2) 1 put s1,2 D   1  D  2

put s2,  4 E  8  E   2 1 put s3,18 C  9  C  2 11 fromeq.(1) A  1    2 A 3 22 35 from eq.(2) B= -7+18+  8  3  1  2 22

1 3 2 1 1 2 y(t)  L 2    ssss2(s  3) 2(  1)  2 11 y t  3  2 t  e32t  e t  2. e t 22 d2 y dy dy 3. Using Laplace Transform Solve 2 2  3yt  sin , given that y 0 when t=0 dt dt dt d2 y dy Sol: Given equation is 2  3yt  sin . dt2 dt LLLLy11 t   2 y 1  t    3 y t     sin t  1 s21LLL y t sy0  y 0  2 s y t  y 0  3. y t                 2 s 1

2 1 s 23 s  L  y t    2 s 1

1  L yt      22 s1  s  2 s  3 

1 yt  L1  s1  s  3  s2  1  

Now consider

MATHEMATICS - I

1 A B Cs D    2 s1  s  3  s2  1  s1 s  3 s  1

A s3  s22  1  B s  1  s  1  Cs  D  s  1  s  3   1    

Comparing both sides s3,we have 1 put s1,8 A  1  A  8 1 put s 3,  40 B  1  B  40 11 ABCC  00     8 40 5  1  4  1 C    40 40 10 3 1 1 3ABCDD  2   0      8 40 5 15  1  8  8  1 D    40 40 5 1 1 1 1 s  1 8 40 10 5 yt  L    2 s1 s  3 s  1 

11 1  1  1  1  1  1 s  1  1  1  LLLL           8s 1  40  s  3  10  s22  1  5  s  1  1 1 1 1 y t   ett  e3 cos t  sin t 8 40 10 5 dx 4. Solve xx sin t, 0   2 dt dx Sol: Given equation is x sin t dt LLLx1  t   x t  sin t       s.LLx t    x 0   x  t    s22  s.LLx t    2  x t    s22  s 12 L  x  t     s22

MATHEMATICS - I

 1  2 xt  L   22s 1 ss1    111  2LL    s 1 ss1  22   (By using partial fractions) s   2 2 2 t 1 1 1   1   2e L  2 2  2 2 s1 s  s  

tt   1 2e 2 e  2 cos t  2 . sin t 1 1   1    5. Solve D22 n  x  asin  nt   given that x=Dx=0, when t=0 Sol: Given equation is D22 n  x  asin  nt   x11 t  n 2 x  t   asin  nt  

LLLx11 t    n 2  x t    asin nt cos  a cos nt sin 

sxtsx2LLLL   0   x 1  0   nxta 2      cos sin nta   sin  cos nt 

22 ns s  n L  x t    acos2 2  a sin . 2 2 s n s n ns  L x t    acos22 a sin s2 n 2   s 2 n 2 

(By using convolution theorem I –part, partial fraction in II-part)

t  1 1ad sin 1  1 nacos .sin nx . sin n t  x  dx  L  0 n n2 ds 2 2 sn 

aacost sin 1 cosnt  2 nx   cos nt dx  t sin nt 0   22nn aacos t cosn t  2 x   cos nt dx  sin t sin nt 0   22nn

t acos 1 at sin .sinn t  2 x   x cos nt  sin nt 2n 2 n0 2 n acos sin nt at sin  tcos nt  sin nt 2n 2 n 2 n acos sin nt at 2 cos cosnt  sin sin nt  22nn

MATHEMATICS - I

acos sin nt at 2 cos  nt  22nn 6. Solve using L.T given that y (0) = y1 (0) = 1. − Sol: Given equation − is + = − Applying L.T on both− sides we+ get = − 2 1 s L[y] –s y (0) – y (0)} – 4{s L[y] –− y (0)} + 3L{y}+ = = ⇒ (s { 2 + 4s +3) L{y} –s-1-4 = + ⇒ (s2 + 4s +3) L{y} = +s +5 + ⇒ (s2 + 4s +3) L{y} = + + s + 5 ⇒ L{y} = + + y = + ++ + ++ − − + Let us consider [+ ++] + [ ++] ] − − [+ ++] = [+ +

= + + + = + + + + + + + = − + + + + + = − − [+ + + + +] = − − = [+ + + + +] − − − − [+] + [+ ] + [+]

− − − − [ ] = − + + − −−→ + + + + − + − + − [ ++] == [+ −]+ [ + −] ] − − − − − + [ −] + [ − − + − From[ ++ (1) &] (2)= − − −→ + − − − − − ∴ = − + + +

MATHEMATICS - I

7. Solve using L.T. given x (0) =1, x ( ) = -1. 흅 Sol: Given + = L [ + = ] + [] = [] ⇒( [] − − + [} = + ⇒( + []=− − = + ⇒ = + [] + + + X=[] + + + + + + − − − = [[ + +] + [ +] + [ +] − = [ + − +] + + 푖 − − = +] − [ +] + + ------푖 Given x ( −) = -1. + + 푖 → 휋 + + 휋 휋 휋 휋 휋 ∴ − = -1= - − + 푖 ⇒ − + − = − + x = = From (1) ∴ + + 푖

Using L.T given y (0) =1, Sol: Given . − + − = = , = − − + − = [ ] − [ ] + [ ] − [] = [ ] ⇒ { [] − − − } − { [] − − } + {[] − } − [] = [ ] ⇒ − L[y]-+ − [] − − + + + − = − [ ] ⇒ − + = − = −

MATHEMATICS - I

L[y] = + ⇒ − − −

[] = + −

− − − − − − = = [− ] + [− ] − [− ] − − − = [− ] + [− ] − [− ] − − − = [ ] + − − [ ] − Consider ! − ! + [− ] − W.K.T [− ] = = − − [− ] = [ ] !

− [ ] = = + = + + + − = + + ∴ = ! − ! − + +