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.
Example: 1. dy = sinx + 3
dx
2. y`` + 2y` - 6y = ex
3. d x + 2 d y - dy - y = 0.
3
2
dy 3
dx 2
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
or y' ( x) or y'
dx
Similarly the second order differential equation can be
written as
d y
or y' ' ( x) or y' '
2
dx 2
and in general for
n th
differential equation we have
d y
or y ( x) or y
n
( n)
(n )
dx n
The order of an ordinary differential equation is the order
of the highest derivative that appears in the differential
equation.
1
Example: 1.
2.
dy
+ y tan x = sin 2x.
dx
2
x2 d 2y - 4x dy + 6y = x-1.
dx
dx
2
-x
3. y`` - 4y` + 4y = 5x + e .
4. y``` - 3y`` + 3y – y = 0.
(first order)
(second order)
(second order)
(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.
Example 2. Form ODE from the function y = x2 +
Solution:
y = x2 +
A
x
A
x
.
. Multiply with x, then
yx = x3 + A. Differenciate with respect to x,
→ y + x dy = 3x2 is the first order ODE.
dx
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.
3
y``e2x + 4e2x y` + 4e2xy = 9Ae3x …. (iv)
3y`e2x + 6e2xy = 9Ae3x
Or:
3(iii):
______________________________________________ _
y``e2x + y`e2x - 2e2xy
e2x(y``+ y` - 2y) = 0.
Thus the solution is:
= 0
But e2x ≠ 0
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.
Examp. 5. Given d y + dy - 6y = 0. Show that:
2
dx 2
2x
dx
(a) y = e is the solution.
(b) y = 5e2x + 4e-3x is the solution.
(c) y = xe2x is not the solution.
Solution:
(a) y = e2x…(i) thus
dy
dx
= 2e …(ii) and
2x
d2y
=
dx 2
4e2x …(iii)
Substitute (i), (ii) dan (iii) into the given diff. eq. hence
d2y
dx 2
+
dy
dx
- 6y = 4e2x + 2e2x – 6e2x = 0.
It is shown that y = e2x is the solution.
(b) y = 5e2x + 4e-3x
dy
= 10e2x – 12e-3x
dx
4
→
d2y
dx 2
d2y
dx 2
= 20e2x + 36e-3x
+
dy
dx
- 6y = 20e2x + 36e-3x + 10e2x – 12e-3x
-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
- y = (x + 3)e2x, and hence determine the
dx
value of A given that y = 4 when x = 0.
y = Aex + (x + 2)e2x
dy
= Aex + 2(x + 2)e2x + e2x
Solution:
dx
= Aex + (2x + 5)e2x
→
dy
dx
- 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
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  B sin 3x
x
2
x d 2y + 2 dy +
dx
dx
Example 8. Show that y =
solution for
is the general
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 dy + y = -3A sin3x + 3B kos3x … (ii)
dx
2
x d 2y + dy + dy
dx
dx
dx
2
x d 2y + 2 dy =
dx
dx
= -9A kos3x – 9B sin3x.
-9(A kos3x + B sin3x) … (iii)
Substitute (i) into (iii), thus:
7
2
y
dx
2
xd
+ 2 dy + 9xy = 0. (shown)
dx
Substitute y( ) = -3 into (i) → -3 = -A or A = 3
y`( ) = 0 into (ii) → y = -3B
-3 = -3B or B = 1.
The particular solution: y = 3kos3x  sin 3x .
x
Example 9. Show that y = Ax3 +
B
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.
y = Ax3 +
B
x3
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 + B is the general solution.
x3
8
Substituting y(2) = 1 or y = 1 when x = 2
into diff. equation y = Ax3 + B we get
x3
8 = 64A + B…(iv)
1 = 8A + 1 B or
8
Substituting y`(1) = 0 or y`= 0 when x = 1
into xy`= 6Ax3 – 3y we get
xy`= 6Ax3 – 3(Ax3 + B )
x3
xy` = 3Ax3 -
3B
x3
0 = 3A – 3B → A = B … (v)
Fron simuntaneous equation (iv) dan (v), thus
64A + A = 8 → A = B =
Particular equation: y =
8
(x3
65
+
1
x3
8
65
).
9
2. First Order Ordinary Differential Equation
(ODE)
dy
dx
General Form:
Example: a)
b)
dy
dx
dy
dx
= f(x,y)
= 2y + sin x.
=
x 2  (1  x) y
.
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
= u(x).v(y) or
dy
v( y )
= u(x).dx
hence, integration both sides:
∫
dy
v( y )
= ∫ u(x) dx.
10
Example 1. Solve the equation: (x + 2) dy = y.
dx
(x + 2) dy =
dx
∫ dy = ∫ dx
x2
y
Solution:
y
ln|y| = ln|x+2| + C
ln| y | = ec = A
x2
y = A(x+2).
Example 2.
Solve the equation: ex
dy
dx
+ xy2 = 0.
Solution:
dy
+ xy2 = 0.
dx
dy
= - ∫ xe-xdx.
2
y
1
= -[x ∫e-xdx - ∫{e-xdx} d ( x) dx.
dx
y
1
= -xe-x -∫-e-xdx
y
1
= -xe-x – e-x + C.
y
1
= -(x+1) e-x + C.
y
ex
∫
-
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 = x dx.
2
y
- dy =
y
-∫ dy =
y
x 1
{(x – 1) +
∫(x – 1)dx
-ln|y| =
x2
2
1
}dx.
x 1
+ ∫ dx
x 1
- x + ln|x + 1| + C.
ln|y(x + 1)| = x – ½ x2 – C.
y(x + 1) = ex-1/2 x
2
-C
2
y(x + 1) = A.ex-1/2 x , where A = e-C
y = 2 when x = 0, thus: 2 = A.
The solution is:
y=
2
.
x 1
ex- ½ x
2
12
2.1.1. Substitution Method.
Example 4. Solve the equation:
dy
dx
=
x  y 1
x y5
which satisfied the condition y(1) = 1.
Solution
: Subsitute z = x + y
dz
dy
 1+
thus
→
dx
dz
dx
dz
dx
dx
-1 =
z 5
z 3
∫(1 +
=
z 1
z 5
z 1
z 5
+1=
dy
dx
=
2z  6
z5
dz
dx
=
-1
2( z  3)
z5
dz = 2 dx.
2
)
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:
x dy + y = 2x((1 + x2y2).
dx
13
Solution: Substitute z = xy, hence
dz
dy
x  y
dx
dx
→
dz
dx
∫
= 2x(1 + z2)
= ∫ 2xdx.
dz
1 z2
-1
tan z = x2 + C.
z = tan(x2 + C)
xy = tan(x2 + C).
y=
2.2
tan( x 2  C )
x
HOMOGENEOUS EQUATION.
Consider the differential equation
dy
dx
= f(x, y).
If: f(λx, λy) = f(x, y) for each   , hence
dy
= f(x, y) is called a homogeneous equation.
dx
Example: i).
dy
dx
=
f(λx,
ii).
dy
dx
xy
x  y2
2
= f(x, y)
(x)(y )
2 ( xy)
λy) =
= 2 2 2
( x ) 2  ( y ) 2
 (x  y )
= 2 xy 2 = f(x, y) [homogeneous].
x y
= 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
dx
y = x.v, hence
= x dv + v
dx
Example 6. Solve the differential equation
dy
= xy with condition y(0) = 2.
x2  y2
dx
Solution: By using substitution y = xv and
Thus: x dv + v =
dx
x dv =
dx
v
1 v2
x( xv)
x  ( xv) 2
2
-v=
∫ ( 1  v ) dv = - ∫
2
v
1
 2v 2
3
=
= x dv + v.
dx
v
1 v2
v  v(1  v 2 )
1 v2
dx
x
dy
dx
=-
v3
1 v2
dx.
+ ln |v| = -ln|x| + C.
ln |xv| =
1
2v 2
y = A.e
+ C.
[v = y ]
x 2 /2y 2
, where A = eC
x
Then y(0) = 2 , hence A = 2.
The solution is: y = 2ex
2
/2y 2
15
Example 7. Solve the differential equation
dy
= 2 x  y with condition y(3) = 1.
dx
Solution:
x  2y
2 x   y
=  (2 x  y ) = 2 x  y
x  2y
 ( x  2 y)
x  2y
xv and dy  x dv  v , hence
dx
dx
+ v = 2 x  xv = 2  v .
x  2 xv
1  2v
2
= 2  v - v =  2(v  1) .
1  2v
2v  1
f(λx, λy) =
Substitute y =
x dv
dx
x dv
dx
= f(x, y).
∫( 2v  1 )dv = ∫-2 dx
∫{
v2 1
3
}dv
2(v  1)
x
2 dx .
x
1
2(v  1)
+
1
ln|v
2
+ 1| + 3 ln|v – 1| = -2ln|x| + C
= -∫
2
ln|v + 1| + 3ln|v – 1| = -4ln|x| + 2C
(v + 1)(v – 1)3.x4 = A , where A = e2C
( y  x )( y  x )3.x4 = A
x
x
(y + x)(y – x)3 = A
The condition y(3) = 1 → A = -32.
The solution is: (y + x)(y – x)3 + 32 = 0.
16
Example 8. Solve: x dy - y = x2 – y2, with condition
dx
y = 1 when x = 1.
Solution:
Substitute y = xv, hence:
x(x dv + v) – xv = x√ 1 - v2 .
dx
x dv = √ 1 – v2
∫
dx
dv
1 v2
= ∫
dx
x
sin-1 v = ln|x| + C
sin-1( y ) = ln|x| + C
x
The condition y = 1 when x = 1, thus
sin-1 (1) = 0 + C → C =  .
2
Solution: sin-1( y ) = ln|x| +
x

2
17
2.3 Linier differential Equations.
a(x) dy + b(x).y = c(x).
Note:
dx
dy
+ b( x) .y = c( x)
dx
a( x)
a( x)
dy
+ p(x).y = q(x)
dx
or:
where p(x) =
and
b( x )
a( x)
q(x) =
c( x)
a( x)
This is the general form of a linier differential equations.
The Method of Solution.
dy
dx
i)
Write to the general form :
+ 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) dy + u(x).p(x).y = u(x).q(x).
dx
v)
Write
d
{u(x).y}
dx
= u(x).q(x).
vi) ∫ d(u(x).y = ∫ u(x).q(x)dx.
vii)
u(x).y = ∫ u(x).q(x)dx.
dy
+y=
dx
dy
- y =
dx
(1 + x2) dy
dx
Example: i). x
ii).
iii).
x3
2ex
- xy = x(1 + x2)
18
Solution i).
p(x) =
1
x
x
dy
dx
dy
dx
+ y = x3
+
y
x
= x2
→ ∫p(x)dx = ∫ 1 dx = ln x.
x
Integrating factor: u(x) = e∫p(x)dx = eln x = x.
y.x = ∫x.x2 dx = ∫x3dx
= 1 x4 + C
→
y=
Example ii).
4
1 3
x
4
dy
dx
+
C
x
.
- 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)
dy
dx
∫p(x)dx
dy
dx
- xy = x(1 + x2)
x
).y = x
1 x2
= ∫-( x 2 )dx = ln(1
1 x )
ln(1+x 2 ) 1 / 2
∫p(x)dx
-(
u(x) = e
=e
(1+x2)-1/2.y = ∫(
hence
= (1+x2)-1/2
)dx. Substitute z = (1+x2)
x
(1  x 2 )1 / 2
∫( x2 1 / 2 )dx
(1  x )
2 1/2
(1+x2)-1/2.y = (1 + x )
+ x2)-x/2
= (1 + x2)1/2 + C
+ 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.
Condition of an Exact Equation:
M
N

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.
a) M dx + N dy = 0. Test for exactness:
b) Write
u
x
u
y
M
N

y
x
= M …….. (i)
= 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:
i) dy + 3y = e2x
ii)
dx
dy
dx
+ y = x2
iii) sin x
iv) sin x
dy
dx
dy
dx
[ y = 1 e2x+ Ce-3x]
5
[ y = x2- 2x + 2 + Ce-x]
+ 2y kos x = kos x
[ysin2x = A- 1 kos2x]
- y kos x = cot x. [y =
4
- 1 kosek
2
x+Csin x]
2. Show that these equations is exact and solve.
i)
(y3 -
1
x
) dy +
y
x2
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:
an(x) d y + an-1(x)
n
dx
n
d n 1 y
+
dx n 1
…+ a1(x)
dy
dx
+ 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
dny
dx n
+ an-1
d n 1 y
dx n 1
+ … + a1
dy
dx
+ 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) d y + 20y = 0
2
dx 2
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) d y + 2 dy = ln x
Non HDE with variable coefficient.
2
dx 2
x dx
22
3.1 The Method of Solution for A Homogeneous
Differentiel Equation.
Consider a second order linier differential equation:
a
d2y
dx 2
+b
+ cy = 0 where a, b, c constant. …… (3)
dy
dx
If y = emx is the solution, hence
dy
= memx and d y = m2emx
2
dx 2
dx
Substitute into (3), hence
a d y + b dy + cy = 0 can be written as:
2
dx 2
2 mx
dx
am e + 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).
23
a) If b2 – 4ac > 0 hence m1 ≠ m2. Then y1 = em1x and
y2 = em2x are the solutions of the homogeneous equation.
Then the general solution written as:
y = A em1x + B em2x
{ A, B constants}.
b) If b2 – 4ac = 0 hence m1 = m2. The characteristic
equation has only one root, m = - b .
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 )
Hence y1 = eαx kos 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.
25
3.2 Non Homogeneous Linier Equations.
a
d2y
dx 2
+b
dy
dx
+ cy = f(x)
Example: Solve the equation:
Solution: f(x) = 10 e
-2x
d2y
dx 2
4 dy + 3y = 10 e-2x.
dx
x
has e expression.
Let Ce-2x is the solution.
Thus: y = Ce-2x ;
d2y
dx 2
dy
dx
= -2Ce-2x ;
d2y
dx 2
= 4Ce-2x.
- 4 dy + 3y = 10e-2x. Substitute :
dx
-4(-2Ce-2x) + 3Ce-2x = 10e-2x.
15Ce-2x = 10e-2x.
C = 2/3.
Hence y = 2 e-2x satisfied the given equation and
4Ce
-2x
3
is called the particular integral.
The other solution which could be obtain from
homogenous equation d y - 4 dy + 3y = 0.
2
dx 2
dx
Characteristic equation: m – 4m + 3 = 0.
→ m = {1, 3}.
The solution of HDE: yc = Aex + Be3x.
2
General Solution: y = Aex +Be3x + 2 e-2x
3
(A, B constants).
26
Definition:
i) The general solution of equation: a d
2
y
dx
2
+ b dy + cy = 0
dx
is yc , called complementary function..
ii) The solution of : a d y + b dy + cy = f(x) is yp, called
2
dx 2
dx
pacticular integral.
Teorem:
If yc is the complementary function for diff. equation
a d y + b dy + cy = 0 and yp is the particular integral for
2
dx 2
dx
non homogenous equation
d2y
a 2 +b dy
dx
dx
+ 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) = Anxn + An-1xn-1 + … + A1x + Ao .
(An , An-1 , … , A1 , Ao constants).
Suppose: yp = Bnxn + Bn-1xn-1 + … + B1x + B0. ……(ii).
(Bn , Bn-1 , … , B1 , Bo constants).
27
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= 5.
6a + 2b = 0
2
→ b = - 15 .
2
2a + 3b + 2c = 0 → c =
→ yp = 5 x2 2
15
x
2
+
35
.
4
35
.
4
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 + 5 x2 2
15
x
2
+
35
.
4
28
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 .
29
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= 1
2
Then: yp = 1 e3x.
2
Consider:
Charac.eq:
y``- y`- 2y = 0.
m2 – m – 2 = 0
m = {2, -1}
Thus : yc = Ae2x + Be-x
The genenal solution is:
y = Ae2x + Be-x + 1 e3x.
2
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
Case c:
yp = P cos αx + Q sin αx
yp` = -αP sin αx + αQ cos αx
yp``= -α2P cos αx – α2Q sin αx.
30
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.
→ 5p = 1 → p =
1
5
5q = 0 → q = 0
Then yp = 1 cos 2x.
5
The general solution: y = A cos 3x + B sin 3x +
1
5
cos 2x.
Exercises: Solve the equation.
a) y``+ y` - 6y = 52 cos2x.
b) y``- y`- 2y = cos x+ 3 sin x.
31
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).
(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.
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
yp2 = - 2 cos x + 1 sin x = 1 (sin x – 2kos x).
5
5
5
Hence: y = e-x(Acos x + Bsin x)+ 1 (x-1)2 + 1 (sin x – 2cosx)
2
5
Case e: f(x) = g(x).v(x)
f(x)
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
Yp
xr(Bnxn + Bn-1xn-1 + … + B1x + Bo).eαx
xr(Bnxn + Bn-1xn-1 + … + B1x + B0).cos βx
xr(Bnxn + Bn-1xn-1 + … + B1x + B0).sin βx
xr.eαx(p cos βx + q sin βx)
xr(Bnxn + Bn-1xn-1 + … + B0).eαxsin βx
xr(Bnxn + Bn-1xn-1 + … + B0).eαxcos βx
r is the smallest non negative interger.
33
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
2x
+ B sin
2x
– ½ sin 2x).
34
3.3 The Method of Variation of Parameters.
This method can be used in solving non homogeneous
differential equation:
a d y + b dy + cy = f(x), (a, b, c constans) and
2
dx 2
dx
f(x) = tan x, cot x, sec x, cosec x,
1
xn
, 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.
y
y
iii) Find Wronskian: W =
.
1
2
'
1
'
2
y
iv) Obtain: u = - ∫ y
v)
f ( x)
dx
aW
2
y
+ A and v = ∫ y
f ( x)
dx
aW
1
+ B.
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.
iii) W =
cos x
sin x
 sin x
cos x
iv) u = - ∫
v=∫
v)
y 2 f ( x)
dx
aW
y1 f ( x)
dx
aW
= cos2x + sin2x = 1.
= -∫ sin x. cot x dx = - sin x + A.
1
= ∫cos x.cot x dx = ∫ cos
2
x
dx
sin x
= ∫(cosec x – sin x)dx
= ln[cosec x – cot x] + cos x + B.
General solution: y = uy1 + vy2.
y = (-sinx+A)kosx+(ln[cosecx–cotx]+cosx+B)sinx.
36
3.4 Euler’s Equation.
A differential equation in the form of
anx
n
dny
dx n +
an-1x
n-1
d n 1 y
dx n 1 +
… + a1x dy + a0x = f(x)
dx
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 :
2
ax
d2y
dx 2
+ bx dy + cy = f(x) [a, b and c constants] … (1)
dx
The method of solution.
Substitute x = et, or, equivalent t = ln x and
dy dy dt dy 1
 .  .
dx dt dx dt x
→
or x dy =
dx
dy
dt
dt 1

dx x
. … (2)
d
dy
d dy
(x ) =
( )
dx dx
dx dt
2
2
x d 2y + dy = d ( dy ) dt = d 2y . 1
dx
dt dt dx
dx 2
dt x
d y
2
x2 dx 2 = d 2y - x dy
[ x dy = dy
dx
dx
dt
dx
d2y
2
x2 dx 2 = d 2y - dy … (3)
dt
dx
[ dt  1 ]
dx
x
]
Substitute (2) and (3) into (1) and then
a( d
2
y
2
dt
2
a d 2y
dt

dy
dt
) + b dy + cy = f(et) or
+ (b –
dt
a) dy
dt
+ cy = f(et) … (4)
37
(4) is the Euler’s equation with constant coefficients.
Example: Solve the equation x
2
d2y
dx 2 -
2x dy - 4y = x2.
dx
Solution: a = 1, b = -2, c = -4.
Substitute x = et, then t= ln x and
→
d2y
a 2 + (b
dt
d2y
- 3 dy
2
dt
dt
yc = Ae4t
dy
dt
t
=
1
x
- a) dy + cy = f(e ).
dt
- 4y = e2t.
+ Be-t
yp = ke2t, yp` = 2ke2t , yp``= 4te2t
4ke2t – 6ke2t – 4ke2t = e2t.
k=- 1
yp = -
6
1 2t
e
6
→ y = Ae4t + Be-t - 1 e2t and substitute et=x then
y = Ax4 +
B
x
6
2
- 1x .
6
38
4.0 LAPLACE TRANSFORMS.
Let f(t) be a fuuction defined in [0 , ∞).
The integral  e f (t )dt …..
(1) ,
is called Laplace Transforms for f(x), if
that integral convergent.
Definition:

 st
0
Notation: £{f(x)} where £ is an operator.

 e f (t )dt is improper integral.
Then  e f (t )dt = lim  e f (t )dt .
 st
0

T
 st
 st
0
0
T→∞
(1) depends on parameter S, then
£{f(t)} =  e f (t )dt = F(S).

 st
0
Generally: £{f(t)} = F(S)
£{g(t)} = G(S)
£{y(t)} = Y(S)
Example: Show that £{1} =

Solution. : £{1} =  e
0
 st
1dt =
1
S
.
T
lim
T 
 st
 e dt
= lim
T 
0
= lim [- 1 e
s
 sT

T 
1
]
s
a) If S < 0, then 2) → £{1} = ∞
b) If S = 0, then 2) → £{1} = ∞
c) If S > 0, then 2) → £(1) = - 1 e-sT +
S
Ł(1) =
1
S
1
S
e  st
S
]
=∞
=0+
T
0
… (2)
1
S
.
39
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} =  e
 st

.a dt
= a e
0
 st
 st
= a[ e ]
dt
s
0
£{a} =
a
S
a = 5 → £{5} =
b) £{t} =  e
s
→ £{ 1 } =
a=
1
3
 st
= t∫e-stdt - ∫{∫e-stdt} d (t ) dt
t dt
3
1
3S
dt
 st
= t. e ]
s

0
-∫
 st
= 0 + 1[e ]
s
£{t} =
c) £{tn} =  e
 st
a
s
1
S
5
S
0

0
= a[ 1 ] =
, s>0
Substitute: a = 1 → £{1} =


1
S2
s
e  st
s

0
=
1 1
1
( )= 2
s s
s
, S>0
n
= tn ∫e-stdt - ∫{ ∫e-stdt} d (t ) dt
t n dt
dt
0
 st
= tn. e ]
s
=
→ £{tn} =
n
S
0+
e  st
.n.tn-1dt
0
s
n -st n-1
∫e t dt = n £{tn-1}.
S
S

-∫
£{tn-1}.
40

£{tn-1} =  e
 st
.t n1dt
0
= tn-1 ∫e-stdt - ∫{∫e-stdt} d (t
=
t n1e  st
s
]
= 0 +
Then: £{tn-1} =
Thus: £{tn-2} =
.
.
2
n 1
s
n2
s

0
n 1
s
n-2
dt.
dt.
£{t }
£{tn-3}
£{t}
£{t}
£{1}
£{1}
dt
n-2
)
£{t }
£{t } =
2
s
= 1
s
1
=
s
n-1
-
e  st
∫ (n-1)t
s
n-2
n 1
→ £{tn} = ( n ) £{t }
=
=
s
( n )( n  1 ) £{tn-2}
s
s
( n )( n  1 )( n  2 ) £{tn-3}
s
s
s
.
.
.
= ( n )( n  1 )( n  2 ) …( 1 ) £{1}
s
=
s
n! 1
( n )(
s
s
£{tn} =
n!
s n 1
s
s
)
n = 0, 1, 2, …
n = 0 → £{t0} = £{1} =
1
S
;
n = 3 → £{t3} =
6
s4
41

d) £{eat} =  e
 st at
e dt
= ∫e-t(s-a)dt =
0
£{eat} =
If:
1
,
sa
e  ( s a )t
]
 ( s  a)

0
=
1
,
S a
s>0.
s>0
a = 0, then £{1} =
1
S
.
a = 3, then £{e3t} =
a = -2, then £{e-2t}
1
.
S 3
= 1 .
S 2
e) Let f(t) = cos at. Then:
£{cos at} =  e kos at dt

 st
0
= cos at ∫e-stdt - ∫{∫e-stdt} d (cos at)dt.
dt
 st
 st
kos at e
] 0 - ∫ e (-asin at)dt
s
s
1
a
= - [sin at ∫e-st dt - ∫{∫e-stdt} d (sin at)dt]
s
s
dt
 st
 st

= 1 - a {sin at. e ] 0 - ∫ e (a cos at)dt}
s
s
s
s
1
a
a -st
= - { 0 + ∫e cos at dt}
s
s
s
2
= 1 - a 2 £{cos at}
s
s
2
(1+ a 2 )Ł{cos at} = 1
s
s
=

Ł{cos at} =
Ł{cos 2t} =
s
s  22
2
=
s
s  a2
2
s
s 4
2
;
,
s>0
Ł{cos ½ t} =
4s
4s 2  1
42

f) £{sin at) =  e-stsin at dt
0
= sin at ∫e-stdt - ∫{∫e-stdt} d (sin at) dt
dt
=
=
=
=
→
Notice:
sin at.e
s
 st
]
 st
- ∫ e .a cos at dt

s
 st
0 + a {cos at. e ] 0
s
s
a
1
a -st
{ - ∫e sin at dt
s
s
s
2
a
- a 2 £{sin at}
2
s
s
0
£{sin at} =
d
dt
a
s  a2
s
s >0
2
(sinh t) = cosh t and
 st
- ∫ e (-a sin at)dt}
d
dt
(cosh t) = sinh t.
By the same calculation we get:
Ł{sinh at} =
Ł{kosh at} =
a
,
s  a2
s
,
2
s  a2
2
s0
s0
Example:
Using the definition of the Laplace transformation,
determine £{f(t)}, if:
1
t,
5
0≤t<5
1,
t≥5
f(t) =
Solution:
43
5
£{f(t)} =  e
0
=
=
=
=
=
 st
1
. t dt
5

+ e
 st
.1dt
5
1
[t ∫e-stdt - ∫{∫e-stdt} d (t)dt]
5
dt
 st
 st
5
1
e
e
e 5 s
{ t. ] 0 - ∫ dt } +
5
s
s
s
5 s
 st
5 s
1 5e
- 1 e 2 ] 50 + e
5 s
5 s
s
5 s
5 s
5 s
- e - 1 { e 2 - 12 } + e
5
s
s
s
s
1
-5s
( 1 - e ).
5s 2
+
e  st
s
]

5
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 ( 4! ) - 1
=
s5
s  (4)
48
- 1
5
s4
s
44
Example: If cosh at = 1 (eat + e-at), determine £{cosh at}.
2
Solution: £{cosh at} = £{ 1 (eat + e-at)}
2
= 1 £{eat} + 1 £{e-at}
=
2
1
2
2
(
1
)
sa
s
s  a2
=
2
+ 1(
2
1
)
sa
.
Exercise: If sinh at = 1 (eat – e-at) shows that
2
₤{sinh at} =
a
s  a2
2
.
Example: Find the Laplace transform of f(t) = sin 3t.kos 5t.
Solution: f(t) = sin 3t.kos 5t
= 1 {sin(3t + 5t) + sin(3t – 5t)}
=
=
₤{f(t)} =
=
=
=
2
1
{sin 8t – sin(-2t)}
2
1
{sin 8t – sin 2t}
2
₤{ 1 (sin 8t – sin 2t)}
2
1
₤{sin 8t} - 1 ₤{sin 2t}
2
2
1
8
1
( 2 ) - ( 22 )
2 s  64
2 s 4
2
3s  48
.
2
( s  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)} =  e
 st
. 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
→ ₤{e f(t)} = F(s – a).
at
Examples.
a) Find the Laplace transform for f(t) = t4e3t.
₤{t4} =
4!
s 41
=
24
s5
= F(s)
₤{t4e3t} = F(s – 3) =
24
( s  3) 5
.
b) Find the Laplace transform for f(t) = 2e4tsin 4t.
₤{sin 4t} =
4
s  16
2
= F(s).
₤{2e4tsin 4t}= 2 ₤{e4tsin 4t} = 2 F(s – 2)
4
= 2[
]
=
( s  4) 2  16
8
.
2
s  8s  32
46
Theorem. If ₤{f(t)} = F(s), then for n = 1, 2, 3, …
₤{tn.f(t)} = (-1)n d [F(s)].
n
ds n
Example: Find the Laplace transform for f(t) = t2sin 2t .
Solution : ₤{sin 2t) = 2 = F(s).
₤{t2sin 2t} =
=
s2  4
2
2
(-1)2 d 2 [ 2 2 ] = d 2 [2(s2 + 4)-1]
s 4
ds
ds
d
2
-2
[-2(s + 4) (2s)]
ds
2
-2
2
-3
= -4(s + 4) – 4s(-2)(s + 4) (2s)
=  4(s  4)  16s
2
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.
₤{f(t)} = F(s)
If f(t) = a, then ₤{a} =
Notice:
a
s
→ ₤-1{ a } = a.
s
Examples:
a) ₤-1{ 4 } = 4, because ₤{4} = 4 .
s
s
b) ₤-1{
c)
d)
e)
f)
1
} = e4t, because ₤{e4t} = 1 .
s4
s4
2
2
₤-1{ 2 }= ₤-1{ 2 2 } = sin 2t.
s 4
s 2
3
-1
-1
₤ {
} = ₤ { 1 } = e5/3 t
3s  5
s  5/3
₤-1{ 24s } = ₤-1{ s 49 } = cos 7 t.
2
4s  49
s2 
4
6
6
/
4
₤-1{ 2 } = ₤-1{ 2
} = sin 3 t.
2
4s  9
s  9/ 4
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:
a) ₤-1{ 12 } = ₤-1{6( 2 )} = 6 ₤-1{ 2! } = 6 t2.
s3
s3
b) ₤-1{
2
}=
s3
c) ₤-1{
4
}
s 9
d) ₤-1{
2s
}
16 s 2  9
e) ₤-1{
2s  5
}
s 2  25
2
₤-1{2(
s3
1
}
s3
= ₤-1{ 4 (
3
=
2
16
=2
3
)}
s 9
2
₤-1{
= ₤-1{
= 2 ₤-1{
₤-1{
4
3
s
}
s  9 / 16
2
2s
}+
s  25
₤-1{ 2 s }
s  25
2
=
1
8
=
₤-1{
+
1
}
s3
= 2 e-3t.
3
}
s 9
2
= 4 sin 3t.
3
cos 3 t.
4
5
}
s  25
₤-1{ 2 5 }
s  25
2
= 2 cosh 5t + sinh 5t.
f) ₤-1{
3s  5
}
16 s 2  9
= ₤-1{
=
=
=
3s
} + ₤-1{ 52 }
2
16 s  9
16 s  9
3
s
₤-1{ 2
} + 5 ₤-1{ 2 1 }
16
16
s  9 / 16
s  9 / 16
3
kosh 3 t + 5 . 4 ₤-1{ 2 3 / 4 }
16
4
16 3
s  9 / 16
3
cosh 3 t + 5 sinh 3 t.
16
4
12
4
49
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{
b) ₤-1{
1
}
s4
= e4t ₤-1{ 1 } = e4t.1 = e4t.
3s
( s  1) 4
} = ₤-1{ 3(s  1)  3 }
s
=
( s  1) 4
₤-1{ 3 3 }
( s  1)
- ₤-1{
3
( s  1) 4
}
= 3 ₤-1{
=
=
=
c) ₤-1{
8s  13
}
s  4s  5
2
2
} - 1 ₤-1{ 6 4
3
2
2
( s  1)
( s  1)
3 -t -1 2
e ₤ { 2 } - 1 e-t₤-1{ 63 }
2
2
s
s
3 -t 2 1 -t 3
e t - e t
2
2
1 -t
e (3t2 – t3).
2
}
= ₤-1{ 8(s  2)  3 }
=
=
( s  2) 2  9
₤-1{ 8(s 22) } - ₤-1{ 3 2 }
( s  2)  9
( s  3)  9
8e-2t₤-1{ 2 s } - e-2t ₤-1{ 2 3 }
s 9
s 9
-2t
-2t
= 8e cosh 3t – e sinh 3t
= 8e-2t( e  e ) – e-2t( e  e )
3t
3t
2
=
1
(7et
2
3t
3t
2
+ 9e-5t).
50
d) ₤-1{
1
}
s ( s  2)
= ₤-1{- 1 } + ₤-1{
=
=
=
e) ₤-1{
3s  1
}
s ( s 2  1)
2s
- 1 ₤-1{ 1 } +
2
s
1
1 2t
- + e
2
2
1 2t
(e – 1).
2
1
}
2( s  2)
1 -1
₤ { 1 }
2
s2
= ₤-1{ 1 }+ ₤-1{  s  3 }
=
s
₤-1{ 1 }
s
-
s2 1
₤-1{ 2 s }+
s 1
3₤-1{
1
}
s 1
2
= 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.
2.
3.
4.
5.
y` + y = kos t, if y(0) = 0
y` + 3y = 13 sin 2t , y(0) = 6.
y` + y = te-2t , y(0) = 0
y`` - 4y = 4e2t , y(0) = 0 and y`(0) = 5.
y`` + 2y` - 3y = t , y(0) = 2 and y`(0) = 1.
52
SIRI
Definsi : Siri ialah suatu baris susunan nombor yang mempunyai sifat yang tetap.
Contoh: a)
b)
1, 2, 3, … , n-1
1
, 1, 1, … , 1
an = n – 1.
an = 1 .
1, -2, 3, -4 , …
1
, 2, 3, …
an = (-1)n+1n
an = n .
2
c)
d)
2
3
3
4
4
n
n
n 1
Siri Kuasa (Power Siries).
Definisi: Siri kuasa ialah siri yang berbentuk:

(1)  c x = c0 + c1x + c2x2 + … + cnxn + … atau
n
n 0
(2)  c
n
n
( x  a) n
= 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 disekitar lengkungan a dan termasuk a. Maka f adalah suatu siri
Taylor disekitar a yang ditakrif sebagai:
53


k 0
f ( k ) (a)
(x
k!
- a)k = f(a) + f `(a)(x - a) +
+
f ( n ) (a)( x  a) n
n!
f ``( a)( x  a) 2
2!
+…+
+…
(3)
Jika a = 0, maka


k 0
f ( k ) (0)
k!
xk = f(0) + f `(0)x +
f `` (0) x 2
2!
+ …+
f ( n ) (a) x n
n!
+ … (4)
(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
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.
55
(b). y = A sin nx.
Amplitude = A; period =
360 0
n
=
2
n
, 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
X
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:
f(x) =
3,
0<x<4
0,
4<x<6
f(x + 6)
57
Example 2.
Y
5
X
0
8
16
The function define:
5
x
8
,
0<x<8
f(x) =
f(x + 8)
Example 3.
Y
2
0
f(x) =
2
6
x,
- x + 3,
2
8
12
X
0<x<2
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,
cn = a  b and αn = arc tan( a ).
2
n
2
n
n
bn
For convenience in calculation, we write A0 =
1
a0
2
, and
then, putting n = 1, 2, 3, …the hole Fourier siries becomes:
f(x) = 1 a0 + a1cos x + a2cos 2x + a3cos 3x + …+ ancos nx
2
+ b1sin x + b2sin 2x + b3 sin 3x + …+ bnsin nx + ..
or

f(x) = 12 a0 +  (ancos nx + bnsin nx)
n 1
n – positive integer.
59
To find a0.
Integrate f(x) with respect to x from - π to π, then:

 f ( x) dx =


1
2
a
0



n 1


+  {  ancos nx dx +  bnsin nx dx}


dx


a0x ]  + Σ {0 + 0} = 1 a0 { π – (-π)
2
= a0π.

→
a0 = 1  f(x) dx
=
1
2



To find an .
Multiply f(x) by cos mx and integrate from -π to π.

 f(x)cos

mxdx= 1
2

 a0cos mx dx+




n 1


 {  ancos nx cos mx dx +  bnsin nx cos mx dx}


(i) 1 ∫ a0cos mx dx =
2
1
sin
2m
mx ] =
 ]
1
{sin
2m
mπ- sin(-mπ)}
= 0.
(ii) ∫ancos nx cos mx dx =
∫an 1 {cos(n + m)x + cos (n – m) dx}
=
2
an
sin(n
2(n  m)
+ m)x ] +

an
sin(n
2(n  m)
– m)x ]

= 0 , if n ≠ m.
If n = m then:
60
∫ ancos2nx dx = an ∫ 1 (cos 2nx + 1)dx
=
=
2
a n sin 2nx
{
+ x} ]
2n
2
an
{ 0 + π – (-π)}
2
= an π.
(iii) ∫ bnsin nx cos mx dx
= bn 1 ∫{sin (n + m)x + sin (n – m)x} dx
2
=-
bn
2(m  n)
kos(n + m)x ] 
bn
kos(n
2(n  m)
– m)x ]

= 0 , if n ≠ m
If n = m, then:
∫ bn sin nx cos nx dx =
bn
2
=-
∫ sin 2nx dx
bn
cos
4n
2n ] = 0.


So that  f(x) cos nx dx = an π


→
an =
1


 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 }
n 1
=
1
a0(0)
2
+ 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
 { 2 ∫sin 2nx dx + ∫ bn sin nx dx }
n 1
=0+0+
=
bn
2
[x-
bn
∫ (1 –
2
sin 2nx 
]
2n
cos 2nx) dx
= bnπ.
→
bn =
1


 f(x) sin nx dx

Example.
Determine the Fourier siries to represent the priodic
function shown.
a)
Y
π
X
0
2π
b)
4π
Y
4
-3π/2
|
-π
- π/2
0
π/2
|
π
3π/2
X
62
Solution:
a)
a0 = π ; an = 0 ; bn = - 1 .
n
f(x) = ½ π – { sin x + ½ sin 2x + 1/3 sin 3x + …}
b) a0 = 4 ; an =
8
n
sin
n
2
; bn = 0.
f(x) = 2 + 8/π{ sin x – 1/3 cos 3x + 1/5 cos 5x - … }
63
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
a
-a
The graph of even function
is therefore symmetrical
about the Y-exis.
2
0
a
X
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
-a
P
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.
64
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.
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.
65
Two usefulfacts emerge from odd and even functios.
a) Even function.
Y
-a
0
0
a
a
0
a
X
 f(x) dx =  f(x) dx →
a
a
a
0
 f(x) dx = 2  f(x) dx.
b) Odd function.
Y
X
-a
0
0
a
a
0
a
 f(x) dx = -  f(x) dx →
a
 f(x) dx = 0
a
66
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.

0
Proof: Since f(x) is even,  f(x) dx =  f(x) dx.


a) a0 =
b) an =
1


 f(x) dx =

1

0
2


 f(x) dx
0

 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 = 2  f(x).kon nx dx.

c) bn =
1

0

 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 = 1  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,
f(x) = π – x ,
-π < x < 0
0<x<π
67
f(x + 2π).
Solution:
Y
π
-π
π
0
X
f(x) is an evev function.



1
2
2
1 2]
a0 =  f(x) dx =  (π – x) dx = [πx - x

an =
=
=
=
=
=
1

2
= π.
 f(x).cos nx dx.

f(x) =
2
2
0
[f(x)cos nx is even).


(π – x).cos nx dx
 0
2
{∫ π cos nx dx - ∫ x cos nx dx}

2
{  sin nx ]0 - x sin nx ]0 - 12 cos nx ]0 }
n

n
n
2


{ sin nπ – 0 - sin nx + 0 - 12 (cos nπ –

n
n
n
- 2 2 (cos nπ – 1).
n
bn = 0.

2


0
1)}
n = 0, 2, 4, … then (cos nπ – 1) = 0.
n = 1, 3, 5, … then (cos nπ – 1) = -2.
If
If
f(x) =


+(
+
(why).
2
)(-2) Σ cos nx.
 n2
4
{cos x + 1 cos 3x
9

+
1
cos
25
5x + … }.
68
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.


a) a0 =
b) an =
1

 f(x) dx = 0


1

0

 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

So, if f(x) is odd, ao = 0. an = 0 and bn = 2  f(x)sin x dx.

0
Example: Determine the Fourier siries for the function
shown.
Y
6
-π
π
0
X
6
Solution: The function can be written as follows:
f(x) =
-6,
6,
-π < x < 0
0<x<π
69
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 =
=
1

2

 f(x).sin nx dx =


 6 sin nx dx =
0
2

12
n
 f(x).sin nx dx
0
(1- kos nπ).
If n = 0, 2, 4, … (1 – kon nπ) = 0 → bn = 0.
24
If n = 1, 3, 5, … (1 – kos nπ) = 2 → bn =  n
→
f(x) =
24

{sin x +
1
3
sin 3x + 1 sin 5x + … }
5
70
Exercises.
Determine the Fourier siries of the following functions..
11.
x

, 0 < x < 2π
f(x) =
f(x + 2π).
2.
3.
f(x) =
3,
-2 < x < 0
-5 ,
0<x<2
f(x + 4).
f(x) =
π + x , -π < x < 0
π–x, 0<x<π
f(x + 2π).
4. f(x) =
5.
6.
0 , -π < x < 0
x, 0<x<π
f(x + 2π)
f(x) =
x,
0 < x < π/2
π – x , π/2 < x < π
f(x + π).
f(x) =
-1 ,
-1 < x < 0
2x ,
0<x<1
f(x + 2).
71
-π < x < π
x2 ,
7.
f(x) =
f(x + 2π).
7-
8.
3x

, -π < x < π
f(x) =
f(x + 2π).
1 – x2,
9.
-1 < x < 1
f(x) =
f(x + 2).
10. f(x) =
x 
2
x 
2
,
-π < x < 0
,
0<x<π
f(x + 2π).
72
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
f(x) = 2x, -π<x<π
adalah fungsi genap.
2π-
-π
0
π
X
b). Jika f(x) = 2x, 0<x<π simetri terhadap titik asalan 0,
73
maka f(x) = 2x, -π<x<π adalah suatu fungsi ganjil dan
siri Fourier hanya mengandungi ungkapan sinus sahaja.
Y
2π
f(x) = 2x, -π<x<π
adalah fungsi ganjil.
-π
X
π
0
2π
c) Jika f(x) = 2x, 0<x<π dan tidak dinyatakan samada
fungsi genap atau fungsi ganjil, maka siri Fourier mengandungi kedua-dua ungkapan iaitu sinus dan kosinus.
Y
2π
-π
0
π
X
f(x)= 2x. –π<x<π
bukan fungsi genap
atau ganjil.
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.
74
Penyelesaian:
Kerana siri yang akan dinyatakan adalah mengandungi ungkapan cos, maka f(x) adalah fungsi genap.
Y
2π
y=2x
-π
a0 =
an =
=
1

1


X
π
0
f ( x)dx


=
2



f ( x)dx
=
0
2
f ( x) cos nxdx =
 

4 x sin nx 
{
+ cos2nx
]
0
n

n
2


 2xdx
0
= 2 (x2) ] = 2π

0

 2 x cos nx dx
0

]}=
an = 0, jika n genap dan
an = - 8 , jika n ganjil.
0
4
n 2
(cosnx – 1)
 n2
bn = 0, kerana f(x) fungsi genap. Maka:
f(x) =
a0
2

+  {ancosnx + bnsinnx}
n 1
f(x) = π - 8 {cosx + 1 cos3x +

9
1
cos5x
25
+…}
Contoh: f(x) ditakrif sebagai berikut:
x+1, 0<x<π.
f(x)=
f(x+2π).
Nyatakan siri sin separoh julat bagi fungsi tersebut.
Penyelesaian: Siri yang akan dinyatakan hanya mengandungi ungkapan sinus, maka f(x) adalah
fungsi ganjil dan simetri terhadap titik 0.
75
Y
π+1
-π
π
0
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

=
=



2

0
0

{  x sin nxdx +  sin nxdx }

0
0
2  x cos nx 
{
+ sin 2nx ]0 - cos nx
]
0
n
n

n

] =
0
2
{1-(π+1)cosnπ}.
n
cos nx = 1, untuk n genap ataupun ganjil. Maka:
bn = 2 (1-π-1) = - 2 , jika n genap dan
n
n
bn = 2 (1+π+1) = 4  2 , jika n ganjil. Maka:
n
n
f(x) = 4  2 {sinx + 1 sin3x + 1 sin5x + …}

3
5
-2{ 1 sin2x + 1 sin4x + 1 sin6x + …}.
2
4
6
76
Functions with period T.
If y=f(x) is defined in the range (- T
T
2 2
,
), 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.=
If
T
= 2π rad.
→

=
360 0
= 57018`.
2
2
rad. and T
T
=
2

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
f(t) =
1
a0
2

+  {ancos n t + bnsin n t }.
n 1
77
With the new variable
 
a0 = 2  f(t)dt =   f(t)dt.
T
T
an =
bn =
2
T
2
T
2 /

0
0
T
 f(t)cos n  t dt =
0
T
 f(t)sin n  t dt =
0
2 / 

T
 f(t)cos n  t dt.
0
2 / 


 f(t)sin n  t dt.
0
Example: Determine the Fourier siries for the periodic
function defined by
2(1+t),
0,
f(t+2).
f(t) =
Solution:
-1 <t <0
0< t <1
Y
2
-1
0
X
1

f(t) = 1 ao +  {ancosn  t + bnsinn  t}
2
n 1
T = 2.
a0 = 2  f (t ) dt =
T
T /2
T / 2
2
2
2
1
0
1
1
1
0
 f (t ) dt =  2(1+t)dt +  0 dt
0
= {2t + t } ] = -(-2 + 1) = 1.
1
an =
2
T
T /2
 f(t)cosn  t dt =
T / 2
2
2
1
 f(t)cosn  t dt
1
=  2(1+t)dt + 0 = 2{(1+t) sin n t +
n
0
1
cos n t
n 2 2
0
}]
1
78
=
2
n 2
2
(1 – cos n  ).
 T = 2π and T
2
(1 – cos nπ).
2
n 2
Now
an =
= 2 , then 2  = 2π →

= π.
If n is even → an = 0,
If n is odd → an = 4 .
n 2 2
bn =
2
T
T /2
0
T / 2
1
2
 f(t) sin n  t dt = 2 {  2(1+t) sin n  t dt + 0 }
= 2{(1 + t)  cos n +
1
0
sin n  t } ]
1
n
n
= 2{(1 – 0)(  cos 0 ) – (1 – 1)(  cos n )+ 21 2 (sin
n
n
n
= 2{- 1 + 21 2 sin(-n  )}, but  = π. Then:
n
n
2
bn = - .
n
2
2
0 – sinn  )}
So the first few temrs of the Fourie siries
f(t) =
1
2
+
-
4
(cos  t + 1 cos 3  t +

2
(sin  t

2
9
1
cos
25
5  t + …)
+ 1 sin 2  t + 1 sin 3  t + … ).
2
3
79
Siri Separuh Julat Kalaan T.
a. Fungsi Genap.
y = f(t),
Y
0 <t <
T
2
f(t) = f(t + T)
simetri terhadap Y.
-T/2
0
X
T/2
Jika y = f(t) adalah fungsi genap, maka bn = 0.
f(t) = 1 a0 +  an cos nωt dimana

2
n 1
a0 =
4
T
T /2
 f(t) dt dan an =
0
4
T
T /2
 f(t) cos nωt dt.
0
b. Fungsi Ganjil.
Y
-T/2
X
0
T/2
y = f(t), 0 < t < T
f(t) = f(t + T).
Simetri terhadap titik O.
a0 = 0 ; an = 0.

f(t) =  bn sin nωt.
n 1
bn =
4
T
T /2
 f(t)sin nωt dt.
0
80
Contoh: Diberi f(t) = 4 – t , 0 < t < 4.
Y
4
-4
0
X
4
Bina suatu fungsi yang simetri terhadap paksi Y.
f(t) menjadi suatu fungsi genap.
ωT = 2π dan T = 8.
a0 = 2  f(t)dt = 4  f(t)dt = 4  (4 – t)dt
4
4
T
=
an =
T
4
4
8
0
0
2
1
{4t - t } ]04 = 4.
2
2
4
2
f(t) cos nωt dt
T 4
=
4
4
8
4
 (4 – t) cos nωt dt
0
4
= 1 {  4cos nωt dt -  t.cos nωt dt
2
0
0
= 1.
4 sin n t
]0
n
=
sin 4nω -
=
2
2
n
4
1
2 n 2 2
-
4
1 sin n t 4
1
t
cos
n

t
+
]
]
0
0
2
n
2 n 2 2
2
sin 4nω + 12 2 (cos 4nω
n
2n 
– 1)
(cos nωt – 1).
Tetapi: ωT = 2π dan T = 8, maka ω = 1 π.
4
Maka: cos 4nω = cos nπ. → an =
1
2 n 2 2
(cos nπ – 1).
Jika n genap maka: an = 0, dan
Jika n ganjil maka: an = - 1 .
n 2 2
bn = 0, kerana f(t) adalah fungsi genap.

f(t) = 1 a0 +  an cos nωt =
2
n 1
f(t) = 2 +
1

2
(cos ωt + 1 cos 3ωt +
9
1
cos
25
5ωt + … ).
81
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 ; a n = 0 .
ωT = 2π dan T = 4 . Maka ω = 1 π.
2

f(t) =  bn sin nωt.
n 1
bn =
2
T
T /2
 f(t) sin nωt dt =
T / 2
= -(3 + t)
cos n t 2
]0
n
+
4
T
2
 (3 + t).sin nωt dt
0
sin n t 2
]
n 2 2 0
1
{5 cos 2nω – 3 cos 0} + 21 2 {sin 2nω – sin 0}
n
n
1
{3 – 5 cos 2nω} + 21 2 sin 2nω. [gantikan ω = 1 π]
2
n
n
1
{3 – 5 cos nπ} + 21 2 sin nπ.
n
n
n ganjil, maka bn = 8
n
n genap, maka bn = - 2 .
n
==
bn =
Jika
Jika
f(t) =
2

{4 sin ωt - 1 sin 2ωt + 4 sin 3ωt - 1 sin 3ωt + … }
2
3
4
dimana ω = 1 π.
2
82