Lecture notes on Ordinary Differential Equations-(Unit-I) Khanday M. A.
Ordinary Differential Equations Introduction: Ordinary differential equations-ODE‟s serve as a tool in mathematical models for many exciting “real-world” problems. It has applications not only in science, but it also plays a key role in many areas in engineering, social sciences, biomedical engineering and other areas such as economics, political sciences etc. We believe that many problems of future technologies will also be described by means of differential equations. Physical problems have motivated the development of much of mathematics and this is especially true of differential equations. Due to differential equations mathematics and physical problems come closer and closer. An understanding of the required mathematics will aid in the solution of physical problems, and an understanding of the physical model often makes the mathematics easier. Additional applications of differential equations are (heat, radioactive decay, orthogonal trajectories, chemical reactions, economics, chemical diffusion, and vibrating strings).The increasing availability of technology (including graphing and programmable calculators, computer algebra systems, and powerful personal computers) has caused many to question the existing syllabuses in university courses in differential equations. However, we believe that the importance of applications will continue to motivate the study of differential equations.
In the first two chapters of this course, we deal with the introduction of ordinary differential equations. Singular solutions, p- discriminant and c- discriminant of the differential equations, Initial value problems of the first order ordinary differential equations (ODE), Integration in Series, Simultaneous and Total differential equations.
In many branches of science and engineering we come across equations, which contain an independent variable and the derivative of the dependent variable with respect to the independent variable is called a differential equation. Examples:- dy (i) log x dx
3 d 2 y dy (ii) 4 3x dx2 dx (iii) ∂z/∂x + ∂z/∂y = k, where k is a constant.
M.A/M.Sc. Mathematics 3rd Semester University of Kashmir, Srinagar 1 Lecture notes on Ordinary Differential Equations-(Unit-I) Khanday M. A.
Differential equations involving only one independent variable are called ordinary differential equations. Differential equations, which involve partial differential coefficients with respect to more than one independent variables, are called Partial differential equations. ORDER AND DEGREE OF A DIFFERENTIAL EQUATION The order of a differential equation is the order of the highest derivative involved in the differential Equation. The degree of the differential Equation is the power of the highest order derivative involved in the equation, when the equation has been made rational and integral as far as the derivatives are concerned. Examples:-In the given below differential equation
5 2 d 2 y dy 2 2 K 1 dx dx The order is two and the degree is also two. Linear and Non- Linear Differential Equations A differential equation is said to be a linear differential equation if it can be expressed in the form. d n y d n1 y dy P P . . . . P P y Q 0 dxn 1 dxn1 n1 dx n where P0 , P1 , P2 ,..., Pn1 , Pn and Q are either constants or functions of independent variable x. Thus, if a differential equation when expressed in the form of a polynomial involves the derivatives and dependent variable in the first power and there are no product of these, and also the coefficient of the various terms are either constants or the functions of the dependent variable, then it is said to be a linear differential equation, otherwise it is a non- linear differential equation. It follows from the above definition that a differential equation will be non- linear if, (i) Its degree is more than one. (ii) Any of the differential exponent has exponent more than one. (iii) Exponent of the dependent variable is more than one.
M.A/M.Sc. Mathematics 3rd Semester University of Kashmir, Srinagar 2 Lecture notes on Ordinary Differential Equations-(Unit-I) Khanday M. A.
(iv) Products containing dependent variable and its differential coefficients are present. Solution of a Differential Equation Any relation between the dependent and independent variables which, when substituted in the differential equation reduce it an identity is called a solution of the differential equation. A solution of the differential equation does not contain the derivative of the dependent variable with respect to the independent variable or variables. Complete Primitive or the General Solution The solution of an ordinary differential equation of nth order containing n arbitrary constants is called the complete primitive or the General solution. Particular Integral:-Any solution which obtained from the complete primitive by giving particular values to the arbitrary constants is called a Particular Integral. Exercises:Solve the following differential Equations. dy (i) P 2 Px Py xy 0 where P . dx d (ii) Dy z 0 and (D 1)y (D 1)z 0, where D . dx
Solutions:(i) We have
We can write (P + x) (P+y) = 0 Implies either (P + x) = 0 or (P+y) = 0 i.e., dy/dx = -x or dy/dx = - y or dy = - x dx or dy/y = - dx Integrating both sides we get 2 y = - x /2 + C1 or log y = - x + C2 2 Thus (y + x /2 + C1 )(log y + x –C2) is the required solution of the differential equation. (ii) Dy – z = 0 ------(i) (D-1) y – (D+ 1) z= 0 ---- (ii) Operating equation (i) by (D +1), we get on subtraction, (D2 + D –D + 1) y = 0 or (D2 + 1) y = 0 Now the auxiliary equation is m2 + 1 = 0
M.A/M.Sc. Mathematics 3rd Semester University of Kashmir, Srinagar 3 Lecture notes on Ordinary Differential Equations-(Unit-I) Khanday M. A.
Implies m = ± ί Therefore y = A Cos x + B Sin x is the complementary function. Also from (i) Z = Dy or Z = D( A Cos x + B Sin x) = -A Sin x + B Cos x, This is the required complete solution of the given differential Equation. Home Assignments Solve the following Differential Equations 1. ( D + 1) y = Z + ex and ( D + 1) Z = y + ex 2. P2 + 2 xp – 3x2 = 0, where p = dy/dx 3. x + yp2 = P ( 1 + xy) 4. p2 – x5 = 0. 5. y = p Sin p + Cos p. 6. p2 – 2p Cosh x + 1 = 0. 7. p3 – p ( x2 + xy + y2) = 0 8. t dx/dt + y = 0 and t dy/dx + x = 0. Hint: Take t = ep, so that dp/dt = 1/t and dx/dt = (dx/dp).(dp/dt) = 1/t. dx/dp
Double Point:A point on a curve is called a double point if two branches of the curve passes through it, It is called a triple point if three branches of the curve passes through it. In general a point is called a multiple point of the kth order if k branches of the curve pass through the point.
Classification of Double Points: Since two branches of the curve passes through a double point, there must be two tangents to the curve at the point, one to each of the two branches. If the two tangents are real and different, the double point is called a Node. If the two tangents are real and co-incident, it is called a Cusp. And if the tangents are imaginary so that there are no real points on the curve of the double point. Envelope:The envelope of the family is the locus of ultimate intersection of consecutive curves of the family. i.e., If f ( x, y, α ) = 0 is a family of curves corresponding to
M.A/M.Sc. Mathematics 3rd Semester University of Kashmir, Srinagar 4 Lecture notes on Ordinary Differential Equations-(Unit-I) Khanday M. A. different values of α. Then the ultimate intersection of curves f (x, y,α ) = 0 and f ( x, y, α + ∂α ) = 0 is called the envelope of the family. e.g. If x Cos α + y Sin α = p is the equation of the straight line, then x Cox (α + ∂α ) + y Sin (α + ∂α ) = p Solving these equations x = p Cos (α + ½ ∂α ) / Cos (½ ∂α) , y = pSin (α + ½ ∂α )/ Cos ( ½ ∂α ) When ∂α → 0, we have the ultimate point of intersection x = p Cos α , y = p Sin α Eliminating α between these equations, we have x2 + y2 = p2as the equation of the circle. Singular Solution:-If an infinite system of curves which all touch to a fixed curve, which we call their envelope represent a Complete solution of certain differential equation of the first order, then the envelope represents a solution of the differential equation. For every point of the envelope x, y and p have the same value for the envelope and the curves of the family that touches it there. Such a solution is called a Singular solution. 2 2 Discriminant:-The discriminant of the quadratic equation ax +bx+c = 0, a(≠0) is b – 4ac. If Φ (x, y, c) = 0 is the solution of the differential equation f(x, y, p) = 0 The p-discriminant is obtained by eliminating p between f(x, y, p) = 0 and ∂f/∂p = 0. and the c- discriminant is obtained by eliminating c between Φ (x, y, c) = 0 and ∂ Φ /∂c = 0. Extraneous Loci:-Let Φ (x, y, c) = 0 be the primitive of the differential equation f(x,y, p) = 0. If ψ (x, y) = 0 is its singular solution, then ψ (x, y) = 0 is a factor in both between p- discriminant and c- discriminant. Each discriminant have other factors which correspond to other loci associated with the primitive. In general these loci do not satisfy the differential equation. Therefore these are sometimes called extraneous loci. Tac- Locus:-p- discriminant gives equal value of p, but these values may belong to two curves of the system that are not consecutive, i.e. these curves have different c‟s. Locus of such points is called the tac- locus. For example, consider a family of circles, all of equal radii, whose centers lie on a straight line. the two circles , which are not consecutive, touch at p i.e, have same values
M.A/M.Sc. Mathematics 3rd Semester University of Kashmir, Srinagar 5 Lecture notes on Ordinary Differential Equations-(Unit-I) Khanday M. A. of p but the tangent at p is not the direction of the line of centers which is the locus of points of contact of the circles. The line of centers is the tac- locus.
Tac locus
Nodal Locus:-c- discriminant gives equal values of c but these values may belong to the nodes( i.e. double points with distinct tangents) which are also ultimate points of intersection of the consecutive curves, Locus of such points is called Node- Locus.
Cusp- Locus:-c-discriminant gives equal values of c but these values may belong to the cusps( i.e. double points with coincident tangents) which are also ultimate points of intersection of the consecutive curves. Their locus is called the Cusp Locus.
Note:- The p- discriminant equated to zero may include the envelope E, as a factor once, Cusp locus C, once and the tac- locus T, twice and the c- discriminant equated to zero may include envelope E, once the cusp locus C thrice and the Node locus N twice. i.e., p- discriminant ~ ECT2 = 0. and c- discriminant ~ EC3N2 =0 The singular solution is obtained as a common factor from the c and p discriminate and it must satisfy the differential equations. Examine the equation for Singular Solution 4x p2 = (3x - a)2 Solution:- From the given equation, we have p = dy/dx So that we can write p = ± (3x -a)/ 2√x or we can write dy = ± (3x -a)/ 2√x dx Integrating both sides, we get dy (3x a)/ 2 x dx or y = c ± ( x 3/2 - a√x) or (y –c)2 = x( x3-a)2 …(I) which is the complete solution of the given equation. Now, we have p- discriminant, x(3x- a)2 = 0. Also, (I) can be written as, c2 -2 yc + y2 – x(x3-a)2 = 0
M.A/M.Sc. Mathematics 3rd Semester University of Kashmir, Srinagar 6 Lecture notes on Ordinary Differential Equations-(Unit-I) Khanday M. A.
Therefore c-discriminant can be calculated as 4y2 – 4.1. (y2 –x(x3-a)2 = 0, or 4y2 -4y2 + 4x(x3-a)2 =0, or 4x (x3-a)2 = 0 or x (x3-a)2 = 0 Now comparing p- discriminant and c- discriminant with ECT2 and EC3N2, it clearly follows that x= 0 is common and thus is the Singular Solution of the given equation. The factor x3-a =0 which occurs in the c-discriminant and does not occur in p- discriminant is the Node locus. Also, 3x –a =0 is the Tac- locus. Examine for Singular Solution xp2 – 2 yp + 4x = 0 Solution.We have the given equation is quadratic in p, therefore
2 dy 2y 4y 2 16x 2 y y 2 4x 2 y y P = = = = 4 dx 2x x x x This is of homogeneous nature. Put y = vx , so that dy/dx = v+ x dv/dx Thus we have,
v + x dv/dx = v ± v2 4 dv dx or v 2 4 x Integrating both sides, we have
2 v v log 1 log cx 2 2
v v 2 4 or = cx 2 4
y y 2 4x 2 or = cx 2x 4x 2 or y + y 2 4x2 = 2cx 2 or (y 2cx 2 ) 2 y 2 4x 2 or c 2 x 2 cy 1 0 is the required primitive of the given differential equation.
M.A/M.Sc. Mathematics 3rd Semester University of Kashmir, Srinagar 7 Lecture notes on Ordinary Differential Equations-(Unit-I) Khanday M. A.
Now p discriminant is given by 4y2 – 16x2 = 0, or y2 = 4x2 Also, we have c-discriminant is given by y2 – 4x2 =0 or y2 = 4x2 Thus the Singular Solution is y2 = 4x2 Examine the equation y2 -2pxy + p2(x2 -1) = m2, for singular solution. Solution: The given equation can be written as (x2 -1) p2 – 2pxy + (y2 –m2) =0 …(I) or (px –y)2 = p2 + m2 or y px p2 m2 which is of the Claurits form Hence the general solution is
y cx c2 m2 or y cx2 c 2 m 2
2 2 2 2 or c x 1 2cxy y m 0 ---- (II) Hence from (I) and (II) both p- discriminant and c- discriminant are 4x 2 y 2 4(x 2 1)( y 2 m 2 ) 0 or y 2 m2 x 2 m2 This is the required singular solution. Home Assignments Examine for Singular Solution ( If any)
(i) y px p 2
(ii) xp 2 (x a) 2 0
(iii) 4p 2 9x 0
(iv) 4 p 2 (x 2) 1
M.A/M.Sc. Mathematics 3rd Semester University of Kashmir, Srinagar 8 Lecture notes on Ordinary Differential Equations-(Unit-I) Khanday M. A.
Exercise: Reduce the equation x 2 p 2 py(2x y) y 2 0 to Claurites form by putting u = y and v = xy and find its complete primitive and also its singular solution. Solution: We have u y and v xy du dy dv dy So that and x y dx dx dx dx dv dy x y dv xp y Now dx dx du du dy p dx dx dv or p px y du y dv p Where p1 p1 x du Putting the value of p in the given equation, we have
2 2 2 y y 2 x 2 (2x y) y 0 ( p1 x) p1 x
2 2 or x (2x y)( p1 x) ( p1 x) 0
2 p1 xy yp1 0
2 or xy p1 y p1
2 or v p1u p1 , which is of Claurits form of differential equation.. Hence the general solution is v cu c2 or c 2 yc xy 0 Therefore c-discriminant is y 2 4xy 0 or y(y 4x) 0 Also from the given equation p-discriminant is y 2 (2x y) 2 4x 2 y 2 0 y 2 (y 2 4xy) 0 or y.y 2 (y 4x) 0
M.A/M.Sc. Mathematics 3rd Semester University of Kashmir, Srinagar 9 Lecture notes on Ordinary Differential Equations-(Unit-I) Khanday M. A.
Now y(y 4x) 0occurs both in c and p discriminate and y = 0 and (y 4x) = 0 satisfy the given differential equation. Hence y = 0 and (y 4x) 0is the Singular Solution. Exercise: Solve the differential equation for the Singular Solution.
2 4xx 1x 2p2 3x2 6x 2 0 …(i)
2 Solution: Here p-discriminant is xx 1x 23x2 6x 2 0
2 2 2 3x 6x 2 Also from (i) p 4xx 1x 2
3x 2 6x 2 3x 2 6x 2 or p 3 2 2 xx 1x 2 2 x 3x 2x
3x 2 6x 2 or dy dx , 3 2 2 x 3x 2x Integrating both sides we get y xx 1x 2 + c or y c2 xx 1x 2 or c 2 2cy y 2 xx 1x 2 Now c- discriminant is 4y 2 4y 2 xx 1x 2 0 or xx 1x 2 0 Thus xx 1x 2 0 , is a Singular Solution. Also comparing the two discriminate to ECT2≈ 0and EC3T2 ≈ 0 , we have seen that the 1 tac- locus is 3x2 -6x + 2 = 0 or 1 3 1 1 i.e., 1 ...and1 is a tac- locus of Imaginary points of contact. 3 3 Examine for Singular Solution of the following differential equation yp 2 2xp y 0
Solution: Put y2 = v, Differentiating both sides w.r.t. x we get
M.A/M.Sc. Mathematics 3rd Semester University of Kashmir, Srinagar 10 Lecture notes on Ordinary Differential Equations-(Unit-I) Khanday M. A.
dv 2yp ( say ) dx Implies that p , 2y Substitute the value of p in the given equation we have
2 2 x y 2x y 0 or y 0 2y 2y 4y y 2 or 4y 2 4x 2 0 Implies y 2 x 0 4 2 or v x , 4 This is of Claurits form of differential equation Therefore, the complete primitive is given by v 2cx c 2 ( by replacing θ by 2c ) i.e., y 2 2cx c 2 is the complete solution of the given differential equation. Now p-discriminant of the given differential equation is yp 2 2xp y 0 is x 2 y 2 0 and c- discriminant is given x 2 y 2 0
Hence is the required Singular Solution of the given differential equation. Examine for Singular Solution (if any) by using a suitable substitution. 3xp 2 6yp x 2y 0
Solution: We have 3xp 2 6yp x 2y 0 ------(i)
Put x 3y v and x2 u , Differentiating both sides, we get dx 3dy dv and 2xdx du dx 3dy dv Therefore 2xdx du dx 3dy 1 3p i.e., or 2xdx 2xdx 2x 2x
M.A/M.Sc. Mathematics 3rd Semester University of Kashmir, Srinagar 11 Lecture notes on Ordinary Differential Equations-(Unit-I) Khanday M. A.
1 2x or 1 3p 2x or p 3 Now from (i)
2 1 2x 1 2x 3x 6y x 2y 0 3 3
1 4x 2 2 4x 1 2x 3x 6y x 2y 0 9 3
1 4x 2 2 4x x 2y1 2x x 2y 0 3 4x3 3 4x 2 x 12 xy 3x 0 or 1 x 2 2 x 3y 0 Implies that using the substitution, We can write, 1 2u v 0 1 or v u This is of Claurites form of differential equation Therefore the complete primitive of the differential equation is 1 v cu c 1 i.e., x 3y cx 2 or c 2 x 2 1 cx 3y c or x 2c 2 x 3yc 1 0 Now c- discriminant is x 3y2 4x 2 0 or x 2 6xy 9y 2 4x 2 0 ; 3y 2 2xy x 2 0 Also p- discriminant is 3y 2 2xy x 2 0
Hence is the required Singular Solution of the given differential equation.
M.A/M.Sc. Mathematics 3rd Semester University of Kashmir, Srinagar 12 Lecture notes on Ordinary Differential Equations-(Unit-I) Khanday M. A.
Solve by using the suitable substitution or otherwise for Singular Solution of the differential equation p 3 4xyp 8y 2 0
Solution: Put y v 2 , Differentiating both sides w.r.t. x we get dy dv dv 2v or p = 2 v θ where θ = dx dx dx Therefore from the given differential equation, we have 2v 3 4xy2v 8y 2 0 or on simplification, we have v x 3 It is of Claurits form of differential equation. Therefore, the complete primitive is
2 v kx k 3 or y k 2 x k 2 or y cx c2 Now p- discriminant can be calculated by taking partial derivative of the function p 3 4xyp 8y 2 f (say) Differentiating partially w.r.t. p, we get f 4 0, or 3p 2 4xy 0 or p 2 xy p 3
4 Now p xy 4xyp 8y 2 0 3 8 or pxy 8y 2 0 or px 3y 3 4 or p 2 x 2 9y 2 or x 2 xy 9y 2 3 or 4x 3 27 y Also in a similar way, the c- discriminant of the solution is
Therefore, the required singular solution of the given differential equation is 4x3 27 y 0
M.A/M.Sc. Mathematics 3rd Semester University of Kashmir, Srinagar 13 Lecture notes on Ordinary Differential Equations-(Unit-I) Khanday M. A.
Home Assignments Examine for singular Solution (if any) of the following differential equation. 1. p 2 2xp 3x 2
2. p 2 2 px3 4x 2 y 0
3. x 2 y px yp 2 4. a 2 x 2 p 2 2xyp b 2 y 2 0
5. xp 2 x a2 0
6. xp y2 p 2 1
7. yp 2 2xp y 0
8. y xp 2 ap
9. y px a 2 p 2 b2
10. p 2 2xp y 0 11. p logpx y 12. SinpxCosy CospxSiny p
Find the differential equation if the complete primitive is given. Also find the Singular Solution (if any). c 2 2cy x 2 1 0
Solution: Differentiating w.r.t. x we get x 2cp 2x 0 or c p Therefore, from the given differential equation x 2 x 2 y x 2 1 0 p 2 p or p 2 1 x 2 2xyp x 2 0 Now p- discriminant is 4x 2 y 2 x 2 1 0 Also, c- discriminant is
M.A/M.Sc. Mathematics 3rd Semester University of Kashmir, Srinagar 14 Lecture notes on Ordinary Differential Equations-(Unit-I) Khanday M. A.
4y 2 x 2 1 0 Thus, the singular solution is a circle x 2 y 2 1 and x 0 is a tac-locus. Home Assignment Find the differential equations if the complete primitive is given by the following. 1. x 2 y 2 2cx c 2 cos2 0
2. c 2 x yc 1 xy 0
3. x 2 y 2 2cxy c 2 1 0 4. y 2 2cx 2 y c 2 x 4 x3 0 Exercise: Reduce xyp 2 x 2 y 2 1p xy 0 to Clairaut’s form by the substitution X= x2 ; Y= y2. Hence,show that the equation represents a family of conics touching the four sides of a square. Solution:Let X = x2 ; Y= y2 So that we can write on differentiation. 2 x dx = dX and 2 y dy = dY or dY y x dY p or p where dX x y dX
x 2 x Therefore xy 2 X Y 1 xy 0 2 y y
x 2 or y 2 2 1 X Y 1 2 y X or Y 2 1 X Y 1 Y or X 2 Y X Y or X 1 Y 1 or Y X 1
M.A/M.Sc. Mathematics 3rd Semester University of Kashmir, Srinagar 15 Lecture notes on Ordinary Differential Equations-(Unit-I) Khanday M. A.
This is of Clarautis form of differential equation. Therefore its complete primitive is C Y CX or Y1 C C1 CX C 1 C C 2 X 1 X Y C Y 0 Thus c- discriminant is 1 X Y 2 4XY 0 or 1 X 2 Y 2 2X 2Y 2XY 4XY 0 or 1 X 2 Y 2 2X 2Y 2XY 0 or we can write x 4 y 4 2x 2 2y 2 2x 2 y 2 1 0 This represents the family of conics touching the four sides of a square.
Exercise: Find the Singular Solution of the differential equation. a 2 x 2 p 2 2xyp b 2 y 2 0 Solution:- We can write the differential equation as a 2 p 2 x 2 p 2 2xyp b 2 y 2 0 or xp y2 a 2 p 2 b 2 or y px a 2 p 2 b2 It is of Calirauits form of differential equation Thus the complete primitive is given by y cx a2c2 b2 Now p- discriminant is 4x 2 y 2 4a 2 x 2 b 2 y 2 0 or a 2 y 2 b 2 x 2 a 2 b 2 orwe can write this as x 2 y 2 1 a 2 b 2 Similarly the c-discriminant is
M.A/M.Sc. Mathematics 3rd Semester University of Kashmir, Srinagar 16 Lecture notes on Ordinary Differential Equations-(Unit-I) Khanday M. A. x 2 y 2 1 a 2 b 2
Hence the singular solution of the given differential equation is
This represents the equation of the ellipse. Miscellaneous Methods for Equations of the Second and Higher Orders: Here we reduce equations of the second order to the equations of the first order. We shall show that the order can always be so reduced if the equation (i) does not contain y explicitly (ii) does not contain x explicitly (iii) is homogeneous Case I. If y is absent; i.e., If y does not occur explicitly in an equation of the second dp order, we obtain an equation containing only , p and x, so of the first order. dx dy dp Case II. If x absent, In this case we write p for and for y we write p dx 2 dy
dp dy dp dp Since p y dy dx dy dx 2 Examples for Solution
2 1. y2Cos x 1 dp Solution:- Here y is absent, we can simply write y p and y 1 2 dx Therefore we have dp dp Cos2 x 1 or Sec 2 x dx dx or dp Sec2 xdx Integrating both sides we get p tan x a Where „a‟ is a constant. Implies dy (tanx a)dx Integrating again, we have y logCosx ax b log Secx ax k
M.A/M.Sc. Mathematics 3rd Semester University of Kashmir, Srinagar 17 Lecture notes on Ordinary Differential Equations-(Unit-I) Khanday M. A.
This is the required solution of the given differential equation.
2 2. yy2 1 y1 dy Solution:- Here x is absent, thus we can write p y dx 1 dp and y p 2 dy
dp 2 So that y p 1 p dy dp 1 2 p dy or y p p 2 1 or dp 2 dy 2 1 p y or log 1 p 2 log ay where „a‟ is a constant. or p 2 1 a 2 y 2 or we can write this as in terms of dy as dy 1 a 2 y 2 dx Integrating we get by putting ay = Cos θ, So that ady = -Sinθdθ dy 1 Sin d 1 dx = x b 1 a 2 y 2 a Sin a or θ = -a ( x + b) Implies that Cos-1 ay = -a (x + b) or ay = Cos ( ax + c)
2 2 Solve y1 y3 y1 2y2 by using above exercise
Solution: Here both x and y are absent
dp d dp dp dp y1 p , y2 p and y3 p dy dx dy dx dy
2 2 d 2 p dp dy dp dy d 2 p dp d 2 p dp or y p = p P p 2 3 2 dxdy dy dx dy dx dydy dy dy dy Now from the given equation
M.A/M.Sc. Mathematics 3rd Semester University of Kashmir, Srinagar 18 Lecture notes on Ordinary Differential Equations-(Unit-I) Khanday M. A.
2 2 y1 y3 y1 2y2 , we have
2 2 2 2 d p dp 2 dp pp p p 2p 2 dy dy dy
2 2 d 2 p dp dp or p 1 2 2 dy dy dy
2 d 2 p dp or p 1 2 dy dy This gives
2 pp2 1 p1 which is similar to above example. Therefore we have from above the solution in terms of p is pa Cosay c 1 dy or p Cosay c or a dx a Cosay c Implies a Secay cdy dx Sec(ay + c) + tan(ay + c) Þ alog = x + d a or x log Secay c tanay c d is the required solution of the given differential equation. Solve the following
(i) xy3 y2 12 x
x (ii) yn 2yn1 e
3 1 y 2 2 (iii) Integrate and interpret geometrically 1 k y2
Solution: (i) Multiplying each term by x 2 , we have
3 2 3 x y3 x y2 12 x
M.A/M.Sc. Mathematics 3rd Semester University of Kashmir, Srinagar 19 Lecture notes on Ordinary Differential Equations-(Unit-I) Khanday M. A.
dz 1 Let x e z , z log x dx x dy dy dz 1 dy dx dz dx x dz 1 dy y , 1 x dz d Let D dz
Therefore xy1 Dy
d 1 dy 1 dy 1 d 2 y 1 dy 1 d 2 y dz Also, y2 dx x dz x 2 dz x dxdz x 2 dz x dz 2 dx 1 or y D 2 y Dy or x 2 y DD 1y 2 x 2 2
3 Similarly, x y3 DD 1D 2y Therefore from the given differential equation we have DD 1D 2y DD 1y 12e3z or DD 1D 2 1y 12e3z
or DD 12 12e3z
Now the auxiliary equation is mm 12 0 i.e., m = 0, 1, 1 Therefore the Complementary function of the equation is A B Cze z A B C log xx Also the particular integral is 12 1 e3z 12 e3z e3z x 3 DD 12 3.22 Therefore the complete solution of the equation is y A B C log xx x3
x (ii) yn 2yn1 e
Solution:- We know as from above that
2 n y1 Dy , y2 D y , . . . , yn D y
M.A/M.Sc. Mathematics 3rd Semester University of Kashmir, Srinagar 20 Lecture notes on Ordinary Differential Equations-(Unit-I) Khanday M. A.
Therefore from the given equation, we have D n 2D n1 y e x Now the Auxiliary equation is mn 2mn1 mn1 m 2 0 n1times i.e., m 2,m 0,0,0,...0, Therefore the Complementary function is ae2xe b cx dx2 ... hxn2 e0x ae2x b cx dx2 ... hxn2 Also the particular Integral is 1 1 e x e x e x D n 2D n1 1n 21n1 Hence the complete solution of the integral is y = e x ae2x b cx dx2 ... hxn2
3 1 y 2 2 (iii) 1 k ; where k is constant. y2 dp Solution:- We know that y p , y 1 2 dx Therefore from the given equation, we have
3 1+ p2 2 ( ) dp 2 3/2 = k Þ k = 1+ p dp dx ( ) dx dp or we can write k 3/ 2 dx , Integrating both sides we get 1 p 2 dp k dx ; Put p = tanθ ; So that dp = Sec2θdθ 2 3/ 2 1 p
Sec2 Therefore k d dx p 1 p2 Sec3 x a x a p Implies that Sin , or 1 k k 1 p 2
M.A/M.Sc. Mathematics 3rd Semester University of Kashmir, Srinagar 21 Lecture notes on Ordinary Differential Equations-(Unit-I) Khanday M. A.
p 2 x a2 x a2 p Implies that 2 2 or 1 p k k 2 x a2 x as or we have p k 2 x a2
Let k 2 x a2 t 2 So that x a tdt Then we have x a2 dy dx k 2 x a2 tdt i.e., y b t t or y k 2 x a2 b or y b k 2 x a2
y b2 k 2 x a2 or we can write this as x a2 y b2 k 2 which is the required equation of circle with centre (a,b) and radius k given by
3 1 y 2 2 1 k y2 Home Assignments
1. xy2 y1 4x
2 2. yy2 y1 Homogeneous Equations If x and y are regarded as of dimension 1.
y1 is of dimension zero
y2 is of dimension -1
y3 is of dimension -2
M.A/M.Sc. Mathematics 3rd Semester University of Kashmir, Srinagar 22 Lecture notes on Ordinary Differential Equations-(Unit-I) Khanday M. A. we define a homogeneous equation, the equation in which all the terms are of same
2 degree, e.g., x y2 xy1 y 0 is called Cauchy‟s Homogeneous equation. Solve the following homogeneous differential equations.
2 (i). xyy2 xy1 3yy1 .
2 (ii) x y2 xy1 y 0.
2 (iii) x y2 xy1 5y 0.
2 2 2 2 (iv) 2x yy2 y x y1 .
2 2 2 2 2 (v) 2x yy2 4y x y1 2xyy1 [Hint, using substitution y= z ] dy dt 1 dy Solution:- (i) We know that y , 1 dt dx x dt dy 1 dy 1 d dy y 1 , 2 dx x 2 dt x dx dt 1 dy 1 dt d 2 y x 2 dt x dx dt 2 1 dy 1 d 2 y x 2 dt x 2 dt 2 Substitution in and multiplying by x, we get
2 d 2 y dy dy dy y 3y 2 dt dt dt dt
2 d 2 y dy dy i.e., y 4y dt 2 dt dt This is an equation where t is absent dy By putting q , we can obtain easily dt yq 2y 2 b, 1 Giving t c logy 2 b 4 Hence y 2 b e4tc ae4 Which is the required solution of the equation. (ii)
M.A/M.Sc. Mathematics 3rd Semester University of Kashmir, Srinagar 23 Lecture notes on Ordinary Differential Equations-(Unit-I) Khanday M. A.
Solution:- Put x et so that t = log x dy dt 1 dy So that we have y , 1 dt dx x dt dy 1 dy 1 d dy y 1 , 2 dx x 2 dt x dx dt 1 dy 1 dt d 2 y and x 2 dt x dx dt 2 1 dy 1 d 2 y x 2 dt x 2 dt 2 or we can write d 2 y dy x 2 y 2 dt 2 dt Thus the question becomes,
d 2 y dy dy y 0 2 dt dt dt d 2 y dy or 2 y 0 dt 2 dt d dy Let D and dx dt dy Therefore, we have xDy x y dx dy dx dy dy Because, y x dt dt dx dx i.e., xDy y
Similarly, x 2 D 2 y 1y Therefore from the given equation we have 1y y y 0 or 12 y 0 Now the auxiliary equation is m 12 0 or m 1,1 Also the complementary function is given by
M.A/M.Sc. Mathematics 3rd Semester University of Kashmir, Srinagar 24 Lecture notes on Ordinary Differential Equations-(Unit-I) Khanday M. A.
a btet a blog xx Since the particular integral is zero, therefore the complete solution of the given differential equation is y a blog xx ax bxlog x
2 (iii) x y2 xy1 5y 0. ( Try yourself )
2 2 2 2 (iv) 2x yy2 y x y1 dv Solution:- Let y vx, So that we have y v x 1 dx
d 2v dv d 2v y 2v x 2 x And 2 1 2 2 dx dx dx Therefore from the given equation we have
2 dv d 2v dv 2x 2 vx 2 x v 2 x 2 x 2 v x 2 dx dx dx
2 dv d 2v dv or 2 vx 2vx2 x 2 2 dx dx dx dt 1 Put x et so that t= log x and dx x dv dv dt 1 dv . . dx dt dx x dt d 2v 1 d 2v dt 1 dv 1 d 2v dv Also 2 2 2 2 2 dx x dt dx x dt x dt dt Therefore we have
2 dv d 2v dv dv 2v 2v 2 dt dt dt dx
2 d 2v dv 2v 2 dx dx
dv dp d 2v dp dv Now for p; , dx dt dt 2 dv dt dp dp i.e., p , dt dv
M.A/M.Sc. Mathematics 3rd Semester University of Kashmir, Srinagar 25 Lecture notes on Ordinary Differential Equations-(Unit-I) Khanday M. A.
Thus, we can write dp dp 2vp p 2 or 2v p dv dv dp dv or we can write this as 2 p v Integrating both sides we get 2log p log cv dv or b v, where b c dt dv Therefore, we have bdt v or 4v bt a2 or 4y xAlog x B2 is the required solution of the given differential equation.
2 2 2 2 (v) 2x yy2 4y x y1 2xyy1 dy dy dy dz dz Solution:- Let y =z2, so that 2z, and y 2z dz 1 dx dz dx dx
2 dy d 2 z and y2 2 2z dx dx2 Substitute these things in the given differential equation
2 2 2 2 2x yy2 4y x y1 2xyy1 gives
2 2 dz d 2 z dz dz 2x 2 z 2 2 2z 4z 4 x 2 2z 2xz 2 2z 2 dx dx dx dx
d 2 z dz implies that x 2 z x 2 dx dx d x 2 D 2 z xDz z 0, where D dx
2 2 x D xD 1z 0, …(1) As before, put x et dz 1 dz dz Implies that or xDz dx x dt dt
M.A/M.Sc. Mathematics 3rd Semester University of Kashmir, Srinagar 26 Lecture notes on Ordinary Differential Equations-(Unit-I) Khanday M. A.
d 2 z 1 d 2 z dz Also, we have 2 2 2 dx x dt dt d d For and D dt dx dy dy dt 1 we have y Dy y 1 dx dt dx x or xDy y and x 2 D 2 y 1y Thus, we have from the equation (1), 2 2 1z 0 or 12 z 0
Now the Auxiliary equation is m 12 0 , m = 1, 1 Also the complementary function is z a btet a blog xx
Hence y z 2 a blog xx2 is the required solution of the given equation. Home Assignments
2 1. x y2 xy1 y 2log x
4 3 2 2. x y3 2x y2 x y1 xy 1
2 2 3. x y2 2xy1 20 y (x 1)
2 1 4. x 2 D2 3xD 1y x 1
2 x 5. x y2 4xy1 2y e
2 6. y2 x Sinx y Sec 2 y tan y 7. 2
An equation of the type y2 f y This form of differential equations occurs frequently in Dynamics especially in problems on motion under a force directed to a fixed point and of magnitude depending solely on the distance from that fixed point.
Here we multiply each side by 2y1 as it acts as an integrating factor for . Thus we get
M.A/M.Sc. Mathematics 3rd Semester University of Kashmir, Srinagar 27 Lecture notes on Ordinary Differential Equations-(Unit-I) Khanday M. A.
2y1 y2 2y1 f y Integrating, we get dy y 2 2 f y dx 2 f ydy 1 dx Which represents the equation of energy d 2 x Applying the method to the equation of the SHM, we get from p 2 x , dt 2 dx d 2 x dx 2 p 2 x dt dt 2 dt Integrating with respect to t, we get
2 dx p 2 x 2 Cons tant p 2 a 2 x 2 , dt dt 1 1 Hence, , dx p a 2 x 2 On integration we get 1 x t Sin1 Cons tant p a or x aSin pt Solve the following
3 1. y2 y y , given that y1 0, when y = 1
2 y 2. y2 e , given that y = 0 and y1 1, when x =0
2 3. y2 Sec tan y, given that y = 0 and y1 1, when x = 0 d 2 x ga 2 dx 4. , given that x = h and 0 when t = 0. dt 2 x 2 dt
2 d u p 2 3 du 5. u , in the two cases(i) P = μu ; (ii) P = μu given that 0 , d 2 h 2u 2 d 1 when u , where μ, h and c are constants. c
Solution:- 1. Multiplying both sides by an integrating factor 2y1 , we get,
3 2y1 y2 2y1 y yy1
M.A/M.Sc. Mathematics 3rd Semester University of Kashmir, Srinagar 28 Lecture notes on Ordinary Differential Equations-(Unit-I) Khanday M. A.
d or we can write this as y 2 2y 3 yy dx 1 1 Integrating both sides, we get y 2 2 y3 y dy 1
4 2 4 y y y 2 = 2 c y c 4 2 2 y 4 or y 2 y 2 c 1 2
Now using the initial conditions, we have, for y1 0, when y = 1 1 0 1 c or c = 1/2 2
4 4 2 2 2 2 y 2 1 y 2y 1 y 1 Therefore we get y y 1 2 2 2 2 y 2 1 2dy Therefore we have y1 dx 2 y 2 1 Integrating both sides, we have dy 2 dx y 2 1
Implies that,. 2Coth1y x c x c or Coth1 y 2 x c or y Coth 2
2 y 2. y2 e , given that y = 0 and y1 1, when x =0
Solution:- Multiplying both sides by the integrating factor 2 y1 , we get
2 y 2y1 y2 2y1e Now integrating both sides, we get y 2 2 e2 y dy 2e2 y c 1
2 2 y i.e., y1 e c Now using the initial conditions, we get
M.A/M.Sc. Mathematics 3rd Semester University of Kashmir, Srinagar 29 Lecture notes on Ordinary Differential Equations-(Unit-I) Khanday M. A.
y = 0 and y1 1, when x =0 1 e0 c Or c = 0
2 2 y y Therefore y1 e or y1 e or e y dy dx, Integrating both sides we get, e y x a Now again using final condition, we have from y = 0 when x = 0, Therefore a = -1 Thus, we have e y 1 x or y logx 1 Hence, y logx 1
2 3. y2 Sec tan y, given that y = 0 and y1 1, when x = 0
Solution:- Multiplying both sides by the integrating factor 2 y1 , we get
2 2y1 y2 2Sec tan yy1 , Integrating both sides, we get y 2 2 Sec2 tan y y dy tan2 y c 1 1
2 2 i.e., y1 tan y c Using initial conditions, we get, 1 tan 2 0 c or c = 1
2 2 2 or y1 1 tan y Sec y or y1 Secy dy thus dx , on integration we get, Secy or Cosydy dx Siny x a or y Sin1 x a Again using final conditions y = 0 , x = 0 , we get 0 Sin1 0 a or a 0
M.A/M.Sc. Mathematics 3rd Semester University of Kashmir, Srinagar 30 Lecture notes on Ordinary Differential Equations-(Unit-I) Khanday M. A.
Therefore, we get y Sin 1 x is the required solution of the given differential equation.
d 2 x ga 2 dx 4. , given that x = h and 0 when t = 0. dt 2 x 2 dt dx Solution:- Multiplying both sides by the integrating factor 2 , we get dt dx d 2 x ga2 dx 2 2 dt dt 2 x 2 dt Integrating both sides, we get,
2 dx 1 1 2ga2 dx 2ga2 b dt x 2 x
2 dx 2ga2 b dt x
Using initial conditions x = h and , we get,
2ga 2 2ga2 b 0 or b h h
2 dx 2ga2 2ga2 h x Therefore, 2ga2 dt x h hx
dx h x h x 2 Implies that 2ga k , wherek 2ga . dt hx h x
dx h x i.e., k , dt h x Integrating both sides we get,
x h dx k dt, h x
xdx h kt c, h x Put h – x = z2 So that – dx = 2zdz
M.A/M.Sc. Mathematics 3rd Semester University of Kashmir, Srinagar 31 Lecture notes on Ordinary Differential Equations-(Unit-I) Khanday M. A.
h z 2 Therefore h 2zdz kt c, z or we have 2 h h z 2 dz kt c, z hCos Put dz hSin
2 Therefore, 2 h h hCos2 Sind kt c, 2h hSin2d kt c, 1 Cos2 2h h d kt c, 2
3 h2 d Cos2d kt c, 3 Sin2 h2 kt c, 2
3 h2 SinCos kt c,
3 2 1 z z z h2 Cos 1 kt c, h h h
3 1 z z 2 h2 Cos h z kt c, h h
3 1 h x 1 2 h Cos h xx kt c, h h Using final conditions, we get
3 1 1 h2 Cos 0 0 c, or c 0 h Therefore, we have, h h h x 1 1 Cos h xx t k h h
1 h h x 1 2 or t hCos hx x a 2g h
M.A/M.Sc. Mathematics 3rd Semester University of Kashmir, Srinagar 32 Lecture notes on Ordinary Differential Equations-(Unit-I) Khanday M. A.
d 2u p 5. u , in the two cases d 2 h 2u 2 du 1 (i) P = μu2 ; (ii) P = μu3 , given that 0 , when u , d c d 2u p Solution:- we hav u d 2 h 2u 2 Case I :- let P = μu2 d 2u u 2 Therefore u d 2 h 2u 2 h 2 The above equation can be written as d D 2u u Where D h 2 d or D2 1u h2 Now the auxiliary equation of the differential equation is m2 1 0 or m Therefore the complementary function of this is given by u aCos bSin Also the particular integral of the equation is given by
1 0 2 1 0 P.I u 1 D u D 2 1 h 2 h 2 1 D 2 ... u 0 h 2 h 2 Therefore the Complete primitive is given by u aCos bSin h2 Case II let P = μu3,
Therefore, gives us d 2u u 3 u u d 2 h 2u 2 h 2
M.A/M.Sc. Mathematics 3rd Semester University of Kashmir, Srinagar 33 Lecture notes on Ordinary Differential Equations-(Unit-I) Khanday M. A.
d 2u or 1u d 2 h 2
It is of the form u2 f u du Therefore, multiplying both sides by an integrating factor, 2 we get d du d 2u du 2 2 1u d d 2 d h 2 On integration we have
2 du 2 2 1 udu d h 2
u 2 2 1 a h 2 2
2 du i.e., 1u 2 a d h2 du 1 Now using the initial conditions, 0 , when u , d c Therefore we can have, 1 1 0 1 a or a 1 h 2 c 2 h 2 c 2
2 du 1 1u 2 1 d h 2 h 2 c 2 Therefore 1 1 u 2 h 2 c 2
du 1 c 2u 2 or 1 2 d h c Implies that on integration,
du c 1 d 2 1 c 2u 2 h Put cu = Cost Implies c du = -Sin t dt Therefore
M.A/M.Sc. Mathematics 3rd Semester University of Kashmir, Srinagar 34 Lecture notes on Ordinary Differential Equations-(Unit-I) Khanday M. A.
S int c dt 1 () b 2 c 1 Cos 2t h
or dt 1 b h 2
or t 1 b h 2
Cos 1 cu 1 b h 2 Using boundary conditions, we have Cos 1 1 b or b 0
Therefore, Cos 1 cu 1 i.e., Cu Cos 1 2 2 h h
or Cu Cosh 1 according as or h 2 2 h
Factorization of the Operator: Method of Operational factors, If It is possible to factorize the operation, the method of Operational factors may be applied. Let the linear equation of second order d 2 y dy p Qy R dx2 dx be written as f (D) y =R or f1 D. f 2 Dy R
Which is such that f 2 D operates on y and f1 D on the result and we get the same result as if f D. operates upon y. Examples for solution:-
1. x 1y2 x 1y1 2y 0 Solution:- We can write the given differential equation as x 1D 2 y x 1Dy 2y 0
M.A/M.Sc. Mathematics 3rd Semester University of Kashmir, Srinagar 35 Lecture notes on Ordinary Differential Equations-(Unit-I) Khanday M. A. or x 1D 2 x 1D 2y 0 or, we have x 1D 2 x 1 2D 2y 0 x 1D 2 x 1D 2D 2y 0 x 1DD 1 2D 1y 0
x 1D 2D 1y 0 …(i)
Let D 1y z …(ii) Therefore we have x 1D 2z 0 or xD D 2z 0 dz dz Implies that x 2z dx dx dz 2dx or z x 1 Now integrating both sides we get, log z 2logx 1 log a or log z log ax 12 or z ax 12 Substitute in (ii) D 1y ax 12 dy y ax 12 dx This is a linear differential equation,
dx Now integrating factor is e e x dy Therefore, e x ye x e x x 12 a dx d or ye2 ax 12 e x dx Integrating both sides we get ye x ax 12e x dx ax 12 e x 2x 1e x dx
M.A/M.Sc. Mathematics 3rd Semester University of Kashmir, Srinagar 36 Lecture notes on Ordinary Differential Equations-(Unit-I) Khanday M. A.
ax 12 e x 2x 1e x e x dx ax 12 e x 2x 1e x 2e x b Implies that ye x ax 12 e x 2ax 1e x 2aex b y ax 12 2ax 1 2a bex or y ax 2 1 bex is the required solution of the given differential equation.
2. xy2 x 1y1 y 0 Solution:- We have xD 2 x 1D 1y 0 xD 2 xD D 1y 0 xDD 11D 1y 0
xD 1D 1y 0 …(i)
Let D 1y z …(ii) dz Therefore xD 1z 0 or x z dx dz dx or z x Integrating both sides of the equation, we get log z log ax Or z ax Therefore from (ii), we have D 1y ax dy y ax dx which is a linear differential equation,
dx Therefore the integrating factor is e e x d Thus we have, ye x axe x dx Integrating both sides, we get ye x axe x e x dx
M.A/M.Sc. Mathematics 3rd Semester University of Kashmir, Srinagar 37 Lecture notes on Ordinary Differential Equations-(Unit-I) Khanday M. A.
axex aex b or y ax a bex
Therefore y ax 1 be x is the complete primitive of the differential equation. Home Assignments
2 1. xy2 x 1y1 y x
2 2. xy2 x 1y1 2xy 2x, given that y = 2 and y1 0, when x = 0.
2 2 2 2x 3. x 1y2 4x 3x 5y1 4x 6x 5y e given that y = 1 and y1 2 when x = 0. d 2 y dy 4. x 1 x y e x dx2 dx d 2 y dy 5. x x 2 2y x3 dx2 dx 6 x 3D 2 2x 7D 2y x 32 e x ONE INTEGRAL BELONGING TO THE COMPLEMENTRY FUNCTION When the integral belonging to the equation y2 + Py1 +Qy = 0 ...(I) is known, say y z , then the more general equation of the second order y2 + Py1 +Qy = R ...(2) where P, Q , R are functions of x, can be reduced to one of the first order by the substitution y= vz.
Differentiating, we get, y1 v1z vz1 y2 v2 z 2v1z1 vz2 . Hence (2 ) becomes v2 z v1 2z1 Pz vz2 Pz1 Qz R,
dv1 i.e., z + v1 (2z1 + pz) = R ...(3) dx
Since by hypothesis, z2 Pz1 Qz 0.
Equation is linear equation of the first order in v1.
M.A/M.Sc. Mathematics 3rd Semester University of Kashmir, Srinagar 38 Lecture notes on Ordinary Differential Equations-(Unit-I) Khanday M. A.
Similarly a linear equation of nth order can be reduced to one of the (n-1)th if one integral to the complementary function is known. Examples
x 1. Show that y2 Py1 Qy 0 is satisfied by y e if 1 P Q 0 and by y x if P Qx 0
Solution:- The given differential equation is y2 + Py1 +Qy = 0 ...(1)
x So for , we have y1 y2 e Substituting these in (1), we have e x 1 P Q 0, but e x 0, therefore 1 P Q 0 Thus it follows that is the solution of (1), if . Also, we have for y = x, y1 1, and y2 0 Therefore (1) becomes
Hence, is the solution of (1), if .
x 2 y xy y 8x3 … (1) 2. 2 1 Solution:- Dividing each term by x 2 , we have y y y 1 8x, 2 x x 2 Now comparing the equation with . 1 1 We have P and Q x x 2 Thus it is easy to see that , therefore it follows that y x is a part of the solution, Let y vx be the solution of (1), then we can write
dv dv d 2 v y v x and y 2 x 1 dx 2 dx dx 2 Substituting these in the given differential equation we have
M.A/M.Sc. Mathematics 3rd Semester University of Kashmir, Srinagar 39 Lecture notes on Ordinary Differential Equations-(Unit-I) Khanday M. A.
dv d 2v dv x 2 2 x x v x vx 8x 3 2 dx dx dx
d 2v dv x 3 8x 2 dx dx
dv d 2 v dP Put P , So that dx dx 2 dx Thus our equation becomes dP dP 3 x 3P 8x or P 8 dx dx x Which is a linear differential equation, Now we can easily have a a P 2x or dv 2x dx x 3 x 3 Now integrating both sides, we have a v x 2 b 2x 2 y a But v therefore y x3 bx x 2x is the required solution of the given differential equation. x 2 y x 2 2x y x 2 y x 3e x 3. 2 1 …(1) Solution:- Dividing throughout by x 2 , we have
2 x 2x x 2 x y y y xe 2 x 2 1 x 2
Now comparing this equation with y2 Py1 Qy 0 , we have 2 1 2 P 1 and Q x x x 2 Clearly we observe that P Qx 0 Therefore it follows that y x is a part of the solution Let y vx be the solution of (1), Differentiating both sides, we get
M.A/M.Sc. Mathematics 3rd Semester University of Kashmir, Srinagar 40 Lecture notes on Ordinary Differential Equations-(Unit-I) Khanday M. A.
dv dv d 2 v y v x and y 2 x 1 dx 2 dx dx 2 Substituting in (1) we get
dv d 2v dv x 2 2 x x 2 2x v x x 2 vx x3e x 2 dx dx dx i.e., we can easily found d 2 v dv e x dx2 dx
dv d 2 v dP Put P so that dx dx 2 dx dP We have, P e x dx Now this is a linear differential equation, Therefore P xe x ae x dv But P dx Thus we can write dv xe x ae x dx Integrating both sides, we have v xe x ae x e x b Since y vx, therefore, y x 2e x axe x xe x bx
2x 4. xy2 2x 1y1 x 2y x 2e …………………(1) Solution:- Dividing throughout by x , we have 2x 1 x 2 x 2 y y y e 2x 2 x 1 x x
Comparing this differential equation with y2 Py1 Qy 0 , 2 2 We have, P 2 and Q 1 x x Clearly we can have1 P Q 0 ,
Therefore it follows that y e x , is a part of a solution.
Let y ve x be the solution of (1),
M.A/M.Sc. Mathematics 3rd Semester University of Kashmir, Srinagar 41 Lecture notes on Ordinary Differential Equations-(Unit-I) Khanday M. A.
Differentiating both sides, we get dv dv d 2 v y ve x e x and y ve x 2e x e x 1 dx 2 dx dx2 Substituting in (1) we get
dv d 2v dv xvex 2e x e x 2 x 1 vex e x x 2 vex x 2 e 2x 2 dx dx dx or we have d 2 v 2 dv x 2 e x dx2 x dx x dv d 2 v dP Let P and dx dx2 dx Thus we observe that dP 2 x 2 P e x dx x x 1 Now multiplying both sides by an integrating factor , we have x 2 P e x a or P e x ax2 x 2 x 2 dv or since we have P therefore dv e x ax2 dx dx x 3 Integrating again we have v e x a b 3
3 x x Hence y xe a b is the complete solution of the given differential equation. 3 Home Assignment
2 2x 3 1. x y2 xy1 9y x 2e , given that y x is a solution.
2 2 2. xy2 xCosx 2Sinx x 2y1Sinx 2yxSinx Cosx 0 , given that y x is a solution. d 2 y dy 3. (1 x) x y (1 x) 2 dx2 dx
M.A/M.Sc. Mathematics 3rd Semester University of Kashmir, Srinagar 42 Lecture notes on Ordinary Differential Equations-(Unit-I) Khanday M. A.
d 2 y dy 4. (2x 1) 2 3 2xy 2e x dx2 dx d 2 y dy 5. 1 Cotx yCotx Sin2 x. dx2 dx d 2 y dy 6. (x x 2 ) 1 2x 1 3x x 2 y (1 x) 3 dx 2 dx Picard’s Existence and Uniqueness theorem Existence Theorem: Statement:- The initial Value problem dy f x, y, yx y ....(1) dx 0 0 has at least one solution yx provided the function f x, yis continuous and bounded for all values of x, say f x, y M …(2) and satisfies the Lipschitz Condition f x, y2 f x, y1 M y2 y1 , ...(3) for all values of arguments. Proof:- Consider the iterative sequence
x é ù yn = y0 + ò f ët, yn-1 (t)ûdt, n =1,2,3,... …(4) x0 with y0 t y0 , for the initial value problem (1). In order that the initial value problem
(1) may have a solution, it is necessary that the sequence yn x of functions converge to a limiting function y (x) which is a solution of (1) or of the equivalent integral equation
x y(x) y f t, y t dt, …(5) 0 n1 x0 To ensure the existence of a limiting function y(x) lim yn (x) …(6) n we use the fact that yn may be written as a sum of successive differences:
M.A/M.Sc. Mathematics 3rd Semester University of Kashmir, Srinagar 43 Lecture notes on Ordinary Differential Equations-(Unit-I) Khanday M. A.
n1 yn y0 yi1 yi … (7) i0 this shows that the sequence yn x will converge if the infinite series yi1 yi converges.
It is easy to see that the successive differences yi1 yi satisfy the relation
x y (x) y (x) f t, y t f t, y (t) dt, i1 i i i1 … (8) x0 the equation (8) is true for all integers i 0,1,2,3,... for i 0, we have
x y (x) y f t, y dt, … (9) 1 0 0 x0 The condition ensures that the existence of integral (8) and (9). Consider (9) , we have
x y (x) y f t, y dt , 1 0 0 x0
x M dt , by (2) x0
M x x0 , …(10) Again making use of Lipschitz Condition (3), we get from (8),
x y (x) y (x) f t, y t f t, y dt , 2 1 1 0 x0
x M y (t) y dt , 1 0 x0
x M .M t x dt , 0 by (10) x0
x x 2 M 0 . …(11) 2 We may now proceed by mathematical induction. Therefore (10) and (11) show that we shall have
M.A/M.Sc. Mathematics 3rd Semester University of Kashmir, Srinagar 44 Lecture notes on Ordinary Differential Equations-(Unit-I) Khanday M. A.
x x i y (x) y (x) M i 0 , …(12) i1 i1 i! we must show that the inequality (12) holds when i is replaced by i + 1. For this purpose, we again make use of (8) and (3). We have
x y (x) y (x) f t, y t f t, y (t) dt , i i1 i i1 x0
x M y (t) y (t) dt , i i1 x0
x t x i M.M i 0 dt , i! x0
x x i1 M i1 0 . …(13) (i 1)! the relation (13) establishes the validity of (12) for all values of t. From (13), we see that the absolute values of terms in the series (7) are smaller than the corresponding terms in the Taylor‟s series for the function expM x x0 ..
Now the Taylor series for this function converges for all values of ( x – x0), and so the function yn(x) converges uniformly to a function y(x) for all values x, in any finite interval. Proceeding to the limits as n , we get from (4)
x lim yn (x) y0 lim f t, yn1 (t)dt n n x0
x or y(x) y0 lim f t, yn1 (t)dt , …(14) n x0 Since f x, y is continuous function of both x and y is the range of values considered and since yn(x) converges to y(x) uniformly over the interval. Therefore the following interchanges of limiting operations are justified.
x x lim f t, yn (t)dt lim f t, yn (t)dt n n x0 x0
M.A/M.Sc. Mathematics 3rd Semester University of Kashmir, Srinagar 45 Lecture notes on Ordinary Differential Equations-(Unit-I) Khanday M. A.
x
f t,lim yn (t)dt n x0
x f t, y(t)dt , …(15) x0 Thus from (14) and (15), we get
x y(x) y f t, y t dt, 0 x0 this shows that the iterative sequence (4) converges to a solution of the differential equation problem (1) for all values of x under the given conditions. Thus the theorem is completely established. 2: Uniqueness theorem Statement:- The initial Value problem dy f x, y, yx y .... (1) dx 0 0 has a unique solution yx provided the function f x, yis continuous and bounded for all values of x, say f x, y M and satisfies the Lipschitz Condition f x, y2 f x, y1 M y2 y1 , for all values of arguments. Proof:- Suppose if possible the initial value problem (1) has two distinct solutions y(x) and u(x). Then using equation (5) of the above theorem we see that the difference y(x)- u(x) satisfies the relations
x y(x) u(x) f t, yt f t,u(t)dt ...(2) x0 Since f (x,y) is bounded and satisfies Lipschitz condition, we have f (x, y) M …(3) and …(4) using (2) and (3) , we have
M.A/M.Sc. Mathematics 3rd Semester University of Kashmir, Srinagar 46 Lecture notes on Ordinary Differential Equations-(Unit-I) Khanday M. A.
x y(x) u(x) f t, yt f t,u(t)dt x0
x x | f t, ytdt | | f t,u(t)|dt x0 x0
x x Mdt Mdt x0 x0
2M x x0 i.e., y(x) u(x) 2M x x0 …(5) Again using (2) and (4), we have
x y(x) u(x) M y(t) u(t) dt …(6) x0 Combining (5) and (6), we obtain
x x x 2 y(x) u(x) M.2M (t x )dt 2M 2 0 …(7) 0 2! x0 Employing inequality (7) on the right hand side of (6), we have
x t x 2 x x 3 y(x) u(x) M.2M 0 dt 2M 3 0 …(8) 2! 3! x0 Continuing in this way, we shall obtain
x x n y(x) u(x) 2M n 0 , n 1,2,3,... …(9) n! Now the right hand side of (9) tends to zero as n tends to infinity for all values of x, this shows the solution is unique. Method of Variation of Parameters Here we shall explain the method of finding the complete primitive of a linear equation whose complementary function is known. Let y Ax B x be the complementary function of the linear equation of second order d 2 y dy p Qy R …(i) dx2 dx
M.A/M.Sc. Mathematics 3rd Semester University of Kashmir, Srinagar 47 Lecture notes on Ordinary Differential Equations-(Unit-I) Khanday M. A.
Where A and B are constants and x, xare functions of x. Therefore y Ax B x Satisfies the equation d 2 y dy p Qy R dx2 dx or A '' x B '' x PA ' x B ' x QAx B x 0 or A '' x P ' x Qx B ''x P ' x Q x 0 therefore '' x P ' x Qx=0 …(ii) and ''x P ' x Q x=0 …(iii) Now let us assume that y Ax B x …(iv) is the complete primitive of (i) where A and B are functions of x, so chosen that (i) will be satisfied. dy dA dB Therefore A ' x B ' x x x. dx dx dx Let A and B satisfy the equation dA dB x x 0 …(v) dx dx dy Therefore A ' x B ' x dx
2 d y ' ' dA dB and A ' x B ' x x x dx2 dx dx Since the coefficients of A and B are zero by (ii) and (iii) dA dB Therefore ' x ' x =R …(vi) dx dx From (v) and (vi), we have dA x ' x ' x x R x dx dA R x Therefore, dx ' x xx ' x Integrating we get,
M.A/M.Sc. Mathematics 3rd Semester University of Kashmir, Srinagar 48 Lecture notes on Ordinary Differential Equations-(Unit-I) Khanday M. A.
R x A dx c ' x xx ' x 1 Similarly B can be determined from (v) and (vi). Substituting these values of A and B in (iv) we get the complete primitive of (i). Apply the method of variation of parameters to solve the differential equation d 2 y 1. 2 y Cosecx …(1) dx Solution: The auxiliary equation of the given differential equation is m2 1 0, m i Therefore the complementary functions is y = a Cos x + b Sin x …(2) Let y = A Cos x + B Sin x be the complete primitive of the given differential equation. Where A and B are functions of x. Differentiating (2) with respect to x, we get, y1 A1Cosx B1Sinx ASinx BCosx where A1and B1 are chosen such that A Cosx B Sinx 0 …(3) 1 1
Therefore y1 ASinx BCosx Differentiating again both sides, we get y2 A1Sinx B1Cosx ACosx BSinx Substituting these in (1) we have
A1Sinx B1Cosx ACosx BSinx ACosx BSinx Cosecx Implies that A Sinx B Cosx Cosecx 0 1 1 …(4) From (3) and (4), we have
A1Cosx B1Sinx 0 0
A B 1 1 1 SinxCosecx CosxCosecx Cos 2 x Sin2 x
M.A/M.Sc. Mathematics 3rd Semester University of Kashmir, Srinagar 49 Lecture notes on Ordinary Differential Equations-(Unit-I) Khanday M. A.
A B 1 1 1 1 Cotx 1 Implies that
A1 1, B1 Cotx
Now A A dx dx x c 1 Also, we have B B dx Cotxdx log Sinx b 1 Therefore the complete primitive is y a xcos x b log Sinxsin x .
d 2 y 2. 4y 4 tan 2x … (1) dx2 Solution:- The auxiliary equation of the given differential equation is m2 4 0 m 2i Now the complementary function of the given differential equation is y = a Cos 2x + b Sin 2x …(2) Let y = A Cos 2x + B Sin 2x be the complete primitive of the given differential equation. Where A and B are functions of x. Differentiating (2) with respect to x, we get, y1 A1Cos2x B1Sin2x 2ASin2x 2BCos2x where A1and B1 are chosen such that A Cos2x B Sin2x 0 …(3) 1 1
Therefore y1 2ASin2x 2BCos2x Differentiating again both sides, we get y2 2A1Sin2x 2B1Cos2x 4ACos2x 4BSin2x
Substitute the values of y, y1 and y2 in the given differential equation, we have
2A1Sin2x 2B1Cos2x 4ACos2x 4BSin2x 4ACos2x BSin2x 4tan 2x 2A Sin2x 2B Cos2x 4tan 2x 0 or 1 1 …(4) Solving (3) and (4), we get,
A1Cos2x B1Sin2x 0 0
M.A/M.Sc. Mathematics 3rd Semester University of Kashmir, Srinagar 50 Lecture notes on Ordinary Differential Equations-(Unit-I) Khanday M. A.
A B 1 1 1 4Sin2x tan 2x 4Cos2x tan 2x 2
A1 2Sin2x tan 2x and B1 2Cos2x tan2x Thus we have Sin2 2x 1 Cos 2 2x A A dx 2Sin2x tan 2xdx 2 dx 2 dx 1 Cos2x Cos2x 2Sec2x 2Cos2xdx Therefore, A logSec2x tan 2x Sin2x c
Also B B dx 2Sin2xdx Cos2x d 1 Hence, the complete primitive of the given differential equation is y logSec2x tan 2x Sin2xcos2x Cos2xsin 2x h 2 4. y y 2 1 ex Solution:- The given differential equation can be written as
2 2 D 1y x …(1) 1 e Now the Auxiliary equation is given by m2 1y 0 , m 1,1 Therefore the complementary function is given by y aex bex
Let y Ae x Bex …(2) be the solution of (1), where A and B are constants. Differentiating (2) with respect to x, we get,
x x x x y1 Ae Be A1e B1e
where A1and B1 are chosen such that A e x B e x 0 1 1 …(3) Therefore, we have
x x y1 Ae Be Differentiating again, we have
M.A/M.Sc. Mathematics 3rd Semester University of Kashmir, Srinagar 51 Lecture notes on Ordinary Differential Equations-(Unit-I) Khanday M. A.
x x x x y2 Ae A1e Be B1e Substituting these values in the given differential equation, we have; 2 Aex A e x Bex B ex Aex Bex 1 1 1 e x or we can write, this as
x x 2 A1e B1e x …(4) 1 e Solving (3) and (4) we get
x x A1e B1e 0 0 2 A e x B ex 1 1 1 e x 1 e x1 we get, A and B 1 ex 1 ex 1 1 e x Integrating these, we get
1 1 1 A A dx dx dx ex log 1 ex b 1 x x x x e 1 e e 1 e or A ex log1 ex x b e x Also B B dx log1 e x c 1 1 e x or B log1 e x c Hence the complete primitive of the given differential equation is y Ae x Be x ex log1 e x x be x log1 e x cex
2 2 x 5. x y2 xy1 y x e …(1)
given that the complementary function is ax bx1. Solution:- We have the complementary function is y a bx1.
-1 Let y = Ax + Bx …(2) be the solution of (1), where A and B are functions. Differentiating (2) with respect to x, we get,
M.A/M.Sc. Mathematics 3rd Semester University of Kashmir, Srinagar 52 Lecture notes on Ordinary Differential Equations-(Unit-I) Khanday M. A.
2 1 y1 A A1x Bx B1x Where A and B are Chosen such that, -1 A1x + B1x = 0 …(3) Therefore we have
2 y1 A Bx Differentiating again we have,
3 3 y2 A1 2Bx B1 x …(4) Substituting these in (1), we get
2 3 3 2 1 1 2 x x A1 2Bx B1 x xA A1 x Bx B1 x Ax Bx x e e x x 2e x This gives A and B 1 2 1 2 Integrating we get, e x e x A A dx dx a 1 2 2 x 2e x 1 B B dx dx x 2e x xe x e x b 1 2 2 Hence the complete solution is given by (2) as,
x e 1 1 2 x x x y x a x x e xe e b is the required solution. 2 2 6. y 6y 11 y 6y e 2x …(1) 3 2 1 Solution:- The given differential equation can be written as (D 1)(D 2)(D 3)y e2x Now the Auxiliary equation of the equation is given by (m 1)(m 2)(m 3) 0 implies m 1, 2 ,3, Therefore, the complementary function is y aex be2x ce3x
x 2x 3x Let y Ae Be Ce …(2) is the solution of (1).
x x 2x 2x 3x 3x Therefore y1 Ae A1e B1e 2Be C1e 3Ce
M.A/M.Sc. Mathematics 3rd Semester University of Kashmir, Srinagar 53 Lecture notes on Ordinary Differential Equations-(Unit-I) Khanday M. A.
where A1, B1 and C1 are chosen such that x 2x 3x A1e + B1e +C1e = 0 …(3) Differentiating again, we get
x x 2x 2x 3x 3x y2 Ae A1e 4Be 2B1e 9Ce 3C1e Ae x 4Be 2x 9Ce3x
A e x 2B e2x 3C e3x 0 where 1 1 1 …(4) Differentiating again, we have y = Aex + A ex +8Be2x + 4B e2x +27Ce3x +9C e3x 3 1 1 1 …(5) Substituting these in the given differential equation, we have
x x 2x 2x 3x 3x Ae A1e 8Be 4B1e 27Ce 9C1e 6Aex A e x 4Be2x 2B e 2x 9Ce3x 3C e 1 1 1 x x 2x 2x 3x 3x 11Ae A1e B1e 2Be C1e 3Ce 6Aex Be2x Ce3x e 2x . 6A e x 3B e2x 2C e3x e2x or 1 1 1 …(6) Solving (3), (4) and (6), we have
x 2x 3x A1e B1e C1e 0
x 2x 3x 2x 6A1e 3B1e 2C1e e 0 A B C 1 1 1 1 e2x e3x 0 e x e3x 0 e x e2x 0 e x e2x e3x 2e2x 3e3x 0 e x 3e3x 0 e x 2e2x 0 e x 2e2x 3e3x 3e2x 2e3x e2x 6e x 2e3x e2x 6e x 3e2x e2x 6e x 3e2x 2e3x
A B C 1 or 1 1 1 e 2x e5x e 2x 2e 4x e 2x e3x 2e 6x Therefore, e x 1 A ; B 1 ; C 1 2 1 1 2e 2x Integrating each term we get,
M.A/M.Sc. Mathematics 3rd Semester University of Kashmir, Srinagar 54 Lecture notes on Ordinary Differential Equations-(Unit-I) Khanday M. A.
e x 1 A a ; B x b and C c 2 2e x Using these in (2), we have
e x 1 y ae x b x e2x c e3x x 2 2e or y aex (b x)e 2x ce3x is the required solution.
d 2 y 7. n 2 y Secnx dx2 Solution:- Here the complementary function is y ACosnx BSinnx where A and B are constants,
Let y ACosnx BSinnx …(1) Be the complete primitive of the given differential equation, where A and B are functions of x, dy dA dB AnSinnx BnCosnx Cosnx Sinnx. dx dx dx Now choosing A and B such that dA dB .Cosnx .Sinnx 0 …(2) dx dx dy we have AnSinnx BnCosnx. dx d 2 y dA dB An 2Cosnx Bn 2 Sinnx n Sinnx n Cosnx. dx2 dx dx Substitute in the given differential equation, we have dA dB n Sinnx n Cosnx Secnx …(3) dx dx Now multiplying (2) by n Cos nx and (3) by Sin nx , then subtracting we get, dA n tan nx dx 1 Therefore A log Cosnx c n2 1
M.A/M.Sc. Mathematics 3rd Semester University of Kashmir, Srinagar 55 Lecture notes on Ordinary Differential Equations-(Unit-I) Khanday M. A.
Again multiplying (2) by nSinnx and (3) by Cos nx and adding we get, dB n 1 dx x Therefore B c n 2 Substituting the values of A and B in (1), the complete primitive of the given differential equation is 1 x y c Cosnx c Sinnx Cosnx.log Cosnx Sinnx. 1 2 n2 n Home Assignment Solve the following by the method of Variation of Parameters.
d 2 y dy 1. x 2 x y x 2e x dx2 dx d 2 y dy 2. x 2 2x(1 x) 2(x 1)y x3 dx2 dx d 2 y dy 3. (1 x) x y (1 x) 2 dx2 dx d 2 y dy 4. x (1 x) y e x dx2 dx
M.A/M.Sc. Mathematics 3rd Semester University of Kashmir, Srinagar 56