411. DIFFERENTIAL EQUATIONS.
1.1 Introduction.
Definition: An Ordinary Differential Equation (ODE) is an equation that contains one or several derivatives of an unknown function. dy Example: 1. dx = sinx + 3 2. y`` + 2y` - 6y = ex d 3 x d 2 y dy 3. 3 + 2 2 - - y = 0. dy dx dx
Notation: F(x, y, y`, y``, … ) = 0.
Standard Notations: If y(x) is a function of x , then the first order differential equation can be written as dy y'(x) y' dx or or Similarly the second order differential equation can be written as d 2 y y''(x) y'' 2 or or dx and in general for nth differential equation we have d n y y (n) (x) y (n) n or or dx
The order of an ordinary differential equation is the order of the highest derivative that appears in the differential equation.
1 dy Example: 1. dx + y tan x = sin 2x. (first order) 2 2 d y dy -1 2. x 2 - 4x + 6y = x . (second order) dx dx 3. y`` - 4y` + 4y = 5x2 + e-x. (second order) 4. y``` - 3y`` + 3y – y = 0. (third order)
1.2 How to form ODE.
A differential equation could be formed by eliminating an arbitrary constant from a given function.
Example 1. Form ODE from the function y = Ax + x2. (A constant) Solution: y = Ax + x2 … (i) y`= A + 2x … (ii) → x(ii) : xy` = Ax + 2x2 (i) : y = Ax + x2 ______xy` - y = x2 (iii) xy`- y = x2 . This is a first order differential equation which derived from y = Ax + x2.
2 A Example 2. Form ODE from the function y = x + x . 2 A Solution: y = x + x . Multiply with x, then yx = x3 + A. Differenciate with respect to x, dy 2 → y + x dx = 3x is the first order ODE.
2 Example 3. Form ODE from the function: y = Ax2 + Bx5.
Solution: y = Ax2 + Bx5 …. (i) y`= 2Ax + 5Bx4…. (ii) y``= 2A + 20Bx3…. (iii)
x(ii): xy`= 2Ax2 + 5Bx5 2(i) : 2y = 2Ax2 + 2Bx5 ______- xy`-2y = 3Bx5 …… (iv)
x(iii): xy`` = 2Ax + 20 Bx4 (ii): y` = 2Ax + 5Bx4 ______xy``- y`= 15 Bx4 ….. (v)
x(v): x2y``- xy` = 15Bx5 5(iv): 5xy` - 10y = 15Bx5 ______x2y``- 6 xy`+ 10y = 0 (second order ODE )
Example 4. Form ODE from the function y = Aex + Be-2x
Solution: y = Aex + Be-2x …. (i) e2x(i): ye2x = Ae3x + B. …. (ii)
Differentiating (ii): y`e2x + 2e2xy = 3Ae3x ….(iii)
Differentiating (iii): y``e2x + 2e2xy`+ 2e2xy`+4e2xy = 9Ae3x.
Or: y``e2x + 4e2x y` + 4e2xy = 9Ae3x …. (iv)
3 3(iii): 3y`e2x + 6e2xy = 9Ae3x ______y``e2x + y`e2x - 2e2xy = 0
e2x(y``+ y` - 2y) = 0. But e2x ≠ 0
Thus the solution is: y`` + y` - 2y = 0
1.3. Solution of a Differential Equation.
Definition: If y = F(x) is the solution of an ODE, hence a function F(x) satisfies the given differential equation. d 2 y dy Examp. 5. Given 2 + - 6y = 0. Show that: dx dx (a) y = e2x is the solution. (b) y = 5e2x + 4e-3x is the solution. (c) y = xe2x is not the solution.
Solution: 2 2x dy 2x d y 2x (a) y = e …(i) thus = 2e …(ii) and 2 = 4e …(iii) dx dx Substitute (i), (ii) dan (iii) into the given diff. eq. hence
2 d y dy 2x 2x 2x 2 + - 6y = 4e + 2e – 6e = 0. dx dx It is shown that y = e2x is the solution.
(b) y = 5e2x + 4e-3x dy 2x -3x dx = 10e – 12e
4 2 d y 2x -3x 2 = 20e + 36e dx 2 d y dy 2x -3x 2x -3x → 2 + - 6y = 20e + 36e + 10e – 12e dx dx -30e2x – 24e-3x = 0 y = 5e2x + 4e-3x is the solution.
(c) y = xe2x y` = 2xe2x + e2x y``= 2e2x + 4xe2x + 2e2x = 4xe2x + 4e2x. → y``+ y` - 6y = 4xe2x + 4e2x + 2xe2x + e2x – 6e2x = 5e2x ≠ 0. y = xe2x is not the solution.
Example 6. Find the value of m so that y = emx is the solution of the diffrential equation 2y`` + 5y` - 3y = 0.
Solution: Given y = emx …..(i), thus y`= memx …(ii) and y``= m2emx …(iii) Substitute (i), (ii) and (iii) into the ODE, hence
2y``+ 5y` - 3y = 2m2emx + 5memx – 3emx = emx(2m2 + 5m – 3) = 0. But emx ≠ 0 hence, 2m2 + 5m – 3 = 0 (2m -1)(m + 3) = 0 m = { ½ , -3}.
5 1.4 General & Particular Solution.
Example 7. Show that y = Aex + (x + 2)e2x is the general solution of the differential equation dy 2x dx - y = (x + 3)e , and hence determine the value of A given that y = 4 when x = 0.
Solution: y = Aex + (x + 2)e2x dy x 2x 2x dx = Ae + 2(x + 2)e + e = Aex + (2x + 5)e2x
dy x 2x x 2x → dx - y = Ae +(2x + 5)e – Ae – (x + 2)e = (2x + 5 – x – 2)e2x = (x + 3)e2x. (shown)
Given that y = 4 when x = 0 → y = Aex + (x + 2)e2x 4 = Ae0 + (0 + 2)e0 4 = A + 2 → A = 2 Particular solution: y = 2ex + (x + 2)e2x
6 The particular solution could be obtained by substituting the given condition (y = 4 when x = 0). The conditions are called the initial condition of the differential equation.
Definition: (i) Initial Value Problem (IVP) is a differential equation with initial conditions. (Ex. y = 1 and y`= 2 when x = 0)
(ii) Boundary Value Problem (BVP) is a diff. equation with boundary conditions. (Ex. y = 0 when x = 0 and y`= 2 when x = 1)
Akos3x Bsin 3x Example 8. Show that y = x is the general d 2 y dy solution for x 2 + 2 + 9xy = 0. dx dx And hence obtain the particular solution with condition y() = -3 and y`() = 0.
Solution:
The conditions above are an initial condition (IVP) y = -3 and y`= 0 when x = .
Given: yx = A kos3x + B sin 3x … (i) dy x dx + y = -3A sin3x + 3B kos3x … (ii) d 2 y dy dy x 2 + + = -9A kos3x – 9B sin3x. dx dx dx d 2 y dy x 2 + 2 = -9(A kos3x + B sin3x) … (iii) dx dx Substitute (i) into (iii), thus:
7 d 2 y dy x 2 + 2 + 9xy = 0. (shown) dx dx
Substitute y() = -3 into (i) → -3 = -A or A = 3 y`() = 0 into (ii) → y = -3B -3 = -3B or B = 1. 3kos3x sin3x The particular solution: y = x .
3 B Example 9. Show that y = Ax + x3 is the general solution for x2y`` + xy` - 9y = 0 and hence obtain the particular solution with conditions y(2) = 1 and y`(1) = 0.
Solution: The condition above are a boundary condition (BVP), y(2) = 1 and y`(1) = 0.
3 B 3 6 y = Ax + x3 or x y = Ax + B … (i). Differentiating (i), thus 3x2y + x3y`= 6Ax5 → xy` = 6Ax3 – 3y … (ii). Differentiating (ii), thus xy``+ y`= 18Ax2 – 3y` → xy``= 18Ax2 – 4y` …(iii) Substitute (ii) and (iii) into given diff. equation,
x2y``+ xy`- 9y = 18Ax3- 4y`x + xy`- 9y = 18Ax3 -3(6Ax3- 3y) – 9y = 18Ax3 – 18Ax3 + 9y – 9y = 0 3 B Thus: y = Ax + x3 is the general solution.
8 Substituting y(2) = 1 or y = 1 when x = 2 3 B into diff. equation y = Ax + x3 we get 1 1 = 8A + 8 B or 8 = 64A + B…(iv) Substituting y`(1) = 0 or y`= 0 when x = 1 into xy`= 6Ax3 – 3y we get 3 3 B xy`= 6Ax – 3(Ax + x3 ) 3 3B xy` = 3Ax - x3 0 = 3A – 3B → A = B … (v) Fron simuntaneous equation (iv) dan (v), thus
8 64A + A = 8 → A = B = 65 8 3 1 Particular equation: y = 65 (x + x3 ).
9 2. First Order Ordinary Differential Equation (ODE)
dy General Form: dx = f(x,y)
dy Example: a) dx = 2y + sin x. 2 b) dy = x (1 x)y . dx 2x
There are four types of a first order ODE, i) Separable differentiel equation. ii) Homogeneous differential equation. iii) Linear differential equation. iv) Exact differential equation.
2.1. Separable Differential Equation.
The differential equation: y` = f(x,y) is said to be separable if the equation can be written as the product of a function of x, u(x) and the function of y, v(y). The equation can be wtitten in the form
dy dy dy = u(x).v(y) or v(y) = u(x).dx hence, integration both sides:
dy ∫ v(y) = ∫ u(x) dx.
10 dy Example 1. Solve the equation: (x + 2) dx = y. dy Solution: (x + 2) dx = y dy dx ∫ y = ∫ x 2 ln|y| = ln|x+2| + C y c ln| x 2 | = e = A y = A(x+2).
x dy 2 Example 2. Solve the equation: e dx + xy = 0. Solution:
x dy 2 e dx + xy = 0. dy -x ∫ y 2 = - ∫ xe dx.
1 -x -x d(x) - y = -[x ∫e dx - ∫{e dx} dx dx. 1 -x -x y = -xe -∫-e dx 1 -x -x y = -xe – e + C. 1 -x y = -(x+1) e + C.
11 Example 3. Solve the following differential equation: x2y dx + (x + 1) dy = 0 which satisfied condition y = 2 when x = 0.
Solution:
x2y dx + (x + 1) dy = 0 dy 2 - = x dx. y x 1 dy 1 - y = {(x – 1) + x 1 }dx. dy dx -∫ y = ∫(x – 1)dx + ∫ x 1 2 -ln|y| = x - x + ln|x + 1| + C. 2 ln|y(x + 1)| = x – ½ x2 – C.
y(x + 1) = ex-1/2 x 2 -C
y(x + 1) = A.ex-1/2 x 2 , where A = e-C
y = 2 when x = 0, thus: 2 = A.
The solution is:
2 x- ½ x 2 y = x 1 . e
12 2.1.1. Substitution Method.
dy x y 1 Example 4. Solve the equation: dx = x y 5 which satisfied the condition y(1) = 1.
Solution : Subsitute z = x + y dz dy dy dz dx 1 + dx thus dx = dx - 1 dz z 1 → dx - 1 = z 5 dz z 1 2z 6 2(z 3) dx = z 5 + 1 = z 5 = z 5
z 5 z 3 dz = 2 dx. 2 ∫(1 + z 3 ) dz = ∫2 dx. z + 2ln|z+3| = 2x + C.
2ln|z+3| = 2x – x – y + C
(z + 3)2 = A.ex-y, where A = eC.
y(1) = 1 → (1+1+3)2 = A.e1-1 25 = A
The solution is: (x + y + 3)2 = 25 ex-y .
Example 5. Solve the equation:
dy 2 2 x dx + y = 2x((1 + x y ).
13 Solution: Substitute z = xy, hence dz dy x y dx dx
dz 2 → dx = 2x(1 + z ) dz ∫ 1 z 2 = ∫ 2xdx. tan-1 z = x2 + C. z = tan(x2 + C) xy = tan(x2 + C).
2 y = tan(x C) x
2.2 HOMOGENEOUS EQUATION.
dy Consider the differential equation dx = f(x, y). If: f(λx, λy) = f(x, y) for each , hence dy dx = f(x, y) is called a homogeneous equation.
dy xy Example: i). dx = x 2 y 2 = f(x, y) (x)(y) 2 (xy) f(λx, λy) = (x)2 (y)2 = 2 (x 2 y 2 ) xy = x 2 y 2 = f(x, y) [homogeneous].
dy ii). dx = x – y = f(x, y). f(λx, λy) = λx – λy = λ(x – y) ≠ f(x, y). f(x, y) non-homogeneous.
14 The method of solving a homogenous diff. equation is by using the following substitution.
dy dv y = x.v, hence dx = x dx + v
Example 6. Solve the differential equation dy xy dx = x 2 y 2 with condition y(0) = 2.
dy dv Solution: By using substitution y = xv and dx = x dx + v. dv x(xv) v Thus: x dx + v = x 2 (xv)2 = 1 v 2 dv v v v(1 v 2 ) v3 x = 2 - v = 2 = - 2 dx 1 v 1 v 1 v
1 v 2 dx ∫ 3 dv = - dx. ( v ) ∫ x 1 2v 2 + ln |v| = -ln|x| + C. 1 y ln |xv| = 2v 2 + C. [v = x ] y = A.ex 2 /2y 2 , where A = eC
Then y(0) = 2 , hence A = 2.
The solution is: y = 2ex 2 /2y 2
15 Example 7. Solve the differential equation dy 2x y dx = x 2y with condition y(3) = 1. Solution: 2x y (2x y) 2x y f(λx, λy) = x 2y = (x 2y) = x 2y = f(x, y). dy dv x v Substitute y = xv and dx dx , hence dv 2x xv 2 v x dx + v = x 2xv = 1 2v . 2 x dv = 2 v - v = 2(v 1) . dx 1 2v 2v 1
2v 1 dx ∫( v 2 1 )dv = ∫-2 x 1 3 dx ∫{ 2(v 1) + 2(v 1) }dv = -∫ 2 x .
1 3 2 ln|v + 1| + 2 ln|v – 1| = -2ln|x| + C ln|v + 1| + 3ln|v – 1| = -4ln|x| + 2C
(v + 1)(v – 1)3.x4 = A , where A = e2C y x y x 3 4 ( x )( x ) .x = A (y + x)(y – x)3 = A
The condition y(3) = 1 → A = -32.
The solution is: (y + x)(y – x)3 + 32 = 0.
16 dy 2 2 Example 8. Solve: x dx - y = x – y , with condition y = 1 when x = 1.
Solution: Substitute y = xv, hence: dv 2 x(x dx + v) – xv = x√ 1 - v . dv 2 x dx = √ 1 – v dv dx ∫ 1 v 2 = ∫ x
sin-1 v = ln|x| + C -1 y sin ( x ) = ln|x| + C The condition y = 1 when x = 1, thus
-1 sin (1) = 0 + C → C = 2 .
-1 y Solution: sin ( x ) = ln|x| + 2
17 2.3 Linier differential Equations.
dy Note: a(x) dx + b(x).y = c(x). dy b(x) c(x) dx + a(x) .y = a(x) dy b(x) or: dx + p(x).y = q(x) where p(x) = a(x) c(x) and q(x) = a(x) This is the general form of a linier differential equations.
The Method of Solution.
dy i) Write to the general form : dx + p(x).y = q(x) ii) Determine p(x) and evaluate : ∫ p(x) dx.
iii) Obtain the integrating factor : u(x) = e∫ p(x)dx.
dy iv) u(x) dx + u(x).p(x).y = u(x).q(x). d v) Write dx {u(x).y} = u(x).q(x). vi) ∫ d(u(x).y = ∫ u(x).q(x)dx.
vii) u(x).y = ∫ u(x).q(x)dx.
dy 3 Example: i). x dx + y = x dy x ii). dx - y = 2e 2 dy 2 iii). (1 + x ) dx - xy = x(1 + x )
18 dy 3 Solution i). x dx + y = x dy y 2 dx + x = x 1 1 p(x) = x → ∫p(x)dx = ∫ x dx = ln x. Integrating factor: u(x) = e∫p(x)dx = eln x = x.
y.x = ∫x.x2 dx = ∫x3dx 1 4 = 4 x + C 1 3 C → y = 4 x + x .
dy x Example ii). dx - y = 2e . p(x) = -1 → ∫p(x) dx = ∫(-1) dx = -x. Integrating factor: u(x) = e∫p(x) dx = e-x. e-x.y = ∫e-x.2ex dx = 2x + C → y = 2xex + Cex.
2 dy 2 Example iii): (1 + x ) dx - xy = x(1 + x ) dy x dx - (1 x 2 ).y = x x 2 -x/2 ∫p(x)dx = ∫-(1 x 2 ) )dx = ln(1 + x ) u(x) = e∫p(x)dx = eln(1+x 2 ) 1/ 2 = (1+x2)-1/2 2 -1/2 x 2 (1+x ) .y = ∫( (1 x 2 )1/ 2 )dx. Substitute z = (1+x )
x 2 1/2 hence ∫( (1 x 2 )1/ 2 )dx = (1 + x ) + C (1+x2)-1/2.y = (1 + x2)1/2 + C.
→ y = (1 + x2) + C(1 + x2)1/2
19 2.4. Exact Equations.
General form: M(x,y) dx + N(x,y) dy = 0.
M N Condition of an Exact Equation: y x
Example: i) (2x + 3y2) dx + (6xy + 2y) dy = 0 ii) (3x2y + ey) dx + (x3 + xey – 2y) dy = 0 iii) (2x + y – kos y) dx + (4y + x + sin x) dy = 0.
The method of solution.
M N a) M dx + N dy = 0. Test for exactness: y x u b) Write x = M …….. (i) u y = N ………(ii) c) Inregrate with renpect to x: ∫ du = ∫ M dx u = ∫ Mdx + Q(y) ….. (iii) d) Differentiate (iii) with respect to y.
e) Equate: u(x,y) = A.
Example: Solve the following differential equation.
(6x2 – 10 xy + 3y2) dx + (6xy – 5x2 – 3y2) dy = 0
20 Exercises.
1. Solve the differential equations: dy 2x 1 2x -3x i) dx + 3y = e [ y = 5 e + Ce ] dy 2 2 -x ii) dx + y = x [ y = x - 2x + 2 + Ce ] dy 2 1 iii) sin x dx + 2y kos x = kos x [ysin x = A- 4 kos2x] dy 1 iv) sin x dx - y kos x = cot x. [y = - 2 kosek x+Csin x]
2. Show that these equations is exact and solve.
3 1 y i) (y - x ) dy + x 2 dx = 0 ii) (3x2 – y sin xy) dx – x sin xy = 0 iii) (2x + 3 kos y) dx + (2y – 3x sin y) dy = 0.
21 3. Second Order Linier Differential Equation (LDE)
General Form: d n y d n1 y dy an(x) n + an-1(x) n1 + …+ a1(x) + a0(x)y = f(x) …(1) dx dx dx where the coefficients a0(x), a1(x),…, an(x), f(x) is the function of x and an(x) ≠ 0.
If one of the coefficients is not constant, hence (1) is called a Linear Differential Equation with variable coefficient.
If all of the coefficients are constants, hence (1) could be written as:
d n y d n1 y dy an n + an-1 n1 + … + a1 + a0y = f(x) … (2) dx dx dx
(2) is called a Linier Differential Equation with constant coefficient.
If f(x) in (1) and (2) equal to zero, is called a Homogeneous Differential Equation (HDE). If f(x) ≠ 0, is called Non Homogeneous Diff. Equation.
Examples: d 2 y a) 2 + 20y = 0 HDE with constant coeffficient. dx b) y``- 5y` + 3y = ex Non HDE with constant coefficient. c) x2y``+xy`+(x2-2)y = 0 HDE with variable coefficient.
22 d 2 y 2 dy d) 2 + = ln x Non HDE with variable coefficient. dx x dx
3.1 The Method of Solution for A Homogeneous Differentiel Equation.
Consider a second order linier differential equation:
d 2 y dy a 2 + b + cy = 0 where a, b, c constant. …… (3) dx dx
If y = emx is the solution, hence 2 dy mx d y 2 mx = me and 2 = m e dx dx Substitute into (3), hence d 2 y dy a 2 + b + cy = 0 can be written as: dx dx am2emx + bmemx + cemx = 0. (am2 + bm + c) emx = 0. But emx ≠ 0, hence
am2 + bm + c = 0 ……. (4).
(4) is the quadratic equation and called characteristic equation. The roots of (4) are called the characteristic roots.
Equation (4) has three forms of roots. .
(i) Real and different roots, if b2 – 4ac > 0. (ii) Real and equal roots, if b2 - 4ac = 0. (iii) Two complex roots, if b2 - 4ac < 0.
23 Let m1 and m2 are the characteristic roots of equation (4).
2 m x a) If b – 4ac > 0 hence m1 ≠ m2. Then y1 = e 1 and y2 m x = e 2 are the solutions of the homogeneous equation. Then the general solution written as:
m x m x y = A e 1 + B e 2 { A, B constants}.
2 b) If b – 4ac = 0 hence m1 = m2. The characteristic b equation has only one root, m = - 2a . Then the general solution written as:
y = (A + Bx) emx. {A, B constants} c) If b2 – 4ac < 0 the characteristic equation has two complex roots, m1 = α + βi dan m2 = α – βi . Then the general solution written as :
y = C.e(α + βi)x + D.e(α – βi)x {C, D constans}.
By using Euler formula:
eiθ = kosθ + i sinθ and e-iθ = kosθ – i sinθ, then
y = C.e(α + βi)x + D.e(α – βi)x = eαx { C.eiβx + D.e-iβx } = eαx { C(kos βx + i sin βx) + D(kos βx – i sin βx)} = eαx {(C + D) kos βx + i(C – D) sin βx} = eαx { A kos βx + B sin βx } where A = C + D and B = (C – D)i.
24 Conclution: If characteristic equation has two complex roots, m1 = α + βi and m2 = α – βi , then the general solution could be written as:
y = eαx ( A kos βx + B sin βx )
αx αx Hence y1 = e kos βx and y2 = e sin βx
Exercises: Determine the general solution from the folowing equations:
1). y`` - y` - 6y = 0 2). y`` - 4y = 0 3). y`` - 2y` - 3y =0 with conditions y(0) = 2 and y`(0) = 1 4). y`` - 4y`+ 13y = 0, y(0) = -1, y`(0) = 2.
25 3.2 Non Homogeneous Linier Equations.
d 2 y dy a 2 + b + cy = f(x) dx dx
2 d y dy -2x Example: Solve the equation: 2 - 4 + 3y = 10 e . dx dx Solution: f(x) = 10 e-2x has ex expression.
Let Ce-2x is the solution.
2 -2x dy -2x d y -2x Thus: y = Ce ; = -2Ce ; 2 = 4Ce . dx dx 2 d y dy -2x 2 - 4 + 3y = 10e . Substitute : dx dx 4Ce-2x -4(-2Ce-2x) + 3Ce-2x = 10e-2x. 15Ce-2x = 10e-2x. C = 2/3. 2 -2x Hence y = 3 e satisfied the given equation and is called the particular integral.
The other solution which could be obtain from d 2 y dy homogenous equation 2 - 4 + 3y = 0. dx dx Characteristic equation: m2 – 4m + 3 = 0. → m = {1, 3}. x 3x The solution of HDE: yc = Ae + Be .
x 3x 2 -2x General Solution: y = Ae +Be + 3 e (A, B constants).
26 Definition: d 2 y dy i) The general solution of equation: a 2 + b + cy = 0 dx dx is yc , called complementary function.. d 2 y dy ii) The solution of : a 2 + b + cy = f(x) is y , called dx dx p pacticular integral.
Teorem: If yc is the complementary function for diff. equation d 2 y dy a 2 + b + cy = 0 and yp is the particular integral for dx dx d 2 y dy non homogenous equation a 2 +b + cy = f(x), hence dx dx the general sulution of the non homogenous equation is given by: y = yc + yp.
3.2.1. Method of Undertemined Coefficients.
Consider : ay``+ by` + c = f(x), a ≠ 0. ………. (i).
The basic idea behind this approach is as follows. a) f(x) a polynomial of degree n. b) f(x) an exponential form Ceαx , (α, C constants). c) f(x) = C kosβx or C sin βx, (C, β constants).
n n-1 Case a: f(x) = Anx + An-1x + … + A1x + Ao . (An , An-1 , … , A1 , Ao constants).
27 n n-1 Suppose: yp = Bnx + Bn-1x + … + B1x + B0. ……(ii). (Bn , Bn-1 , … , B1 , Bo constants). (n) Differentiate (ii) for yp`, yp``, … , yp and substituting into (i). Equate the coefficients of corresponding powers of x, and solve the resulting equations for undertemined coefficients, then we get: B1 , B2 , … , B1, Bo .
Example: Solve the diff. equation: y`` + 3y` + 2y = 5x2. Solution: f(x) = 5x2 2 . Suppose: yp = ax + bx + c
yp` = 2ax + b yp`` = 2a
→ y``+ 3y`+ 2y = 5x2. 2a + 3(2ax + b) + 2(ax2+ bx + c) = 5x2. 2ax2 + (6a + 2b)x + (2a + 3b + 2c) = 5x2.
5 Hence: 2a = 5 → a = 2 . 15 6a + 2b = 0 → b = - 2 . 35 2a + 3b + 2c = 0 → c = 4 . 5 2 15 35 → yp = 2 x - 2 x + 4 .
Consider: y``+ 3y`+ 2 = 0. (HDE). Characteristic eq. : m2 + 3m + 2 = 0. (m + 1)(m + 2) = 0. m = {-1, -2}. -x -2x → yc = Ae + Be .
28 -x -2x 5 2 15 35 y = yc + yp. or: y = Ae + Be + 2 x - 2 x + 4 . Example: Solve the equation: y`` - 2y` + y = x2 – 3x.
Solution: f(x) = x2 – 3x. 2 Suppose: yp = ax + bx + c then yp`= 2ax + b and yp``= 2a
y``- 2y`+ y = x2 – 3x. 2a – 2(2ax + b) + ax2 + bx + c = x2 – 3x Hence: a = 1; b = 1; c = 0. 2 → yp = x + x.
Consider: y`` - 2y`+ y = 0 (HDE) Characteristic equation: m2 – 2m + 1 = 0 m = 1 x → yh = (A + Bx).e . General solution: y = (A+Bx)ex + x2 + x.
Case b: f(x) = Ceαx , (C, α constants). Then: ay``+ by`+ cy = Ceαx. ….. (iii)
αx Suppose : yp = k.e , then, αx yp`= αke . 2 αx yp``= α ke . = By substituting yp , yp`, and yp`` into (iii), then
[a(α2k) + b(αk) + ck].eαx = Ceαx.
29 or: aα2k+ bαk + ck = C .
Example: Solve y`` - y` - 2y = 2e3x.
Solution : f(x) = 2e3x 3x Suppose yp = ke . 3x yp`= 3ke 3x yp``= 9ke
y`` - y` - 2y = 2e3x 9ke3x – 3ke3x – 2ke3x = 2e3x 4ke3x = 2e3x 1 k = 2 1 3x Then: yp = 2 e .
Consider: y``- y`- 2y = 0. Charac.eq: m2 – m – 2 = 0 m = {2, -1} 2x -x Thus : yc = Ae + Be
2x -x 1 3x The genenal solution is: y = Ae + Be + 2 e .
Case c: f(x) = C cos αx or C sin αx. (C, α constants) Then ay``+ by` + cy = C cos αx or ay``+ by` + cy = C sinαx For the two expressions, suppose
yp = P cos αx + Q sin αx yp` = -αP sin αx + αQ cos αx
30 2 2 yp``= -α P cos αx – α Q sin αx.
Substituting yp , yp` dan yp`` into the given equation, then equate the coefficient of corresponding sinαx or kosαx.
Example. Find the general sulution of the equation y``+ 9y = cos 2x. Solution. The characteristic equation of the homogeneous equation is m2+ 9 = 0 and its roots are m = ± 3i. The complementary fuction is yc = A cos 3x + B sin 3x.
We choose the particular integral is yp = p cos 2x + q sin 2x yp`= -2p sin 2x + 2q cos 2x. yp``= -4p cos 2x – 4q sin 2x. Substituting in the given equation we get
y``+ 9y = -4p cos 2x – 4q sin 2x + 9(p cos 2x + q sin 2x) = 5p cos 2x + 5q sin 2x = cos 2x.
1 → 5p = 1 → p = 5 5q = 0 → q = 0 1 Then yp = 5 cos 2x.
1 The general solution: y = A cos 3x + B sin 3x + 5 cos 2x.
Exercises: Solve the equation.
31 a) y``+ y` - 6y = 52 cos2x. b) y``- y`- 2y = cos x+ 3 sin x.
Case d: f(x) = f1(x) ± f2(x) ± f3(x) ± … ± fn(x).
For this case, suppose: yp = yp1 + yp2 + yp3 + … + ypn , where yp1 is the particular integral for ay``+ by`+ cy = f1(x) yp2 is the particular integral for ay``+ by`+ cy = f2(x) . Ypn is the particular integral for ay``+ by` + cy = fn(x)
General Solution: y = yc + yp .
Example: Solve the differential equation y``+ 2y`+ 2y = x2 + sin x.
Solution: Characteristic equation: m2 + 2m + 2 = 0 m = -1 ± i. -x yc = e (A cos x + B sin x).
2 (i) Suppose yp1 is particular integral for y``+ 2y`+ 2y = x . 2 Then yp1 = ax + bx + c yp1` = 2ax + b and yp1``= 2a .
2a + 2(2ax + b) + 2(ax2 + bx + c) = x2. → a = ½ , b = -1 , c = ½ . 2 2 yp1 = ½ x – x + ½ = ½ (x – 1) .
32 (ii) yp2 is particular integral for y``+ 2y`+ 2y = sin x . Then yp2 = p cos x + q sin x. yp2`= -p sin x + q cos x. yk2``= -p cos x – q sin x.
y``+ 2y` + 2y = sin x. (-p cos x – q sin x) + 2(-p sin x + q cos x) + 2(p cos x + q sin x) = sin x. (-2p + q)sin x + (p + 2q)cos x = sin x. → -2p + q = 1 p + 2q = 0 p = -2/5 dan q = 1/5 2 1 1 yp2 = - 5 cos x + 5 sin x = 5 (sin x – 2kos x).
-x 1 2 1 Hence: y = e (Acos x + Bsin x)+ 2 (x-1) + 5 (sin x – 2cosx)
Case e: f(x) = g(x).v(x)
f(x) Yp
αx r n n-1 αx Pn(x).e x (Bnx + Bn-1x + … + B1x + Bo).e r n n-1 Pn(x).cosβx x (Bnx + Bn-1x + … + B1x + B0).cos βx r n n-1 Pn(x).sin βx x (Bnx + Bn-1x + … + B1x + B0).sin βx Ceαx.cos βx or xr.eαx(p cos βx + q sin βx) Ceαx sin βx αx r n n-1 αx Pn(x)e sin βx x (Bnx + Bn-1x + … + B0).e sin βx αx r n n-1 αx Pn(x)e cos βx x (Bnx + Bn-1x + … + B0).e cos βx
33 r is the smallest non negative interger.
Example: Find the general solution of the equation y`` - 2y` + 3y = ex sin 2x.
Solution.: Characteristic equation: m2 – 2m + 3 = 0 m = 1 ± i 2 . x yc = e (A cos √2 x + B sin √2 x)
f(x) = ex sin 2x. x yp = e (p cos 2x + q sin 2x). x yp` = e {(p + 2q)cos 2x + (-2p + q)sin 2x}. x yp``= e {(-3p + 4q)cos 2x – (4p + 3q) sin 2x}.
y``- 2y`+ 3y = ex{-2p cos 2x – 2q sin 2x).
→ ex{-2p cos 2x – 2q sin 2x} = ex sin 2x.
Then: p = 0 dan q = - ½ . x Hence: yk = - ½ e sin 2x.
General solution: y = yc + yp or:
y = ex(A kos 2 x + B sin 2 x – ½ sin 2x).
34 3.3 The Method of Variation of Parameters.
This method can be used in solving non homogeneous differential equation: d 2 y dy a 2 + b + cy = f(x), (a, b, c constans) and dx dx 1 f(x) = tan x, cot x, sec x, cosec x, x n , ln x. In this method , the general solution is in the form: y = uy1 + vy2 where u = u(x) and v = v(x) and y1 , y2 are independent solution respectively.
The method of solution as follows.
Given: ay``+ by` + cy = f(x). i) Determine a and f(x). ii) Determine y1 and y2, the independent solution for homogeneous linier equation.
y1 y2 iii) Find Wronskian: W = ' ' . y1 y2
y2 f (x) y1 f (x) iv) Obtain: u = - ∫ aW dx + A and v = ∫ aW dx + B.
v) Hence the general solution is: y = uy1 + vy2.
Example: Solve the following differential equations: (i) y``+ y = cot x. (ii) y`` + 6y`+ 8y = e-2x.
35 Soluton (i): y`` + y` = cotx.
i) a = 1, f(x) = cot x. ii) The characteristic equation is m2 + 1 = 0. thus m = ± i and yc = Acos x + Bsin x hence y1 = cos x and y1` = - sin x. y2 = sin x and y2` = cos x.
cos x sin x 2 2 iii) W = sin x cos x = cos x + sin x = 1.
y2 f (x) sin x.cot x iv) u = - ∫ aW dx = -∫ 1 dx = - sin x + A.
y f (x) cos 2 x v = 1 dx = cos x.cot x dx = dx ∫ aW ∫ ∫ sin x = ∫(cosec x – sin x)dx = ln[cosec x – cot x] + cos x + B.
v) General solution: y = uy1 + vy2.
y = (-sinx+A)kosx+(ln[cosecx–cotx]+cosx+B)sinx.
3.4 Euler’s Equation.
36 A differential equation in the form of
d n y d n1 y dy n n n-1 n1 anx dx + an-1x dx + … + a1x dx + a0x = f(x) where a0, a1, … , an are constans, is known as Euler’s equation of nth order.
A second order Euler’s equation can be written as :
d 2 y dy ax2 dx 2 + bx dx + cy = f(x) [a, b and c constants] … (1)
The method of solution.
dt 1 t Substitute x = e , or, equivalent t = ln x and dx x dy dy dt dy 1 dy dy . . dx dt dx dt x or x dx = dt . … (2)
d dy d dy (x ) ( ) dx dx = dx dt d 2 y dy d dy dt d 2 y 1 dt 1 ( ) x 2 + = = 2 . [ ] dx dx dt dt dx dt x dx x d 2 y d 2 y dy dy dy 2 2 x dx = dx 2 - x dx [ x dx = dt ] d 2 y d 2 y dy 2 2 → x dx = dx 2 - dt … (3) Substitute (2) and (3) into (1) and then
2 d y dy dy t a( 2 ) + b + cy = f(e ) or dt dt dt 2 d y dy t a 2 + (b – a) + cy = f(e ) … (4) dt dt
37 (4) is the Euler’s equation with constant coefficients.
d 2 y dy Example: Solve the equation x2 dx 2 - 2x dx - 4y = x2.
Solution: a = 1, b = -2, c = -4.
t dy 1 Substitute x = e , then t= ln x and dt = x 2 d y dy t a 2 + (b - a) + cy = f(e ). dt dt 2 d y dy 2t → 2 - 3 - 4y = e . dt dt 4t -t yc = Ae + Be
2t 2t 2t yp = ke , yp` = 2ke , yp``= 4te
4ke2t – 6ke2t – 4ke2t = e2t. 1 k = - 6 1 2t yp = - 6 e 4t -t 1 2t t → y = Ae + Be - 6 e and substitute e =x then 4 B 1 2 y = Ax + x - 6 x .
38 4.0 LAPLACE TRANSFORMS.
Definition: Let f(t) be a fuuction defined in [0 , ∞). est f (t)dt The integral 0 ….. (1) , is called Laplace Transforms for f(x), if that integral convergent.
Notation: £{f(x)} where £ is an operator.
est f (t)dt 0 is improper integral. T est f (t)dt est f (t)dt Then 0 = lim 0 . T → ∞ (1) depends on parameter S, then est f (t)dt £{f(t)} = 0 = F(S).
Generally: £{f(t)} = F(S) £{g(t)} = G(S) £{y(t)} = Y(S)
1 Example: Show that £{1} = S . T st T est1dt lim est dt lim e Solution. : £{1} = = T = T ] 0 0 0 S 1 1 lim e sT = T [- s s ] = ∞ … (2) a) If S < 0, then 2) → £{1} = ∞ b) If S = 0, then 2) → £{1} = ∞ 1 -sT 1 1 c) If S > 0, then 2) → £(1) = - S e + S = 0 + S
39 1 Ł(1) = S . Example: Using the definition, determine the Laplace Transforms for the following functions: a) f(t) = a. b) f(t) = t. c) f(t) = tn. d) f(t) = eat. Let a be constant and n – non negative interger.
st 1 a est .a dt est dt e Sulution: a) £{a} = = a = a[ ] 0 = a[ ] = 0 0 s s s
a £{a} = S , s > 0
1 Substitute: a = 1 → £{1} = S 5 a = 5 → £{5} = S 1 1 1 a = 3 → £{ 3 } = 3S
d(t) b) £{t} = est t dt = t∫e-stdt - ∫{∫e-stdt} dt 0 dt st st = t. e ] - ∫ e s 0 s 1 est 1 1 1 ( ) = 0 + [ ] = = 2 s s 0 s s s
1 £{t} = S 2 , S>0
n c) £{tn} = est t n dt = tn ∫e-stdt - ∫{ ∫e-stdt} d(t ) dt 0 dt st st = tn. e ] - ∫ e .n.tn-1dt s 0 s n -st n-1 n n-1 = 0 + S ∫e t dt = S £{t }.
40 n n n-1 → £{t } = S £{t }. £{tn-1} = est .t n1dt 0 n1 = tn-1 e-stdt - { e-stdt} d(t ) dt. ∫ ∫ ∫ dt n1 st st = t e ] - ∫ e (n-1)tn-2dt. s 0 s n 1 n-2 = 0 + s £{t } n-1 n 1 n-2 Then: £{t } = s £{t } n-2 n 2 n-3 Thus: £{t } = s £{t } . . 2 2 £{t } = s £{t} 1 £{t} = s £{1} 1 £{1} = s n n n-1 → £{t } = ( s ) £{t } n n 1 n-2 = ( s )( s ) £{t } n n 1 n 2 n-3 = ( s )( s )( s ) £{t } . . . n n 1 n 2 1 = ( s )( s )( s ) …( s ) £{1} n! 1 = ( s n )( s )
n n! £{t } = s n1 n = 0, 1, 2, …
41 0 1 3 6 n = 0 → £{t } = £{1} = S ; n = 3 → £{t } = s 4
(sa)t e 1 at est eat dt -t(s-a) d) £{e } = = ∫e dt = ] 0 = , s>0. 0 (s a) S a
at 1 £{e } = s a , s > 0
1 If: a = 0, then £{1} = S . 3t 1 a = 3, then £{e } = S 3 . -2t 1 a = -2, then £{e } = S 2 . e) Let f(t) = cos at. Then: £{cos at} = est kos at dt 0 -st -st d = cos at ∫e dt - ∫{∫e dt} dt (cos at)dt. st st = kos at e ] - ∫ e (-asin at)dt s 0 s 1 a -st -st d = s - s [sin at ∫e dt - ∫{∫e dt} dt (sin at)dt] 1 a st st = - {sin at. e ] - ∫ e (a cos at)dt} s s s 0 s 1 a a -st = s - s { 0 + s ∫e cos at dt} 1 a 2 = - 2 £{cos at} s s a 2 1 (1+ 2 )Ł{cos at} = s s
s Ł{cos at} = s 2 a 2 , s > 0
42 s s 4s Ł{cos 2t} = s 2 22 = s 2 4 ; Ł{cos ½ t} = 4s 2 1
f) £{sin at) = e-stsin at dt 0 -st -st d = sin at ∫e dt - ∫{∫e dt} dt (sin at) dt st st = sin at.e ] - ∫ e .a cos at dt s 0 s a st st = 0 + {cos at. e ] - ∫ e (-a sin at)dt} s s 0 s a 1 a -st = s { s - s ∫e sin at dt a a 2 = 2 - 2 £{sin at} s s
a → £{sin at} = 2 2 s > 0 s a
d d Notice: dt (sinh t) = cosh t and dt (cosh t) = sinh t. By the same calculation we get:
a , s 0 Ł{sinh at} = s 2 a 2 s , s 0 Ł{kosh at} = s 2 a 2
Example: Using the definition of the Laplace transformation, determine £{f(t)}, if:
1 5 t, 0 ≤ t < 5
43 f(t) = 1, t ≥ 5
Solution:
5 1 £{f(t)} = est . t dt + est .1dt 0 5 5 1 d st = [t ∫e-stdt - ∫{∫e-stdt} (t)dt] + e ] 5 dt s 5 1 est 5 est e5s = { t. ] - ∫ dt } + 5 s 0 s s 1 5e5s 1 est 5 e5s = - 2 ] + 5 s 5 s 0 s e5s 1 e5s 1 e5s = - - { 2 - 2 } + s 5 s s s 1 -5s = 5s 2 ( 1 - e ).
Theorem: If £{f1(t)} and £{f2(t)} exist, α and β are constants, then:
£{αf1(t) + βf2(t)} = α £{f1(t)}+ β £{f2(t)}
Theorem: £{a1f1(t) + a2f2(t) + … + anfn(t)} = a1£{f1(t)} + a2£{f2(t)} + … + an£{fn(t)},
where f1(t), f2(t), …, fn(t) exist and a1, a2, …, an are constants.
Example: Determine £{f(t)} if f(t) = 2t4 – e- 4t. Solution : £{f(t)} = £{2t4 – e- 4t} = 2 £{t4} – £{e- 4t}
44 4! 1 = 2 ( s 5 ) - s (4) 48 1 = s 5 - s 4
1 at -at Example: If cosh at = 2 (e + e ), determine £{cosh at}. 1 at -at Solution: £{cosh at} = £{ 2 (e + e )} 1 at 1 -at = 2 £{e } + 2 £{e } 1 1 1 1 = 2 ( s a ) + 2 ( s a ) s = s 2 a 2 .
1 at -at Exercise: If sinh at = 2 (e – e ) shows that a ₤{sinh at} = s 2 a 2 .
Example: Find the Laplace transform of f(t) = sin 3t.kos 5t.
Solution: f(t) = sin 3t.kos 5t 1 = 2 {sin(3t + 5t) + sin(3t – 5t)} 1 = 2 {sin 8t – sin(-2t)} 1 = 2 {sin 8t – sin 2t} 1 ₤{f(t)} = ₤{ 2 (sin 8t – sin 2t)} 1 1 = 2 ₤{sin 8t} - 2 ₤{sin 2t} 1 8 1 2 = 2 ( s 2 64 ) - 2 ( s 2 4 ) 3s 2 48 = (s 2 64)(s 2 4) .
45 First-Shift Theorem.
Theorem: If ₤{f(t)} = F(s) and a constant, then ₤{eat.f(t)} = F(s – a).
Proof: ₤{f(t)} = est . f (t)dt = F(s). [definition]. 0 ₤{eat.f(t)} = e-st.eatf(t) dt 0 = e-(s-a)t.f(t) dt. [suppose p = s – a] 0 = e-p.f(t) dt = F(p) = F(s – a). 0 → ₤{eat f(t)} = F(s – a).
Examples. a) Find the Laplace transform for f(t) = t4e3t.
4 4! 24 ₤{t } = s 41 = s 5 = F(s) 4 3t 24 ₤{t e } = F(s – 3) = (s 3)5 . b) Find the Laplace transform for f(t) = 2e4tsin 4t.
4 ₤{sin 4t} = s 2 16 = F(s). ₤{2e4tsin 4t}= 2 ₤{e4tsin 4t} = 2 F(s – 2)
46 4 = 2[ (s 4)2 16 ] 8 = s 2 8s 32 .
Theorem. If ₤{f(t)} = F(s), then for n = 1, 2, 3, … n n n d ₤{t .f(t)} = (-1) n [F(s)]. ds
Example: Find the Laplace transform for f(t) = t2sin 2t . 2 Solution : ₤{sin 2t) = s 2 4 = F(s). 2 2 2 2 d 2 d 2 -1 ₤{t sin 2t} = (-1) 2 [ 2 ] = 2 [2(s + 4) ] ds s 4 ds d 2 -2 = ds [-2(s + 4) (2s)] = -4(s2 + 4)-2 – 4s(-2)(s2 + 4)-3(2s) 4(s 2 4) 16s 2 = (s 2 4) 3 12s 2 16 = (s 2 4)3
47 Invers Laplace Transforms (ILT)
Definition: If ₤{f(t)} = F(s), then Invers Laplace Transforms for F(s) as written as: ₤-1{F(s)} = f(t).
₤-1 is known as operator for invers Laplace transforms.
Notice: ₤{f(t)} = F(s) a If f(t) = a, then ₤{a} = s -1 a → ₤ { s } = a. Examples: -1 4 4 a) ₤ { s } = 4, because ₤{4} = s . -1 1 4t 4t 1 b) ₤ { s 4 } = e , because ₤{e } = s 4 . -1 2 -1 2 c) ₤ { s 2 4 }= ₤ { s 2 22 } = sin 2t. -1 3 -1 1 5/3 t d) ₤ { 3s 5 } = ₤ { s 5/ 3 } = e s 4s 7 -1 -1 49 e) ₤ { 4s 2 49 } = ₤ { s 2 } = cos 2 t. 4 -1 6 -1 6/ 4 3 f) ₤ { 4s 2 9 } = ₤ { s 2 9/ 4 } = sin 2 t.
48 Properties of Invers Laplace Transforms.
Theorem: If ₤-1{F(s)} = f(t) and ₤-1{G(s)} = g(t) and if α and β are constants then:
₤-1{α.F(s) + β.G(s)} = α ₤-1{F(s)} + β ₤-1{G(s)}.
Examples:
-1 12 -1 2 -1 2! 2 a) ₤ { s 3 } = ₤ {6( s 3 )} = 6 ₤ { s 3 } = 6 t .
-1 2 -1 1 -1 1 -3t b) ₤ { s 3 }= ₤ {2( s 3 } = 2 ₤ { s 3 } = 2 e .
-1 4 -1 4 3 4 -1 3 4 c) ₤ { s 2 9 } = ₤ { 3 ( s 2 9 )} = 3 ₤ { s 2 9 } = 3 sin 3t.
-1 2s 2 -1 s 1 3 d) ₤ {16s 2 9 } = 16 ₤ { s 2 9/16 } = 8 cos 4 t.
-1 2s 5 -1 2s -1 5 e) ₤ { s 2 25 } = ₤ { s 2 25 } + ₤ { s 2 25 } -1 s -1 5 = 2 ₤ { s 2 25 } + ₤ { s 2 25 } = 2 cosh 5t + sinh 5t.
-1 3s 5 -1 3s -1 5 f) ₤ {16s 2 9 } = ₤ {16s 2 9 } + ₤ {16s 2 9 } 3 -1 s 5 -1 1 = 16 ₤ { s 2 9/16 } + 16 ₤ { s 2 9/16 }
49 3 3 5 4 -1 3/ 4 = 16 kosh 4 t + 16 . 3 ₤ { s 2 9/16 } 3 3 5 3 = 16 cosh 4 t + 12 sinh 4 t.
First-Shift Theorem (Invers).
If ₤-1{F(s)} = f(t) and a is constant, then:
₤-1{F(s – a)} = eat f(t) 0r ₤-1{F(s – a)} = eat ₤-1{F(s)}.
Examples: -1 1 4t -1 1 4t 4t a) ₤ { s 4 } = e ₤ { s } = e .1 = e .
-1 3s -1 3(s 1) 3 b) ₤ { (s 1)4 } = ₤ { (s 1)4 }
-1 3 -1 3 = ₤ { (s 1)3 } - ₤ { (s 1)4 }
3 -1 2 1 -1 6 = 2 ₤ { (s 1)3 } - 2 ₤ { (s 1)4 } 3 -t -1 2 1 -t -1 6 = 2 e ₤ { s 2 } - 2 e ₤ { s 3 } 3 -t 2 1 -t 3 = 2 e t - 2 e t 1 -t 2 3 = 2 e (3t – t ).
-1 8s 13 -1 8(s 2) 3 c) ₤ { s 2 4s 5 } = ₤ { (s 2)2 9 } -1 8(s 2) -1 3 = ₤ { (s 2)2 9 } - ₤ { (s 3)2 9 } -2t -1 s -2t -1 3 = 8e ₤ { s 2 9 } - e ₤ { s 2 9 }
50 = 8e-2tcosh 3t – e-2tsinh 3t 3t 3t 3t 3t = 8e-2t( e e ) – e-2t( e e ) 2 2 1 t -5t = 2 (7e + 9e ). -1 1 -1 1 -1 1 d) ₤ { s(s 2) } = ₤ {- 2s } + ₤ { 2(s 2) } 1 -1 1 1 -1 1 = - 2 ₤ { s } + 2 ₤ { s 2 } 1 1 2t = - 2 + 2 e 1 2t = 2 (e – 1).
-1 3s 1 -1 1 -1 s 3 e) ₤ { s(s 2 1) } = ₤ { s }+ ₤ { s 2 1 } -1 1 -1 s -1 1 = ₤ { s } - ₤ { s 2 1 }+ 3₤ { s 2 1 } = 1 – cos t + 3sin t.
Applications of Laplace transforms.
Theorem:
If ₤{y(t)} = Y(s), then: ₤{y`(t)} = sY(s) – y(0) ₤{y``(t)} = s2Y(s) – sy(0) – y`(0) ₤{y```(t)}= s3Y(s) – s2y(0) – sy`(0) – y``(0) : . ₤{y(n)(t) = snY(s) – sn-1y(0) – sn-2y`(0) – …- y(n-1)(0).
51 Exercises: By using Laplace transform determine the following equations.
1. y` + y = kos t, if y(0) = 0 2. y` + 3y = 13 sin 2t , y(0) = 6. 3. y` + y = te-2t , y(0) = 0 4. y`` - 4y = 4e2t , y(0) = 0 and y`(0) = 5. 5. y`` + 2y` - 3y = t , y(0) = 2 and y`(0) = 1.
52 SIRI
Definsi : Siri ialah suatu baris susunan nombor yang mem- punyai sifat yang tetap.
Contoh: a) 1, 2, 3, … , n-1 an = n – 1. 1 1 1 1 1 b) 2 , 3 , 4 , … , n an = n . n+1 c) 1, -2, 3, -4 , … an = (-1) n 1 2 3 n d) 2 , 3 , 4 , … an = n 1 .
Siri Kuasa (Power Siries).
Definisi: Siri kuasa ialah siri yang berbentuk:
n 2 n (1) cn x = c0 + c1x + c2x + … + cnx + … atau n0
c (x a)n 2 n (2) n = c0 + c1(x-a) + c2(x-a) + … + cn(x-a) + …
dimana a dan pekali c0, c1, … , cn adalah pemalar. Siri (1) adalah bentuk khusus siri kuasa (2) dengan a = 0.
Siri Taylor dan Siri Mac Laurin.
53 Katalah f adalah suatu fungsi yang dapat dibezakan diseki- tar lengkungan a dan termasuk a. Maka f adalah suatu siri Taylor disekitar a yang ditakrif sebagai:
(k ) 2 f (a) (x - a)k = f(a) + f `(a)(x - a) + f ``(a)(x a) + … + k0 k! 2!
f (n) (a)(x a)n + n! + … (3) Jika a = 0, maka
(k ) f ``(0)x 2 f (n) (a)x n f (0) xk = f(0) + f `(0)x + + …+ + … (4) k0 k! 2! n ! (4) adalah bentuk siri Mac Laurin.
54 Periodic Function.
Definition: A function f(x) is said to be periodic if its function values repeat at regular intervals of the indipendent variable. The regular interval between repetitions is the period of the oscillations. Y
X 0 x
Example: (a). y = sin x.
Y 1
0 π 2π X Graph of y = sinx goes through its complete range of values while x increases from 0o to 360o. The period is therefore 360o or 2π radians and the amplitude, the maximum displacement from the potition of rest, is 1.
55 (b). y = A sin nx. 0 Amplitude = A; period = 360 = 2 , n cycles in 360o. n n
Some examples for periodic function..
Y 4
X 0 6 8 14 16 period = 8 ms Y
3
X 0 2 5 6 8 11 period = 6 ms Y
2
X 0 2 3 5 7 8 10 period = 5 ms
56 Analytical description of a periodic function.
A periodic function can be defined analytically in many cases. Example 1. Y
3
X 0 4 6 10 12
(a) Between x = 0 and x = 4, y = 3, i.e. f(x)= 3 0 < x < 4 (b) Between x = 4 and x = 6, y = 0, i.e. f(x) = 0. 4 < x < 6
So we could define the function by
f(x) = 3 , 0 < x < 4 f(x) = 0 , 4 < x < 6 f(x) = f(x + 6) , that mean the function is periodic with period 6 units.
The function can be written as follows:
3 , 0 < x < 4 f(x) = 0 , 4 < x < 6 f(x + 6)
57 Example 2. Y 5
X 0 8 16 The function define:
5 8 x , 0 < x < 8 f(x) = f(x + 8)
Example 3.
Y
2
0 2 6 8 12 X
x , 0 < x < 2 x f(x) = - 2 + 3, 2 < x < 6 f(x + 6).
58 Fourier Series.
The basic of a Fourier siries is to represent a periodic function by a trigonometrical series of the form f(x) = A0 + c1sin(x + α1) + c2sin(2x + α2) + c3sin(3x + αn) + … + cnsin(nx + αn) + … where: A0 is a constant term. c1, c2, c3, …, cn denote the amplitudes of the compound sine terms. α1, α2, …, αn are constant auxiliary angles.
Note that each sine term: cnsin(nx + αn) = cn{sin nx.cos αn + cos nx.sin αn} = (cn sin αn) cos nx + (cn cos αn) sin nx. = an cos nx + bn sin nx where: an = cn sin αn and bn = cn cos αn, a 2 2 n cn = an bn and αn = arc tan( ). bn 1 For convenience in calculation, we write A0 = 2 a0 , and then, putting n = 1, 2, 3, …the hole Fourier siries becomes: 1 f(x) = 2 a0 + a1cos x + a2cos 2x + a3cos 3x + …+ ancos nx + b1sin x + b2sin 2x + b3 sin 3x + …+ bnsin nx + ..
1 or f(x) = a + (ancos nx + bnsin nx) 2 0 n1
n – positive integer.
59 To find a0.
Integrate f(x) with respect to x from - π to π, then:
1 f (x)dx a dx = 0 + { ancos nx dx + bnsin nx dx} 2 n1 1 ] 1 = 2 a0x + Σ {0 + 0} = 2 a0 { π – (-π) = a0π. 1 → a0 = f(x) dx
To find an .
Multiply f(x) by cos mx and integrate from -π to π.
1 f(x)cos mxdx= a0cos mx dx+ 2 { ancos nx cos mx dx + bnsin nx cos mx dx} n1
1 1 ] 1 (i) 2 ∫ a0cos mx dx = 2m sin mx ] = 2m {sin mπ- sin(-mπ)} = 0.
(ii) ∫ancos nx cos mx dx = 1 ∫an 2 {cos(n + m)x + cos (n – m) dx}
an an = 2(n m) sin(n + m)x ] + 2(n m) sin(n – m)x ] = 0 , if n ≠ m.
If n = m then:
60 2 1 ∫ ancos nx dx = an ∫ 2 (cos 2nx + 1)dx a sin 2nx n ] = 2 { 2n + x}
an = 2 { 0 + π – (-π)} = an π.
(iii) ∫ bnsin nx cos mx dx 1 = bn 2 ∫{sin (n + m)x + sin (n – m)x} dx
bn bn = - 2(m n) kos(n + m)x ] - 2(n m) kos(n – m)x ] = 0 , if n ≠ m
If n = m, then:
bn ∫ bn sin nx cos nx dx = 2 ∫ sin 2nx dx b n ] = - 4n cos 2n = 0. So that f(x) cos nx dx = an π 1 → an = f(x) kos nx dx.
To find bn .
Multiply f(x) by sin mx and integrate from –π to π. 1 f(x) sin mx dx = ∫a0sin mx dx + 2 { ∫ ancos nx sin mx dx + ∫ bnsin nx sin mx dx } n1
1 = 2 a0(0) + Σ { an(0) + bn(0) } = 0 , if m ≠ n.
61 If m = n , then: 1 f(x) sin nx = ∫a0sin nx dx + 2 1 2 { ∫sin 2nx dx + ∫ bn sin nx dx } n1 2
bn = 0 + 0 + 2 ∫ (1 – cos 2nx) dx b sin 2nx n ] = 2 [ x - 2n = bnπ.
1 → bn = f(x) sin nx dx
Example. Determine the Fourier siries to represent the priodic function shown. a) Y
π
X 0 2π 4π
b) Y
4 | | -3π/2 - π - π/2 0 π/2 π 3π/2 X
Solution:
62 1 a) a0 = π ; an = 0 ; bn = - n . f(x) = ½ π – { sin x + ½ sin 2x + 1/3 sin 3x + …}
8 n b) a0 = 4 ; an = n sin 2 ; bn = 0. f(x) = 2 + 8/π{ sin x – 1/3 cos 3x + 1/5 cos 5x - … }
ODD AND EVEN FUNCTIONS.
63 Definition: A function f(x) is said to be even if f(-x) = f(x).
Example: f(x) = x2 is an even function since f(-2) = 4 = f(2) f(-3) = 9 = f(3) Y The graph of even function a 2 is therefore symmetrical about the Y-exis. X -a 0 a y= f(x) = cos x is even function since cos (-x) = cos x.
Definition: A function f(x) is said to be odd if f(-x) = -f(x)Example: f(x) = x3 , is n oddfunction since f(-2) = -8 = - f(2) f(-5) = -125 = -f(5) Y P
-a X 0 a The graph of an odd function is
thus symmetrical about the or Q y = f(x) = sin x is an odd function since sin (-x) = -sin x. Products of odd and even functions.
Theorem: The rules closely resemble the elementary rules of sign. a) (even) x (even) = (even). b) (odd) x (odd) = (even).
64 c) (odd) x (even) = (odd).
Proof : a) Let F(x) = f(x). g(x) , where f(x) and g(x) are even fuctions. Then: F(-x) = f(-x).g(-x) = f(x). g(x) = F(x). → F(-x) = F(x) → F(x) is even.
b) Let F(x) = u(x).v(x) , where u(x) and v(x) are odd functions. Then: F(-x) = u(-x).v(-x) = {-u(x)}. –{v(x)} = u(x).v(x) = F(x). → F(-x) = F(x) → F(x) is even.
c) Let F(x) = r(x).q(x) , r(x) is odd and q(x) is even. Then: F(-x) = r(-x).q(-x) = -r(x).q(x) = - r(x).q(x) = - F(x) → F(x) = - F(x) → F(x) is odd.
Two usefulfacts emerge from odd and even functios.
a) Even function.
65 Y
-a 0 a X
0 a a a f(x) dx = f(x) dx → f(x) dx = 2 f(x) dx. a 0 a 0
b) Odd function. Y
X -a 0 a
0 a a f(x) dx = - f(x) dx → f(x) dx = 0 a 0 a
Theorem: If f(x) is defined over the interval –π < x < π and f(x) is even, then the Fourier siries for f(x) contains cisine terms only. Included in this is a0 which may be regarded as ancos nx with n = 0.
66 0 Proof: Since f(x) is even, f(x) dx = f(x) dx. 0 1 2 a) a0 = f(x) dx = f(x) dx 0 1 b) an = f(x) cos nxdx f(x) and cos nx are even functions then f(x)cos nx is the product of two even functions and therefore itself even. 2 → an = f(x).kon nx dx. 0 1 c) bn = f(x).sin nx dx f(x) is even function and sin nx is odd function. Then f(x).sin nx is an odd function. 1 . → bn = f(x).sin nx dx. . . bn = 0. Therefore, there are no sine terms in Fourier siries for f(x).
Example: Determine the Fourier siries for the following function.
π + x , -π < x < 0 f(x) = π – x , 0 < x < π f(x + 2π).
Solution: Y π
-π 0 π X f(x) is an evev function.
67 1 2 2 1 2] a0 = f(x) dx = (π – x) dx = [πx - x 0 0 2 = π. 1 an = f(x).cos nx dx. [f(x)cos nx is even). 2 = (π – x).cos nx dx 0 2 = {∫ π cos nx dx - ∫ x cos nx dx} 2 x 1 ] ] 2 ] = { n sin nx 0 - n sin nx 0 - n cos nx 0 } 2 1 = { n sin nπ – 0 - n sin nx + 0 - n2 (cos nπ – 1)} 2 = - n2 (cos nπ – 1). If n = 0, 2, 4, … then (cos nπ – 1) = 0. If n = 1, 3, 5, … then (cos nπ – 1) = -2.
bn = 0. (why).
2 f(x) = 2 +( n2 )(-2) Σ cos nx. 4 1 1 f(x) = 2 + {cos x + 9 cos 3x + 25 cos 5x + … }.
Theorem. If f(x) is odd function defined over the interval –π < x < π, then the Fourier siries for f(x) contains sine terms only.
0 Proof: Sincs f(x) is odd function, f(x) dx = - f(x) dx. 0 1 a) a0 = f(x) dx = 0
68 1 b) an = f(x).cos nx dx = 0. [ f(x).cos nx is odd function]. 1 2 c) bn = f(x).sin nx dx = f(x).sin nx dx. 0
2 So, if f(x) is odd, ao = 0. an = 0 and bn = f(x)sin x dx. 0
Example: Determine the Fourier siries for the function shown. Y 6
X -π 0 π
6 Solution: The function can be written as follows:
- 6 , -π < x < 0 f(x) = 6 , 0 < x < π f(x + 2π) We can see that this is an odd function and therefore, a0 = 0 and an = 0. f(x).sin nx is an even function. (why).
1 2 bn = f(x).sin nx dx = 0 f(x).sin nx dx 2 12 = 6 sin nx dx = n (1- kos nπ). 0
69 If n = 0, 2, 4, … (1 – kon nπ) = 0 → bn = 0. 24 n If n = 1, 3, 5, … (1 – kos nπ) = 2 → bn = 24 1 1 → f(x) = {sin x + 3 sin 3x + 5 sin 5x + … }
Exercises.
Determine the Fourier siries of the following functions..
x 1 - , 0 < x < 2π 1. f(x) = f(x + 2π).
3 , -2 < x < 0
70 2. f(x) = -5 , 0 < x < 2 f(x + 4).
π + x , -π < x < 0 3. f(x) = π – x , 0 < x < π f(x + 2π).
0 , -π < x < 0 4. f(x) = x , 0 < x < π f(x + 2π)
x , 0 < x < π/2 5. f(x) = π – x , π/2 < x < π f(x + π).
-1 , -1 < x < 0 6. f(x) = 2x , 0 < x < 1 f(x + 2).
x2 , -π < x < π 7. f(x) = f(x + 2π).
3x 7 - , -π < x < π 8. f(x) = f(x + 2π).
1 – x2, -1 < x < 1 9. f(x) =
71 f(x + 2).
x 2 , -π < x < 0 x 10. f(x) = 2 , 0 < x < π f(x + 2π).
Siri Separoh Julat (Half-range series)
Adakalanya suatu fungsi yang berada dalam julat 2π, ditakrif melalui julat 0 sehingga π sebagai ganti julat –π ke π atau 0 ke 2π. Misal, suatu fungsi f(x) = 2x yang berada dalam kalaan 2π hanya dinyatakan berada diantara x = 0 dan x = π. [0 Y 72 2π X - π 0 π Dalam kes seperti di atas, terdapat tiga keadaan yang perlu diperhatikan. a) Jika f(x), 0 f(x) = 2x, -π Y 2π- f(x) = 2x, -π 2π c) Jika f(x) = 2x, 0 Y 73 2π f(x)= 2x. –π Contoh. Suatu fungsi f(x) ditakrif sebagai berikut: 2x, 0 Penyelesaian: Kerana siri yang akan dinyatakan adalah mengan- dungi ungkapan cos, maka f(x) adalah fungsi genap. Y 2π y=2x X -π 0 π 1 2 2 2 f (x)dx f (x)dx 2xdx 2 a0 = = = = (x )]0 = 2π 0 0 1 2 an = f (x)cosnxdx = 2x cos nx dx 0 4 xsin nx cos nx 4 ] 2 ] 2 = { n 0 + n 0 } = n (cosnx – 1) an = 0, jika n genap dan 8 an = - n2 , jika n ganjil. 74 bn = 0, kerana f(x) fungsi genap. Maka: a0 f(x) = + {ancosnx + bnsinnx} 2 n1 8 1 1 f(x) = π - {cosx + 9 cos3x + 25 cos5x + … } Contoh: f(x) ditakrif sebagai berikut: x+1, 0 Penyelesaian: Siri yang akan dinyatakan hanya mengan- dungi ungkapan sinus, maka f(x) adalah fungsi ganjil dan simetri terhadap titik 0. Y π+1 -π 0 π X -(π+1) a0 = 0 dan an = 0 , kerana f(x) fungsi ganjil. 1 2 2 bn = f (x)sin nxdx = f (x)sin nxdx = (x 1)sin nxdx 0 0 2 = { xsin nxdx + sin nxdx } 0 0 2 xcosnx sin nx cos nx 2 ] 2 ] ] = { n 0 + n 0 - n 0 = n {1-(π+1)cosnπ}. cos nx = 1, untuk n genap ataupun ganjil. Maka: 75 2 2 bn = n (1-π-1) = - n , jika n genap dan 2 4 2 bn = n (1+π+1) = n , jika n ganjil. Maka: 4 2 1 1 f(x) = {sinx + 3 sin3x + 5 sin5x + …} 1 1 1 -2{ 2 sin2x + 4 sin4x + 6 sin6x + …}. Functions with period T. T T , If y=f(x) is defined in the range (- 2 2 ), i.e. has a period T, we can convert this to an interval of 2π. Y 76 f(t) = f(t+T) 0 2π rad. = 3600 1 rad.= 360 = 57018`. → 2 2 2 If T = 2π rad. → = T rad. and T = rad. The angle, x radians, at any time t is therefore x = t and the Fourier siries to represent the function can be expressed as 1 n t n t f(t) = a0 + {ancos + bnsin }. 2 n1 With the new variable 2 T 2 / a0 = f(t)dt = f(t)dt. T 0 0 2 T 2 / an = f(t)cos n t dt = f(t)cos n t dt. T 0 T 0 2 T 2 / bn = f(t)sin n t dt = f(t)sin n t dt. T 0 0 77 Example: Determine the Fourier siries for the periodic function defined by 2(1+t), -1 Solution: Y 2 X -1 0 1 1 f(t) = ao + {ancosn t + bnsinn t} 2 n1 T = 2. 2 T / 2 2 1 0 1 f (t)dt a0 = = f (t)dt = 2(1+t)dt + 0 dt T T / 2 2 1 1 0 2 0 = {2t + t }]1 = -(-2 + 1) = 1. 2 T / 2 2 1 an = f(t)cosn t dt = f(t)cosn t dt T T / 2 2 1 0 sin n t cos n t 0 = 2(1+t)dt + 0 = 2{(1+t) + 2 2 }]1 1 n n 2 = n2 2 (1 – cos n ). Now T = 2π and T = 2 , then 2 = 2π → = π. 2 an = n 2 2 (1 – cos nπ). If n is even → an = 0, 4 If n is odd → an = n2 2 . 2 T / 2 2 0 bn = f(t) sin n t dt = { 2(1+t) sin n t dt + 0 } T T / 2 2 1 78 cosn 1 0 2 2 ] = 2{(1 + t) n + n sin n t } 1 cos0 cosn 1 = 2{(1 – 0)( n ) – (1 – 1)( n )+ n2 2 (sin 0 – sinn )} 1 1 = 2{- n + n2 2 sin(-n )}, but = π. Then: 2 bn = - n . So the first few temrs of the Fourie siries 1 4 1 1 f(t) = 2 + 2 (cos t + 9 cos 3 t + 25 cos 5 t + …) 2 1 1 - (sin t + 2 sin 2 t + 3 sin 3 t + … ). Siri Separuh Julat Kalaan T. a. Fungsi Genap. T Y y = f(t), 0 79 4 T / 2 4 T / 2 a0 = f(t) dt dan an = f(t) cos nωt dt. T 0 T 0 b. Fungsi Ganjil. Y y = f(t), 0 < t < T -T/2 X f(t) = f(t + T). 0 T/2 Simetri terhadap titik O. a0 = 0 ; an = 0. f(t) = bn sin nωt. n1 4 T / 2 bn = f(t)sin nωt dt. T 0 Contoh: Diberi f(t) = 4 – t , 0 < t < 4. Y 4 X -4 0 4 Bina suatu fungsi yang simetri terhadap paksi Y. f(t) menjadi suatu fungsi genap. ωT = 2π dan T = 8. 2 4 4 4 4 4 a0 = f(t)dt = f(t)dt = (4 – t)dt T 4 T 0 8 0 2 = 1 {4t - t }]4 = 4. 2 2 0 80 2 4 4 4 an = f(t) cos nωt dt = (4 – t) cos nωt dt T 4 8 0 1 4 4 = { 4cos nωt dt - t.cos nωt dt 2 0 0 1 4sin n t 4 1 sin n t 4 1 4 ] t ] 2 2 cos n t ] = 2 . n 0 - 2 n 0 + 2n 0 2 2 1 = n sin 4nω - n sin 4nω + 2n2 2 (cos 4nω – 1) 1 = 2n2 2 (cos nωt – 1). 1 Tetapi: ωT = 2π dan T = 8, maka ω = 4 π. 1 Maka: cos 4nω = cos nπ. → an = 2n2 2 (cos nπ – 1). Jika n genap maka: an = 0, dan 1 Jika n ganjil maka: an = - n2 2 . bn = 0, kerana f(t) adalah fungsi genap. 1 f(t) = a0 + an cos nωt = 2 n1 1 1 1 f(t) = 2 + 2 (cos ωt + 9 cos 3ωt + 25 cos 5ωt + … ). Contoh: Diberi 3 + t , 0 < t < 2. f(t) = Y 5 f(t + 4) 3 -2 0 2 X -3 -5 Bina suatu fungsi yang simetri terhadap O. f(t) adalah suatu fungsi ganjil. a0 = 0 ; an = 0 . 81 1 ωT = 2π dan T = 4 . Maka ω = 2 π. f(t) = bn sin nωt. n1 2 T / 2 4 2 bn = f(t) sin nωt dt = (3 + t).sin nωt dt T T / 2 T 0 cos n t 2 sin n t 2 ] 2 2 ] = -(3 + t) n 0 + n 0 1 1 = - n {5 cos 2nω – 3 cos 0} + n2 2 {sin 2nω – sin 0} 1 1 1 = n {3 – 5 cos 2nω} + n2 2 sin 2nω. [gantikan ω = 2 π] 1 1 bn = n {3 – 5 cos nπ} + n2 2 sin nπ. 8 Jika n ganjil, maka bn = n 2 Jika n genap, maka bn = - n . 2 1 4 1 f(t) = {4 sin ωt - 2 sin 2ωt + 3 sin 3ωt - 4 sin 3ωt + … } 1 dimana ω = 2 π. 82