1. differential equations. - unimap portalportal.unimap.edu.my/portal/page/portal30/lecturer...
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1. DIFFERENTIAL EQUATIONS.
1.1 Introduction.
Definition: An Ordinary Differential Equation (ODE) is an equation that contains one or several derivatives of an unknown function. Example: 1.
dx
dy = sinx + 3
2. y`` + 2y` - 6y = ex
3. 3
3
dy
xd + 2 2
2
dx
yd - dx
dy - y = 0.
Notation: F(x, y, y`, y``, … ) = 0. The order of an ordinary differential equation is the order of the highest derivative that appears in the differential equation. Example: 1.
dx
dy + y tan x = sin 2x. (first order)
2. x2 2
2
dx
yd - 4x dx
dy + 6y = x-1. (second order)
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.
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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. Example 2. Form ODE from the function y = x2 +
x
A .
Solution: y = x2 + x
A . Multiply with x, then
yx = x3 + A. Differenciate with respect to x, → y + x
dx
dy = 3x2 is the first order ODE.
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)
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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(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
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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.
Examp. 5. Given 2
2
dx
yd + dx
dy - 6y = 0. Show that:
(a) y = e2x is the solution. (b) y = 5e2x + 4e-3x is the solution. (c) y = xe2x is not the solution.
Solution:
(a) y = e2x…(i) thus dx
dy = 2e2x …(ii) and 2
2
dx
yd = 4e2x …(iii)
Substitute (i), (ii) dan (iii) into the given diff. eq. hence
2
2
dx
yd + dx
dy - 6y = 4e2x + 2e2x – 6e2x = 0.
It is shown that y = e2x is the solution. (b) y = 5e2x + 4e-3x
dx
dy = 10e2x – 12e-3x
2
2
dx
yd = 20e2x + 36e-3x
→ 2
2
dx
yd + dx
dy - 6y = 20e2x + 36e-3x + 10e2x – 12e-3x
-30e2x – 24e-3x = 0 y = 5e2x + 4e-3x is the solution.
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(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}. 1.4 General & Particular Solution.
Example 7. Show that y = Aex + (x + 2)e2x is the general solution of the differential equation
dx
dy - y = (x + 3)e2x, and hence determine the
value of A given that y = 4 when x = 0.
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Solution: y = Aex + (x + 2)e2x
dx
dy = Aex + 2(x + 2)e2x + e2x
= Aex + (2x + 5)e2x →
dx
dy - y = Aex +(2x + 5)e2x – Aex – (x + 2)e2x
= (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 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)
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Example 8. Show that y = x
xBxAkos 3sin3 + is the general
solution for x2
2
dx
yd + 2dx
dy + 9xy = 0.
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) x
dx
dy + y = -3A sin3x + 3B kos3x … (ii)
x2
2
dx
yd + dx
dy + dx
dy = -9A kos3x – 9B sin3x.
x2
2
dx
yd + 2dx
dy = -9(A kos3x + B sin3x) … (iii)
Substitute (i) into (iii), thus:
x2
2
dx
yd + 2dx
dy + 9xy = 0. (shown)
Substitute y(̟) = -3 into (i) → -3̟ = -A or A = 3̟ y`(̟) = 0 into (ii) → y = -3B -3 = -3B or B = 1. The particular solution: y =
x
xxkos 3sin33 +π .
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Example 9. Show that y = Ax3 + 3x
B 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. y = Ax3 +
3x
B or x3y = Ax6 + 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 Thus: y = Ax3 +
3x
B is the general solution.
Substituting y(2) = 1 or y = 1 when x = 2 into diff. equation y = Ax3 +
3x
B we get
1 = 8A + 8
1 B or 8 = 64A + B…(iv)
Substituting y`(1) = 0 or y`= 0 when x = 1 into xy`= 6Ax3 – 3y we get xy`= 6Ax3 – 3(Ax3 +
3x
B )
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xy` = 3Ax3 - 3
3
x
B
0 = 3A – 3B → A = B … (v) From simultaneous equation (iv) dan (v), thus 64A + A = 8 → A = B =
65
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Particular equation: y = 65
8 (x3 + 3
1
x).
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2. First Order Ordinary Differential Equation
(ODE)
General Form: dx
dy = f(x,y)
Example: a)
dx
dy = 2y + sin x.
b) dx
dy = x
yxx
2
)1(2 −− .
There are four types of a first order ODE,
i) Separable differential 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 = u(x).v(y) or )(yv
dy = u(x).dx
hence, integration both sides: ∫
)(yv
dy = ∫ u(x) dx.
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Example 1. Solve the equation: (x + 2)dx
dy = y.
Solution: (x + 2) dx
dy = y
∫y
dy = ∫ 2+x
dx
ln|y| = ln|x+2| + C ln|
2+x
y | = ec = A
y = A(x+2). Example 2. Solve the equation: ex
dx
dy + xy2 = 0.
Solution: ex
dx
dy + xy2 = 0.
∫ 2y
dy = - ∫ xe-xdx.
- y
1 = -[x ∫e-xdx - ∫{∫e-xdx}dx
xd )( dx.
y
1 = -xe-x -∫-e-xdx
y
1 = -xe-x – e-x + C.
y
1 = -(x+1) e-x + C.
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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 -
y
dy = 1
2
+x
x dx.
-y
dy = {(x – 1) + 1
1
+x}dx.
-∫y
dy = ∫(x – 1)dx + ∫1+x
dx
-ln|y| = 2
2x - x + ln|x + 1| + C.
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: y =
1
2
+x. ex- ½ x
2
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2.1.1. Substitution Method. Example 4. Solve the equation:
dx
dy = 51
++
++
yx
yx
which satisfied the condition y(1) = 1. Solution : Subsitute z = x + y =
dx
dz 1 + dx
dy thus dx
dy = dx
dz - 1
→ dx
dz - 1 = 5
1
+
+
z
z
dx
dz = 5
1
+
+
z
z + 1 = 5
62
+
+
z
z = 5
)3(2
+
+
z
z
3
5
+
+
z
z dz = 2 dx.
∫(1 + 3
2
+z) 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 .
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Example 5. Solve the equation: x
dx
dy + y = 2x((1 + x2y2).
Solution: Substitute z = xy, hence y
dx
dyx
dx
dz+=
→
dx
dz = 2x(1 + z2)
∫ 21 z
dz
+ = ∫ 2xdx.
tan-1 z = x2 + C. z = tan(x2 + C) xy = tan(x2 + C).
y = x
Cx )tan( 2 +
2.2 HOMOGENEOUS EQUATION.
Consider the differential equation
dx
dy = f(x, y).
If: f(λx, λy) = f(x, y) for each ℜ∈λ , hence
dx
dy = f(x, y) is called a homogeneous equation.
Example: i).
dx
dy = 22 yx
xy
+ = f(x, y)
f(λx, λy) = 22 )()(
))((
yx
yx
λλ
λλ
− =
)(
)(222
2
yx
xy
−λ
λ
= 22 yx
xy
− = f(x, y) [homogeneous].
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ii). dx
dy = x – y = f(x, y).
f(λx, λy) = λx – λy = λ(x – y) ≠ f(x, y). f(x, y) non-homogeneous. The method of solving a homogenous diff. equation is by using the following substitution. y = x.v, hence
dx
dy = xdx
dv + v
Example 6. Solve the differential equation
dx
dy = 22 yx
xy
+ with condition y(0) = 2.
Solution: By using substitution y = xv and
dx
dy = xdx
dv + v.
Thus: xdx
dv + v = 22 )(
)(
xvx
xvx
+ =
21 v
v
+
xdx
dv = 21 v
v
+ - v =
2
2
1
)1(
v
vvv
+
+− = - 2
3
1 v
v
+
∫ (3
21
v
v+ ) dv = - ∫ x
dx dx.
22
1
v−+ ln |v| = -ln|x| + C.
ln |xv| = 22
1
v + C. [v =
x
y ]
y = A.ex 2 /2y 2
, where A = eC Then y(0) = 2 , hence A = 2. The solution is: y = 2ex
2/2y
2
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Example 7. Solve the differential equation
dx
dy = yx
yx
22+
+ with condition y(3) = 1.
Solution: f(λx, λy) =
yx
yx
λλ
λλ
22+
+ = )2()2(
yx
yx
+
+
λ
λ = yx
yx
22+
+ = f(x, y).
Substitute y = xv and vdx
dvx
dx
dy+= , hence
xdx
dv + v = xvx
xvx
2
2
+
+ = v
v
21
2
+
+ .
xdx
dv = v
v
21
2
+
+ - v = 12
)1(2 2
+
−−
v
v .
∫(
1
122 −
+
v
v )dv = ∫-2x
dx
∫{)1(2
1+v
+ )1(2
3−v
}dv = -∫ 2x
dx .
2
1 ln|v + 1| + 2
3 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 (
x
xy + )(x
xy − )3.x4 = A
(y + x)(y – x)3 = A The condition y(3) = 1 → A = -32. The solution is: (y + x)(y – x)3 + 32 = 0.
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Example 8. Solve: xdx
dy - y = 22 yx − , with condition
y = 1 when x = 1. Solution: Substitute y = xv, hence: x(x
dx
dv + v) – xv = x√ 1 - v2 .
xdx
dv = √ 1 – v2
∫ 21 v
dv
− = ∫
x
dx
sin-1 v = ln|x| + C sin-1(
x
y ) = ln|x| + C
The condition y = 1 when x = 1, thus sin-1 (1) = 0 + C → C =
2
π .
Solution: sin-1(
x
y ) = ln|x| + 2
π
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2.3 Linier differential Equations.
Note: a(x)dx
dy + b(x).y = c(x).
dx
dy + )()(
xa
xb .y = )()(
xa
xc
or: dx
dy + p(x).y = q(x) where p(x) = )()(
xa
xb
and q(x) = )()(
xa
xc
This is the general form of a linier differential equations. The Method of Solution.
i) Write to the general form : dx
dy + 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.
iv) u(x)
dx
dy + u(x).p(x).y = u(x).q(x).
v) Write dx
d {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.
Example: i). x dx
dy + y = x3
ii). dx
dy - y = 2ex
iii). (1 + x2) dx
dy - xy = x(1 + x2)
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Solution i). x dx
dy + y = x3
dx
dy + x
y = x2
p(x) = x
1 → ∫p(x)dx = ∫x
1 dx = ln x.
Integrating factor: u(x) = e∫p(x)dx = eln x = x. y.x = ∫x.x2 dx = ∫x3dx =
4
1 x4 + C
→ y = 4
1 x3 + x
C .
Example ii).
dx
dy - y = 2ex.
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. Example iii): (1 + x2)
dx
dy - xy = x(1 + x2)
dx
dy - (21 x
x
+).y = x
∫p(x)dx = ∫-()1 2x
x
+)dx = ln(1 + x2)-x/2
u(x) = e∫p(x)dx = eln(1+x 2 ) 2/1−
= (1+x2)-1/2 (1+x2)-1/2.y = ∫(
2/12 )1( x
x
+)dx. Substitute z = (1+x2)
hence ∫(2/12 )1( x
x
+)dx = (1 + x2)1/2 + C
(1+x2)-1/2.y = (1 + x2)1/2 + C. → y = (1 + x2) + C(1 + x2)1/2
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2.4. Exact Equations. General form: M(x,y) dx + N(x,y) dy = 0. Condition of an Exact Equation:
x
N
y
M
∂
∂=
∂
∂
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.
a) M dx + N dy = 0. Test for exactness: x
N
y
M
∂
∂=
∂
∂
b) Write x
u
∂
∂ = M …….. (i)
y
u
∂
∂ = N ………(ii)
c) Inregrate with respect 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
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Exercises. 1. Solve the differential equations: i)
dx
dy + 3y = e2x [ y = 5
1 e2x+ Ce-3x]
ii) dx
dy + y = x2 [ y = x2- 2x + 2 + Ce-x]
iii) sin x dx
dy + 2y kos x = kos x [ysin2x = A-4
1 kos2x]
iv) sin x dx
dy - y kos x = cot x. [y = -2
1 kosek x+Csin x]
2. Show that these equations is exact and solve.
i) (y3 - x
1 ) dy + 2x
y dx = 0
ii) (3x2 – y sin xy) dx – (x sin xy)dy = 0
iii) (2x + 3 kos y) dx + (2y – 3x sin y) dy = 0.
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3. Second Order Linier Differential Equation (LDE)
General Form:
an(x) n
n
dx
yd + an-1(x) 1
1
−
−
n
n
dx
yd + …+ a1(x) dx
dy + a0(x)y = f(x) …(1)
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:
an n
n
dx
yd + an-1 1
1
−
−
n
n
dx
yd + … + a1 dx
dy + a0y = f(x) … (2)
(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:
a) 2
2
dx
yd + 20y = 0 HDE with constant coeffficient.
b) y``- 5y` + 3y = ex Non HDE with constant coefficient. c) x2y``+xy`+(x2-2)y = 0 HDE with variable coefficient.
d) 2
2
dx
yd + x
2
dx
dy = ln x Non HDE with variable coefficient.
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3.1 The Method of Solution for A Homogeneous
Differentiel Equation. Consider a second order linier differential equation:
a 2
2
dx
yd + b dx
dy + cy = 0 where a, b, c constant. …… (3)
If y = emx is the solution, hence
dx
dy = memx and 2
2
dx
yd = m2emx
Substitute into (3), hence
a2
2
dx
yd + bdx
dy + cy = 0 can be written as:
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. Let m1 and m2 are the characteristic roots of equation (4).
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a) If b2 – 4ac > 0 hence m1 ≠ m2. Then y1 = em1
x and y2 = em
2x are the solutions of the homogeneous equation.
Then the general solution written as: y = A em
1x + B em
2x { A, B constants}.
b) If b2 – 4ac = 0 hence m1 = m2. The characteristic equation has only one root, m = -
a
b
2 .
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.
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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 cos βx + B sin βx )
Hence y1 = eαx cos βx and y2 = eαx 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.
26
3.2 Non Homogeneous Linier Equations.
a 2
2
dx
yd + b dx
dy + cy = f(x)
Example: Solve the equation: 2
2
dx
yd - 4dx
dy + 3y = 10 e-2x.
Solution: f(x) = 10 e-2x has ex expression. Let Ce-2x is the solution.
Thus: y = Ce-2x ; dx
dy = -2Ce-2x ; 2
2
dx
yd = 4Ce-2x.
2
2
dx
yd - 4dx
dy + 3y = 10e-2x. Substitute :
4Ce-2x -4(-2Ce-2x) + 3Ce-2x = 10e-2x. 15Ce-2x = 10e-2x. C = 2/3. Hence y =
3
2 e-2x satisfied the given equation and
is called the particular integral. The other solution which could be obtain from
homogenous equation 2
2
dx
yd - 4dx
dy + 3y = 0.
Characteristic equation: m2 – 4m + 3 = 0. → m = {1, 3}. The solution of HDE: yc = Aex + Be3x. General Solution: y = Aex +Be3x +
3
2 e-2x
(A, B constants).
27
Definition:
i) The general solution of equation: a2
2
dx
yd + bdx
dy + cy = 0
is yc , called complementary function..
ii) The solution of : a2
2
dx
yd + bdx
dy + cy = f(x) is yp, called
pacticular integral. Teorem:
If yc is the complementary function for diff. equation
a2
2
dx
yd + bdx
dy + cy = 0 and yp is the particular integral for
non homogenous equation a2
2
dx
yd +bdx
dy + cy = f(x), hence
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).
Case a: f(x) = Anx
n + An-1xn-1 + … + A1x + Ao .
(An , An-1 , … , A1 , Ao constants). Suppose: yp = Bnx
n + Bn-1xn-1 + … + B1x + B0. ……(ii).
(Bn , Bn-1 , … , B1 , Bo constants).
28
Differentiate (ii) for yp`, yp``, … , yp(n) 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 Suppose: yp = ax2 + 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. Hence: 2a = 5 → a =
2
5 .
6a + 2b = 0 → b = -2
15 .
2a + 3b + 2c = 0 → c = 4
35 .
→ yp = 2
5 x2 - 2
15 x + 4
35 .
Consider: y``+ 3y`+ 2 = 0. (HDE). Characteristic eq. : m2 + 3m + 2 = 0. (m + 1)(m + 2) = 0. m = {-1, -2}. → yc = Ae-x + Be-2x. y = yc + yp. or: y = Ae-x + Be-2x +
2
5 x2 - 2
15 x + 4
35 .
29
Example: Solve the equation: y`` - 2y` + y = x2 – 3x.
Solution: f(x) = x2 – 3x. Suppose: yp = ax2 + 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. → yp = x2 + x. Consider: y`` - 2y`+ y = 0 (HDE) Characteristic equation: m2 – 2m + 1 = 0
m = 1 → yh = (A + Bx).ex. General solution: y = (A+Bx)ex + x2 + x. Case b: f(x) = Ceαx , (C, α constants). Then: ay``+ by`+ cy = Ceαx. ….. (iii) Suppose : yp = k.eαx, then, yp`= αkeαx. yp``= α2keαx. = By substituting yp , yp`, and yp`` into (iii), then [a(α2k) + b(αk) + ck].eαx = Ceαx. or: aα2k+ bαk + ck = C .
30
Example: Solve y`` - y` - 2y = 2e3x. Solution : f(x) = 2e3x Suppose yp = ke3x. yp`= 3ke3x yp``= 9ke3x y`` - y` - 2y = 2e3x 9ke3x – 3ke3x – 2ke3x = 2e3x 4ke3x = 2e3x k =
2
1
Then: yp = 2
1 e3x.
Consider: y``- y`- 2y = 0. Charac.eq: m2 – m – 2 = 0 m = {2, -1} Thus : yc = Ae2x + Be-x The genenal solution is: y = Ae2x + Be-x +
2
1 e3x.
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 yp``= -α2P cos αx – α2Q sin αx.
31
Substituting yp , yp` dan yp`` into the given equation, then equate the coefficient of corresponding sin αx or cos αx. 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. → 5p = 1 → p =
5
1
5q = 0 → q = 0 Then yp =
5
1 cos 2x.
The general solution: y = A cos 3x + B sin 3x +
5
1 cos 2x.
Exercises: Solve the equation. a) y``+ y` - 6y = 52 cos2x. b) y``- y`- 2y = cos x+ 3 sin x.
32
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. yc = e-x(A cos x + B sin x). (i) Suppose yp1 is particular integral for y``+ 2y`+ 2y = x2. Then yp1 = ax2 + bx + c yp1` = 2ax + b and yp1``= 2a . 2a + 2(2ax + b) + 2(ax2 + bx + c) = x2. → a = ½ , b = -1 , c = ½ . yp1 = ½ x2 – x + ½ = ½ (x – 1)2.
33
(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 yp2 = -
5
2 cos x + 5
1 sin x = 5
1 (sin x – 2 cos x).
Hence: y = e-x(Acos x + Bsin x)+
2
1 (x-1)2 +5
1 (sin x – 2cosx)
Case e: f(x) = g(x).v(x)
f(x) Yp
Pn(x).eαx Pn(x).cosβx Pn(x).sin βx Ceαx.cos βx or Ceαx sin βx Pn(x)eαxsin βx Pn(x)eαxcos βx
xr(Bnx
n + Bn-1xn-1 + … + B1x + Bo).e
αx xr(Bnx
n + Bn-1xn-1 + … + B1x + B0).cos βx
xr(Bnxn + Bn-1x
n-1 + … + B1x + B0).sin βx xr.eαx(p cos βx + q sin βx) xr(Bnx
n + Bn-1xn-1 + … + B0).e
αxsin βx xr(Bnx
n + Bn-1xn-1 + … + B0).e
αxcos βx
r is the smallest non negative interger.
34
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 . yc = ex(A cos √2 x + B sin √2 x) f(x) = ex sin 2x. yp = ex(p cos 2x + q sin 2x). yp` = ex{(p + 2q)cos 2x + (-2p + q)sin 2x}. yp``= ex{(-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 . Hence: yk = - ½ ex sin 2x. General solution: y = yc + yp or: y = ex(A kos 2 x + B sin 2 x – ½ sin 2x).
35
3.3 The Method of Variation of Parameters.
This method can be used in solving non homogeneous differential equation:
a2
2
dx
yd + bdx
dy + cy = f(x), (a, b, c constans) and
f(x) = tan x, cot x, sec x, cosec x, nx
1 , 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.
iii) Find Wronskian: W = '2
'1
21
yy
yy .
iv) Obtain: u = - ∫
aW
xfy )(2 dx + A and v = ∫aW
xfy )(1 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.
36
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.
iii) W = xx
xx
cossin
sincos
− = cos2x + sin2x = 1.
iv) u = - ∫
aW
xfy )(2 dx = -∫1
cot.sin xx dx = - sin x + A.
v = ∫ aW
xfy )(1 dx = ∫cos x.cot x dx = ∫ dxx
x
sin
cos2
= ∫(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.
37
3.4 Euler’s Equation.
A differential equation in the form of
anxn
n
n
dx
yd
+ an-1xn-1
1
1
−
−
n
n
dx
yd
+ … + a1xdx
dy + 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 :
ax22
2
dx
yd
+ bxdx
dy + cy = f(x) [a, b and c constants] … (1)
The method of solution. Substitute x = et, or, equivalent t = ln x and
xdx
dt 1=
xdt
dy
dx
dt
dt
dy
dx
dy 1.. == or x
dx
dy = dt
dy . … (2)
)(
dx
dyx
dx
d = )(dt
dy
dx
d
x2
2
dx
yd + dx
dy = dx
dt
dt
dy
dt
d)( =
xdt
yd 1.
2
2
[xdx
dt 1= ]
x22
2
dx
yd
= 2
2
dx
yd - xdx
dy [ xdx
dy = dt
dy ]
→ x22
2
dx
yd
= 2
2
dx
yd - dt
dy … (3)
Substitute (2) and (3) into (1) and then
a(dt
dy
dt
yd−
2
2
) + bdt
dy + cy = f(et) or
a2
2
dt
yd + (b – a)dt
dy + cy = f(et) … (4)
38
(4) is the Euler’s equation with constant coefficients.
Example: Solve the equation x22
2
dx
yd
- 2xdx
dy - 4y = x2.
Solution: a = 1, b = -2, c = -4. Substitute x = et, then t= ln x and
dt
dy = x
1
a2
2
dt
yd + (b - a)dt
dy + cy = f(et).
→ 2
2
dt
yd - 3dt
dy - 4y = e2t.
yc = Ae4t + Be-t yp = ke2t, yp` = 2ke2t , yp``= 4te2t 4ke2t – 6ke2t – 4ke2t = e2t. k = -
6
1
yp = - 6
1 e2t
→ y = Ae4t + Be-t - 6
1 e2t and substitute et=x then
y = Ax4 + x
B - 6
1 x2.
39
4.0 LAPLACE TRANSFORMS.
Definition: Let f(t) be a fuuction defined in [0 , ∞). The integral ∫
∞−
0)( dttfe st ….. (1) ,
is called Laplace Transforms for f(x), if that integral convergent. Notation: £{f(x)} where £ is an operator. ∫
∞−
0)( dttfe st is improper integral.
Then ∫∞
−
0)( dttfe st = lim ∫ −
Tst dttfe
0)( .
T → ∞
(1) depends on parameter S, then £{f(t)} = ∫
∞−
0)( dttfe st = F(S).
Generally: £{f(t)} = F(S) £{g(t)} = G(S) £{y(t)} = Y(S) Example: Show that £{1} =
S
1 .
Solution. : £{1} = dte st10∫∞
− = lim∞→T
dte
T
st
∫−
0
= lim∞→T
S
e st
−
−
] T
0
= lim∞→T
[-s
es
sT 11+− ] = ∞ … (2)
a) If S < 0, then 2) → £{1} = ∞ b) If S = 0, then 2) → £{1} = ∞ c) If S > 0, then 2) → £(1) = -
S
1 e-sT + S
1 = 0 + S
1
Ł(1) = S
1 .
40
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.
Sulution: a) £{a} = ∫∞
−
0
. dtae st = a ∫∞
−
0
dte st = a[s
e st
−
−
] ∞
0 = a[
s
1 ] = s
a
£{a} =
S
a , s > 0
Substitute: a = 1 → £{1} =
S
1
a = 5 → £{5} = S
5
a = 3
1 → £{3
1 } = S3
1
b) £{t} = ∫∞
−
0
dtte st = t∫e-stdt - ∫{∫e-stdt}dt
td )( dt
= t.s
e st
−
−
] ∞
0 - ∫
s
e st
−
−
= 0 + s
1 [s
e st
−
−
] ∞
0 = )
1(
1
ss=
2
1
s
£{t} =
2
1
S, S>0
c) £{tn} = dtte nst
∫∞
−
0
= tn ∫e-stdt - ∫{ ∫e-stdt}dt
td n )( dt
= tn.s
e st
−
−
] ∞
0 - ∫
s
e st
−
−
.n.tn-1dt
= 0 + S
n ∫e-st tn-1dt = S
n £{tn-1}.
→ £{tn} = S
n £{tn-1}.
41
£{tn-1} = dtte nst
∫∞
−−
0
1.
= tn-1 ∫e-stdt - ∫{∫e-stdt}dt
td n )( 1−
dt.
= s
et stn
−
−−1
] ∞
0 - ∫
s
e st
−
−
(n-1)tn-2dt.
= 0 + s
n 1− £{tn-2}
Then: £{tn-1} = s
n 1− £{tn-2}
Thus: £{tn-2} = s
n 2− £{tn-3} . . £{t2} =
s
2 £{t}
£{t} = s
1 £{1}
£{1} = s
1
→ £{tn} = (s
n ) £{tn-1}
= (s
n )(s
n 1− ) £{tn-2}
= (s
n )(s
n 1− )(s
n 2− ) £{tn-3} .
.
.
= (s
n )(s
n 1− )(s
n 2− ) …(s
1 ) £{1}
= (ns
n! )(s
1 )
£{tn} =
1
!+ns
n n = 0, 1, 2, …
n = 0 → £{t0} = £{1} = S
1 ; n = 3 → £{t3} = 4
6
s
42
d) £{eat} = ∫∞
−
0
dtee atst = ∫e-t(s-a)dt = )(
)(
as
e tas
−−
−−
] ∞
0 =
aS −
1 , s>0.
£{eat} = as −
1 , s > 0
If: a = 0, then £{1} =
S
1 .
a = 3, then £{e3t} = 3
1
−S.
a = -2, then £{e-2t} = 2
1
+S.
e) Let f(t) = cos at. Then:
£{cos at} = ∫∞
−
0
dtatkose st
= cos at ∫e-stdt - ∫{∫e-stdt}dt
d (cos at)dt.
= s
eatkos st
−
−
] ∞
0 - ∫
s
e st
−
−
(-asin at)dt
= s
1 - s
a [sin at ∫e-st dt - ∫{∫e-stdt}dt
d (sin at)dt]
= s
1 - s
a {sin at.s
e st
−
−
] ∞
0 - ∫
s
e st
−
−
(a cos at)dt}
= s
1 - s
a { 0 + s
a ∫e-stcos at dt}
= s
1 - 2
2
s
a £{cos at}
(1+2
2
s
a )Ł{cos at} = s
1
Ł{cos at} =
22 as
s
+, s > 0
Ł{cos 2t} =
22 2+s
s = 42 +s
s ; Ł{cos ½ t} = 14
42 +s
s
43
f) £{sin at) = ∫∞
0
e-stsin at dt
= sin at ∫e-stdt - ∫{∫e-stdt}dt
d (sin at) dt
= s
eat st
−
−.sin ] ∞
0 - ∫
s
e st
−
−
.a cos at dt
= 0 + s
a {cos at.s
e st
−
−
] ∞
0 - ∫
s
e st
−
−
(-a sin at)dt}
= s
a { s
1 - s
a ∫e-stsin at dt
= 2s
a - 2
2
s
a £{sin at}
→ £{sin at} = 22 as
a
+ s > 0
Notice: dt
d (sinh t) = cosh t and dt
d (cosh t) = sinh t.
By the same calculation we get: Ł{sinh at} = 0,
22>
−s
as
a
Ł{kosh at} = 0,22
>−
sas
s
Example: Using the definition of the Laplace transformation, determine £{f(t)}, if:
5
1 t, 0 ≤ t < 5
f(t) = 1, t ≥ 5 Solution:
44
£{f(t)} = ∫ −5
0 5
1. dtte st + ∫
∞−
5
1. dte st
= 5
1 [t ∫e-stdt - ∫{∫e-stdt}dt
d (t)dt] + s
e st
−
−
] ∞
5
= 5
1 { t.s
e st
−
−
] 5
0 - ∫ dt
s
e st
−
−
} + s
e s5−
= 5
1
s
e s
−
−55 - 5
12s
e st−
] 5
0 +
s
e s5−
= -s
e s5−
- 5
1 { 2
5
s
e s−
- 2
1
s} +
s
e s5−
= 25
1
s( 1 - e-5s ).
45
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} = 2 (
5
!4
s) -
)4(1−−s
= 5
48
s -
4
1
+s
Example: If cosh at =
2
1 (eat + e-at), determine £{cosh at}.
Solution: £{cosh at} = £{2
1 (eat + e-at)}
= 2
1 £{eat} + 2
1 £{e-at}
= 2
1 ( as −
1 ) + 2
1 (as +
1 )
= 22 as
s
−.
46
Exercise: If sinh at = 2
1 (eat – e-at) shows that
₤{sinh at} = 22 as
a
−.
Example: Find the Laplace transform of f(t) = sin 3t.cos 5t. Solution: f(t) = sin 3t.kos 5t =
2
1 {sin(3t + 5t) + sin(3t – 5t)}
= 2
1 {sin 8t – sin(-2t)}
= 2
1 {sin 8t – sin 2t}
₤{f(t)} = ₤{2
1 (sin 8t – sin 2t)}
= 2
1 ₤{sin 8t} - 2
1 ₤{sin 2t}
= 2
1 (64
82 +s
) - 2
1 (4
22 +s
)
= )4)(64(
48322
2
++
−
ss
s .
47
First-Shift Theorem.
Theorem: If ₤{f(t)} = F(s) and a constant, then ₤{eat.f(t)} = F(s – a).
Proof: ₤{f(t)} = dttfe st
∫∞
−
0
)(. = F(s). [definition].
₤{eat.f(t)} = ∫∞
0
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).
→ ₤{eat f(t)} = F(s – a). Examples.
a) Find the Laplace transform for f(t) = t4e3t. ₤{t4} =
14
!4+s
= 5
24
s = F(s)
₤{t4e3t} = F(s – 3) = 5)3(
24
−s.
b) Find the Laplace transform for f(t) = 2e4tsin 4t.
₤{sin 4t} = 16
42 +s
= F(s).
₤{2e4tsin 4t}= 2 ₤{e4tsin 4t} = 2 F(s – 2) = 2[
16)4(
42 +−s
]
= 328
82 +− ss
.
48
Theorem. If ₤{f(t)} = F(s), then for n = 1, 2, 3, …
₤{tn.f(t)} = (-1)n n
n
ds
d [F(s)].
Example: Find the Laplace transform for f(t) = t2sin 2t . Solution : ₤{sin 2t) =
4
22 +s
= F(s).
₤{t2sin 2t} = (-1)2 2
2
ds
d [4
22 +s
] = 2
2
ds
d [2(s2 + 4)-1]
= ds
d [-2(s2 + 4)-2(2s)]
= -4(s2 + 4)-2 – 4s(-2)(s2 + 4)-3(2s)
= 32
22
)4(
16)4(4
+
++−
s
ss
= 32
2
)4(
1612
+
−
s
s
49
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) If f(t) = a, then ₤{a} =
s
a
→ ₤-1{s
a } = a.
Examples: a) ₤-1{
s
4 } = 4, because ₤{4} = s
4 .
b) ₤-1{4
1
−s} = e4t, because ₤{e4t} =
4
1
−s.
c) ₤-1{4
22 +s
}= ₤-1{22 2
2
+s} = sin 2t.
d) ₤-1{53
3
−s} = ₤-1{
3/5
1
−s} = e5/3 t
e) ₤-1{494
42 +s
s } = ₤-1{4
492 +s
s } = cos 2
7 t.
f) ₤-1{94
62 +s
} = ₤-1{4/9
4/62 +s
} = sin 2
3 t.
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)}.
50
Examples: a) ₤-1{
3
12
s} = ₤-1{6(
3
2
s)} = 6 ₤-1{
3
!2
s} = 6 t2.
b) ₤-1{
3
2
+s}= ₤-1{2(
3
1
+s} = 2 ₤-1{
3
1
+s} = 2 e-3t.
c) ₤-1{
9
42 +s
} = ₤-1{3
4 (9
32 +s
)} = 3
4 ₤-1{9
32 +s
} = 3
4 sin 3t.
d) ₤-1{
916
22 +s
s } = 16
2 ₤-1{16/92 +s
s } = 8
1 cos 4
3 t.
e) ₤-1{
25
522 −
+
s
s } = ₤-1{25
22 −s
s } + ₤-1{25
52 −s
}
= 2 ₤-1{252 −s
s } + ₤-1{25
52 −s
}
= 2 cosh 5t + sinh 5t. f) ₤-1{
916
532 −
+
s
s } = ₤-1{916
32 −s
s } + ₤-1{916
52 −s
}
= 16
3 ₤-1{16/92 −s
s } + 16
5 ₤-1{16/9
12 −s
}
= 16
3 cosh 4
3 t + 16
5 . 3
4 ₤-1{16/9
4/32 −s
}
= 16
3 cosh 4
3 t + 12
5 sinh 4
3
51
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: a) ₤-1{
4
1
−s} = e4t ₤-1{
s
1 } = e4t.1 = e4t.
b) ₤-1{
4)1(
3
+s
s } = ₤-1{4)1(
3)1(3
+
−+
s
s }
= ₤-1{3)1(
3
+s} - ₤-1{
4)1(
3
+s}
=
2
3 ₤-1{3)1(
2
+s} -
2
1 ₤-1{4)1(
6
+s}
= 2
3 e-t₤-1{2
2
s} -
2
1 e-t₤-1{3
6
s}
= 2
3 e-t t2 - 2
1 e-t t3
= 2
1 e-t(3t2 – t3).
c) ₤-1{
54
1382 −+
+
ss
s } = ₤-1{9)2(
3)2(82 −+
−+
s
s }
= ₤-1{9)2(
)2(82 −+
+
s
s } - ₤-1{9)3(
32 −+s
}
= 8e-2t₤-1{92 −s
s } - e-2t ₤-1{9
32 −s
}
= 8e-2tcosh 3t – e-2tsinh 3t
= 8e-2t(2
33 tt ee −+ ) – e-2t(2
33 tt ee −− )
= 2
1 (7et + 9e-5t).
52
d) ₤-1{)2(
1−ss
} = ₤-1{-s2
1 } + ₤-1{)2(2
1−s
}
= -2
1 ₤-1{s
1 } + 2
1 ₤-1{2
1
−s}
= -2
1 + 2
1 e2t
= 2
1 (e2t – 1).
e) ₤-1{
)1(
132 +
+
ss
s } = ₤-1{s
1 }+ ₤-1{1
32 +
+−
s
s }
= ₤-1{s
1 } - ₤-1{12 +s
s }+ 3₤-1{1
12 +s
}
= 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).
53
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.
54
SIRI
Definsi : Siri ialah suatu baris susunan nombor yang mem- punyai sifat yang tetap. Contoh: a) 1, 2, 3, … , n-1 an = n – 1. b)
2
1 , 3
1 , 4
1 , … , n
1 an = n
1 .
c) 1, -2, 3, -4 , … an = (-1)n+1n d)
2
1 , 3
2 , 4
3 , … an = 1+n
n .
Siri Kuasa (Power Series).
Definisi: Siri kuasa ialah siri yang berbentuk: (1) ∑
∞
=0n
n
n xc = c0 + c1x + c2x2 + … + cnx
n + … atau
(2) ∑ − n
n axc )( = c0 + c1(x-a) + c2(x-a)2 + … + cn(x-a)n + …
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.
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:
55
∑∞
=0
)(
!
)(
k
k
k
af (x - a)k = f(a) + f `(a)(x - a) + !2
))(``( 2axaf − + … +
+ !
))(()(
n
axaf nn − + … (3)
Jika a = 0, maka
∑∞
=0
)(
!
)0(
k
k
k
f xk = f(0) + f `(0)x + !2
)0( 2`` xf + …+ !
)()(
n
xaf nn
+ … (4)
(4) adalah bentuk siri Mac Laurin.
56
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
0 x 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.
57
(b). y = A sin nx.
Amplitude = A; period = n
0360 = n
π2 , n cycles in 360o.
Some examples for periodic function.. Y 4
X 0 6 8 14 16 period = 8 ms Y
3
0 2
5
6
8
11
X
period = 6 ms Y
2 X 0 2 3 5 7 8 10
period = 5 ms
58
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)
59
Example 2. Y 5
X 0 8 16
The function define:
8
5 x , 0 < x < 8
f(x) = f(x + 8) Example 3. Y 2 0 2 6 8 12 X x , 0 < x < 2 f(x) = -
2
x + 3, 2 < x < 6
f(x + 6).
60
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, cn = 22
nn ba + and αn = arc tan(n
n
b
a ).
For convenience in calculation, we write A0 = 2
1 a0 , and
then, putting n = 1, 2, 3, …the hole Fourier siries becomes: f(x) =
2
1 a0 + a1cos x + a2cos 2x + a3cos 3x + …+ ancos nx
+ b1sin x + b2sin 2x + b3 sin 3x + …+ bnsin nx + .. or f(x) =
2
1 a0 + ∑∞
=1n
(ancos nx + bnsin nx)
n – positive integer.
61
To find a0. Integrate f(x) with respect to x from - π to π, then:
∫−
π
π
dxxf )( = 2
1 ∫−
π
π
dxa0 + ∑∞
=1n
{ ∫−
π
π
ancos nx dx + ∫−
π
π
bnsin nx dx}
= 2
1 a0x ]π
π− + Σ {0 + 0} =
2
1 a0 { π – (-π)
= a0π.
→ a0 = π
1 ∫−
π
π
f(x) dx
To find an . Multiply f(x) by cos mx and integrate from -π to π.
∫−
π
π
f(x)cos mxdx=2
1∫−
π
π
a0cos mx dx+
∑∞
=1n
{ ∫−
π
π
ancos nx cos mx dx + ∫−
π
π
bnsin nx cos mx dx}
(i)
2
1 ∫ a0cos mx dx = m2
1 sin mx]]
π
π−=
m2
1 {sin mπ- sin(-mπ)}
= 0. (ii) ∫ancos nx cos mx dx = ∫an
2
1 {cos(n + m)x + cos (n – m) dx}
= )(2 mn
an
+sin(n + m)x ]
π
π− +
)(2 mn
an
−sin(n – m)x ]
π
π−
= 0 , if n ≠ m. If n = m then:
62
∫ ancos2nx dx = an ∫ 2
1 (cos 2nx + 1)dx
= 2
na {n
nx
2
2sin + x}]π
π−
= 2
na { 0 + π – (-π)}
= an π. (iii) ∫ bnsin nx cos mx dx = bn
2
1 ∫{sin (n + m)x + sin (n – m)x} dx
= -)(2 nm
bn
+ kos(n + m)x ]
π
π− -
)(2 mn
bn
−kos(n – m)x ]
π
π−
= 0 , if n ≠ m If n = m, then: ∫ bn sin nx cos nx dx =
2nb ∫ sin 2nx dx
= - n
bn
4cos 2n ]
π
π−= 0.
So that ∫−
π
π
f(x) cos nx dx = an π
→ an = ∫−
π
ππ
1 f(x) cos nx dx.
To find bn . Multiply f(x) by sin mx and integrate from –π to π.
∫−
π
π
f(x) sin mx dx = 2
1 ∫a0sin mx dx +
∑∞
=1n
{ ∫ ancos nx sin mx dx + ∫ bnsin nx sin mx dx }
=
2
1 a0(0) + Σ { an(0) + bn(0) }
= 0 , if m ≠ n.
63
If m = n , then:
∫−
π
π
f(x) sin nx = 2
1 ∫a0sin nx dx +
∑∞
=1n
{ 2
1 ∫sin 2nx dx + ∫ bn sin2nx dx }
= 0 + 0 + 2nb ∫ (1 – cos 2nx) dx
= 2nb [ x -
n
nx
2
2sin ]π
π−
= bnπ.
→ bn = π
1 ∫−
π
π
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
64
Solution: a) a0 = π ; an = 0 ; bn = -
n
1 .
f(x) = ½ π – { sin x + ½ sin 2x + 1/3 sin 3x + …}
b) a0 = 4 ; an = πn
8 sin 2
πn ; bn = 0.
f(x) = 2 + 8/π{ cos x – 1/3 cos 3x + 1/5 cos 5x - … }
ODD AND EVEN FUNCTIONS.
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. -a 0 a
X
y= f(x) = cos x is even function since cos (-x) = cos x.
65
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 0
a
X
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). 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.
66
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. Y
-a 0 a X
∫−
0
a
f(x) dx = ∫a
0
f(x) dx → ∫−
a
a
f(x) dx = 2 ∫a
0
f(x) dx.
67
b) Odd function. Y
X -a 0 a
∫−
0
a
f(x) dx = - ∫a
0
f(x) dx → ∫−
a
a
f(x) dx = 0
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.
Proof: Since f(x) is even, ∫−
0
π
f(x) dx = ∫π
0
f(x) dx.
a) a0 = π
1∫−
π
π
f(x) dx = π
2 ∫π
0
f(x) dx
b) an = π
1∫−
π
π
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.
→ an = π
2 ∫π
0
f(x).cos nx dx.
68
c) bn = π
1 ∫−
π
π
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.
→ bn = π
1 ∫−
π
π
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.
a0 = π
1 ∫−
π
π
f(x) dx = π
2∫π
0
(π – x) dx = π
2 [πx - 2
1 x2]0
π
= π.
an = π
1∫−
π
π
f(x).cos nx dx. [f(x)cos nx is even).
= π
2∫π
0
(π – x).cos nx dx
= π
2 {∫ π cos nx dx - ∫ x cos nx dx}
69
= π
2 { n
π sin nx ]0
π - n
x sin nx ]0
π - 2
1
ncos nx ]
0
π }
= π
2 { n
π sin nπ – 0 - n
π sin nx + 0 - 2
1
n(cos nπ – 1)}
= - 2
2
nπ(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). f(x) =
2
π +(2
2
nπ
− )(-2) Σ cos nx.
f(x) = 2
π + π
4 {cos x + 9
1 cos 3x + 25
1 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.
Proof: Sincs f(x) is odd function, ∫−
0
π
f(x) dx = - ∫π
0
f(x) dx.
a) a0 = π
1∫−
π
π
f(x) dx = 0
b) an = π
1 ∫−
π
π
f(x).cos nx dx
= 0. [ f(x).cos nx is odd function].
c) bn = π
1∫−
π
π
f(x).sin nx dx = π
2 ∫π
0
f(x).sin nx dx.
So, if f(x) is odd, ao = 0. an = 0 and bn = ∫π
π 0
2 f(x)sin x dx.
70
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).
bn = π
1 ∫−
π
πf(x).sin nx dx =
π
2 ∫π
0f(x).sin nx dx
= π
2 ∫π
0
6 sin nx dx = nπ
12 (1- kos nπ).
If n = 0, 2, 4, … (1 – kon nπ) = 0 → bn = 0. If n = 1, 3, 5, … (1 – kos nπ) = 2 → bn = nπ
24
→ f(x) = π
24 {sin x + 3
1 sin 3x + 5
1 sin 5x + … }
71
Exercises.
Determine the Fourier siries of the following functions.. 1 -
π
x , 0 < x < 2π
1. f(x) = f(x + 2π). 3 , -2 < x < 0 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).
72
x2 , -π < x < π 7. f(x) = f(x + 2π). 7 -
π
x3 , -π < x < π
8. f(x) = f(x + 2π). 1 – x2, -1 < x < 1 9. f(x) = f(x + 2).
2
π+x , -π < x < 0
10. f(x) = 2
π−x , 0 < x < π
f(x + 2π).
73
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<x<π]. Tiada keyataan bagaimana fungsi tersebut diantara x = -π dan x = 0. [ -π<x<0]. Y
2π
- π 0 π
X
Dalam kes seperti di atas, terdapat tiga keadaan yang perlu diperhatikan. a) Jika f(x), 0<x<π simetri terhadap paksi Y, maka f(x) = 2x, -π<x<π adalah suatu fungsi genap dan siri Fourier hanya mengandungi ungkapan kosinus sahaja. Y
2π- f(x) = 2x, -π<x<π adalah fungsi genap. X -π 0 π
b). Jika f(x) = 2x, 0<x<π simetri terhadap titik asalan 0, maka f(x) = 2x, -π<x<π adalah suatu fungsi ganjil dan
74
siri Fourier hanya mengandungi ungkapan sinus sahaja. Y 2π
f(x) = 2x, -π<x<π adalah fungsi ganjil.
-π 0 π
X
2π
c) Jika f(x) = 2x, 0<x<π dan tidak dinyatakan samada fungsi genap atau fungsi ganjil, maka siri Fourier me- ngandungi kedua-dua ungkapan iaitu sinus dan kosinus. Y
2π f(x)= 2x. –π<x<π bukan fungsi genap X atau ganjil. -π 0 π
Contoh. Suatu fungsi f(x) ditakrif sebagai berikut: 2x, 0<x<π f(x) = f(x+2π). Nyatakan siri cos separoh julat yang mewakili fungsi tersebut.
75
Penyelesaian:
Kerana siri yang akan dinyatakan adalah mengan- dungi ungkapan cos, maka f(x) adalah fungsi genap. Y
2π y=2x
-π 0 π X
a0 = π
1 ∫−
π
π
dxxf )( = π
2 ∫π
0
)( dxxf = π
2 ∫π
0
2xdx = π
2 (x2)]0
π = 2π
an = π
1 ∫−
π
π
nxdxxf cos)( = π
2 ∫π
0
cos2 nxx dx
= π
4 {n
nxxsin ]0
π + 2
cos
n
nx ]0
π } = 2
4
nπ(cosnx – 1)
an = 0, jika n genap dan an = -
2
8
nπ, jika n ganjil.
bn = 0, kerana f(x) fungsi genap. Maka: f(x) =
20a + ∑
∞
=1n
{ancosnx + bnsinnx}
f(x) = π - π
8 {cosx + 9
1 cos3x + 25
1 cos5x + … }
Contoh: f(x) ditakrif sebagai berikut: x+1, 0<x<π. f(x)= f(x+2π). Nyatakan siri sin separoh julat bagi fungsi tersebut.
76
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.
bn = π
1 ∫−
π
π
nxdxxf sin)( = π
2 ∫π
0
sin)( nxdxxf = π
2∫ +π
0
sin)1( nxdxx
= π
2 { ∫π
0
sin nxdxx + nxdxsin0∫π
}
= π
2 {n
nxxcos− ]0
π + 2
sin
n
nx ]0
π - n
nxcos ]0
π = nπ
2 {1-(π+1)cosnπ}.
cos nx = 1, untuk n genap ataupun ganjil. Maka: bn =
nπ
2 (1-π-1) = -n
2 , jika n genap dan
bn = nπ
2 (1+π+1) = nπ
π24 + , jika n ganjil. Maka:
f(x) = π
π24 + {sinx + 3
1 sin3x + 5
1 sin5x + …}
-2{2
1 sin2x + 4
1 sin4x + 6
1 sin6x + …}.
77
Functions with period T.
If y=f(x) is defined in the range (-2
,2
TT ), i.e. has a period T,
we can convert this to an interval of 2π. Y f(t) = f(t+T)
2π rad. = 3600 → 1 rad.= π2
3600
= 57018`.
If ωT = 2π rad. → ω = T
π2 rad. and T = ω
π2 rad.
The angle, x radians, at any time t is therefore x = ω t and the Fourier siries to represent the function can be expressed as f(t) =
2
1 a0 + ∑∞
=1n
{ancos tnω + bnsin tnω }.
With the new variable
a0 = T
2∫T
0
f(t)dt = π
ω ∫ωπ /2
0
f(t)dt.
an = T
2∫T
0
f(t)cos nω t dt = T
ω ∫ωπ /2
0
f(t)cos nω t dt.
bn = T
2 ∫T
0
f(t)sin nω t dt = π
ω ∫ωπ /2
0
f(t)sin nω t dt.
78
Example: Determine the Fourier siries for the periodic function defined by 2(1+t), -1 <t <0 f(t) = 0, 0< t <1 f(t+2). Solution: Y
2
-1 0 1
X
f(t) = 2
1 ao + ∑∞
=1n
{ancosnω t + bnsinnω t}
T = 2.
a0 = T
2dttf
T
T
∫−
2/
2/
)( = 2
2∫−
1
1
)( dttf = ∫−
0
1
2(1+t)dt + ∫1
0
0 dt
= {2t + t2}]0
1− = -(-2 + 1) = 1.
an = T
2 ∫−
2/
2/
T
T
f(t)cosnω t dt = 2
2∫−
1
1
f(t)cosnω t dt
= ∫−
0
1
2(1+t)dt + 0 = 2{(1+t)ω
ω
n
tnsin + 22
cos
ω
ω
n
tn }]0
1−
= 22
2
ωn(1 – cos nω ).
Now ω T = 2π and T = 2 , then 2ω = 2π → ω = π. an =
22
2
ωn(1 – cos nπ).
If n is even → an = 0, If n is odd → an =
22
4
ωn.
79
bn = T
2∫
−
2/
2/
T
T
f(t) sin nω t dt = 2
2 { ∫−
0
1
2(1+t) sin nω t dt + 0 }
= 2{(1 + t)ω
ω
n
ncos− +22
1
ωnsin nω t }]
0
1−
= 2{(1 – 0)(ωn
0cos− ) – (1 – 1)(ω
ω
n
ncos− )+22
1
ωn(sin 0 – sinnω )}
= 2{- ωn
1 + 22
1
ωnsin(-nω )}, but ω = π. Then:
bn = - ωn
2 .
So the first few temrs of the Fourier series f(t) =
2
1 + 2
4
ω(cos ω t +
9
1 cos 3ω t + 25
1 cos 5ω t + …)
- ω
2 (sin ω t + 2
1 sin 2ω t + 3
1 sin 3ω t + … ).
80
Siri Separoh Julat Kalaan T.
a. Fungsi Genap.
Y y = f(t), 0 <t < 2
T f(t) = f(t + T) simetri terhadap Y. -T/2 0 T/2
X
Jika y = f(t) adalah fungsi genap, maka bn = 0. f(t) =
2
1 a0 + ∑∞
=1n
an cos nωt dimana
a0 = T
4 ∫2/
0
T
f(t) dt dan an = T
4 ∫2/
0
T
f(t) cos nωt dt.
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) = ∑
∞
=1n
bn sin nωt.
bn = T
4 ∫2/
0
T
f(t)sin nωt dt.
81
Contoh: Diberi f(t) = 4 – t , 0 < t < 4. Y
4
-4 0 4
X
Bina suatu fungsi yang simetri terhadap paksi Y. f(t) menjadi suatu fungsi genap. ωT = 2π dan T = 8.
a0 = T
2∫−
4
4
f(t)dt = T
4∫4
0
f(t)dt = 8
4∫4
0
(4 – t)dt
= 2
1 {4t - 2
2t }]4
0 = 4.
an = T
2∫−
4
4
f(t) cos nωt dt = 8
4∫4
0
(4 – t) cos nωt dt
= 2
1 { ∫4
0
4cos nωt dt - ∫4
0
t.cos nωt dt
= 2
1 . ]4
0
sin4
ω
ω
n
tn - ]4
0
sin
2
1
ω
ω
n
tnt + ]
4
022cos
2
1tn
nω
ω
= ωn
2 sin 4nω - ωn
2 sin 4nω + 222
1
ωn(cos 4nω – 1)
= 222
1
ωn(cos nωt – 1).
Tetapi: ωT = 2π dan T = 8, maka ω = 4
1 π.
Maka: cos 4nω = cos nπ. → an = 222
1
ωn(cos nπ – 1).
Jika n genap maka: an = 0, dan Jika n ganjil maka: an = -
22
1
ωn.
bn = 0, kerana f(t) adalah fungsi genap. f(t) =
2
1 a0 + ∑∞
=1n
an cos nωt =
f(t) = 2 + 2
1
ω(cos ωt +
9
1 cos 3ωt + 25
1 cos 5ωt + … ).
82
Contoh: Diberi 3 + t , 0 < t < 2. f(t) = 5
Y 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 . ωT = 2π dan T = 4 . Maka ω =
2
1 π.
f(t) = ∑∞
=1n
bn sin nωt.
bn = T
2∫
−
2/
2/
T
T
f(t) sin nωt dt = T
4∫2
0
(3 + t).sin nωt dt
= -(3 + t) ]2
0
cos
ω
ω
n
tn + ]2
022
sin
ω
ω
n
tn
= - ωn
1 {5 cos 2nω – 3 cos 0} + 22
1
ωn{sin 2nω – sin 0}
= ωn
1 {3 – 5 cos 2nω} + 22
1
ωnsin 2nω. [gantikan ω =
2
1 π]
bn = ωn
1 {3 – 5 cos nπ} + 22
1
ωn sin nπ.
Jika n ganjil, maka bn = ωn
8
Jika n genap, maka bn = - ωn
2 .
f(t) =
ω
2 {4 sin ωt - 2
1 sin 2ωt + 3
4 sin 3ωt - 4
1 sin 3ωt + … }
dimana ω = 2
1 π.
83