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1

Second Oder Linear
Differential Equation

2

Introduction
 The general form of second order
ODE
𝑑2 𝑦
𝑑𝑦
𝑎2 𝑥
+ 𝑎1 𝑥
+ 𝑎0 𝑥 𝑦 = 𝑓(𝑥)
2
𝑑𝑥
𝑑𝑥
But in this subject proposed, the
coefficient of 𝑎2 𝑥 , 𝑎1 𝑥 and 𝑎0 𝑥 are
treated as constant and become as
𝑑2 𝑦
𝑑𝑦
𝑎 2+ 𝑏
+ 𝑐𝑦 = 𝑓 𝑥
𝑑𝑥
𝑑𝑥

3

𝑑2 𝑦
𝑑𝑦
𝑎 2+ 𝑏
+ 𝑐𝑦 = 𝑓 𝑥
𝑑𝑥
𝑑𝑥
 If the 𝑓(𝑥) is zero, the second order ODE said to be
homogenous
...

𝑑2 𝑦
𝑑𝑦
𝑎 2+ 𝑏
+ 𝑐𝑦 = 0
𝑑𝑥
𝑑𝑥
Or 𝑎𝑦 ′′ + 𝑏𝑦 ′ + 𝑐𝑦 = 0

 The characteristic equation of the homogenous
equation is
𝑎𝑚2 + 𝑏𝑚 + 𝑐 = 0

5

 The characteristic equation is quadratic equation
and solve the quadratic equation and it gives two
roots (𝑚 values) in the type of :
Roots of the Characteristic
Equation
1 Real and Distinct
2 Real but Repeated
3 Complex Conjugates, 𝑎 ±
𝑏𝑖

General solution
𝑦 = 𝐴𝑒 𝑚1 𝑥 + 𝐵𝑒 𝑚2 𝑥
𝑦 = 𝐴𝑒 𝑚1 𝑥 + 𝐵𝑥𝑒 𝑚1 𝑥
𝑦
= 𝑒 𝑎𝑥 ( 𝐴 cos 𝑏𝑥

6

Step to Solve Homogenous
Equation

 Step1: Write the homogenous 2nd order ODE into
characteristic equation (𝑎𝑚2 + 𝑏𝑚 + 𝑐 = 0)

 Step 2: Solve the characteristic equation and get its
roots ( real distinct, real repeated or complex roots)
 Step 3: Substitute the corresponding 𝑚 roots value
into general solution
...


Hence, the solution of this ODE is
𝑦 = 𝐴𝑒 −4𝑥 + 𝐵𝑒 2𝑥
(refer to the Table 2
...

Hence, the solution of this ODE is
𝑦 = 𝐴𝑒 1+
take derivatives of 𝑦
𝑦 ′ = 1 + 3 𝐴𝑒

3 𝑥

1+ 3 𝑥

+ 𝐵𝑒 (1−

3)𝑥

+ 1 − 3 𝐵𝑒

1− 3 𝑥

Given y 0 = 0
𝑦 0 = 𝐴+ 𝐵 =0
𝐴 = −𝐵

9
and 𝑦 ′ 0 = 3

𝑦′ 0 = 1 + 3 𝐴 + 1 − 3 𝐵 = 3
1+ 3 𝐴+ 1− 3 𝐵 = 3
Using 𝐴 = −𝐵 in [1]
−2 3𝐵 = 3
1
𝐵=−
2
1
𝐴=
2
Hence the particular solution is
1
𝑦= 𝑒
2

1+ 3 𝑥

−

1 (1−
𝑒
2

3)𝑥

Or
1

𝑦=2 𝑒

1+ 3 𝑥

− 𝑒 (1−

3)𝑥

1

Or 𝑦 = 2 𝑒 𝑥 𝑒

3𝑥

− 𝑒−

3𝑥

1

Exercise
10

1
...

=0
3
...
2𝑦 ′′ + 𝑦 ′ = 0 , 𝑦 0 = 3; 𝑦 ′ 0 = 2

5
...


𝑑2 𝑦
𝑑𝑦
− 2 − 2𝑦 = 0,
𝑑𝑥 2
𝑑𝑥
𝑦 ′′ + 2𝑦 ′ + 4𝑦 = 0

𝑦 0 = 0;

𝑦′ 0 = 3

11

Solving Nonhomogeneous
Equations

𝑑2 𝑦
𝑑𝑦
𝑎 2+ 𝑏
+ 𝑐𝑦 = 𝑓 𝑥
𝑑𝑥
𝑑𝑥
The solution of nonhomogeneous equations is given by
𝑦 𝑥 = 𝑦ℎ 𝑥 + 𝑦 𝑝 (𝑥)

𝑦ℎ (𝑥)=Solution of homogenous equation
𝑦 𝑝 (𝑥)= Solution of the particular equation
There are two methods of finding the particular solution,
𝑦 𝑝 (𝑥)
a) The method of undetermined coefficients
b) The method of variational of parameters

12

Method of Undetermined
Coefficient
...
2 in page 33)

Step 3: Find 𝑦 ′𝑝 and 𝑦 ′′
...

Step 4: Find the sum of 𝑦ℎ and 𝑦 𝑝
...


13
Form of 𝒇(𝒙)

𝛼 𝑛 𝑥 𝑛 + 𝛼 𝑛−1 𝑥 𝑛−1 + ⋯
+ 𝑎1 𝑥 + 𝛼0
𝛼 𝑛 𝑥 𝑛 + 𝛼 𝑛−1 𝑥 𝑛−1 + ⋯
+ 𝑎1 𝑥 + 𝛼0

Roots

𝒚 𝒑 (𝒙)

𝑚1 ≠ 0 and 𝑚2 ≠ 0
𝑚1 = 0 or 𝑚2 = 0
(either one)

𝑎 𝑛 𝑥 𝑛 + 𝑎 𝑛−1 𝑥 𝑛−1
+ ⋯
...


𝑑2 𝑦
= 6𝑥 2
𝑑𝑥 2
𝑦 ′′ − 𝑦 ′ −

3
...


𝑦 ′′ + 2𝑦 ′ + 2𝑦 = sin 𝑥

5
...


𝑦 ′′ − 2𝑦 ′ + 2𝑦 = 2𝑥 2 − 1;

1
...

 Step 2: Calculate the Wronskian value,
𝑦1 𝑦2
′
′
𝑤 = 𝑦 ′ 𝑦 ′ = 𝑦1 𝑦2 − 𝑦2 𝑦1
1
2
 Step 3: Calculate 𝑢 and 𝑣:

𝑢=−

𝑦2 𝑓(𝑥)
𝑎𝑊

𝑑𝑥 and v =

𝑦1 𝑓(𝑥)
𝑎𝑊

 Step 4: Get the general solution for 𝑦(𝑥)
𝑦 𝑥 = 𝑢𝑦1 + 𝑣𝑦2

𝑑𝑥

23

Example:

Find the General solution for

𝑦 ′′

− 2𝑦 ′

+ 𝑦=

Solution:
Homogeneous solution:
𝑦 ′′ − 2𝑦 ′ + 𝑦 = 0
Characteristic eqn: 𝑚2 − 2𝑚 + 1 = 0
𝑚−1 𝑚−1 =0
𝑚=1
𝑦ℎ 𝑥 = 𝐴𝑒 𝑥 + 𝐵𝑥𝑒 𝑥
⇒ 𝑦1 = 𝑒 𝑥 ; 𝑦2 = 𝑥𝑒 𝑥
Wronskian value:
𝑦1 𝑦2
𝑒𝑥
𝑥𝑒 𝑥
𝑤 = 𝑦′ 𝑦′ = 𝑥
𝑒
𝑒 𝑥 + 𝑥𝑒 𝑥
1
2
𝑤 = 𝑒 𝑒 𝑥 + 𝑥𝑒 𝑥 − 𝑥𝑒 𝑥 ∙ 𝑒 𝑥
𝑤 = 𝑒 2𝑥

𝑒𝑥
1+𝑥 2

Example:
24
𝑤 = 𝑒 2𝑥

𝑣=

Next, find 𝑢 and 𝑣:

𝑢=−

𝑦2 𝑓 𝑥
𝑑𝑥
𝑎𝑊

𝑒𝑥
𝑒
1 + 𝑥2
𝑑𝑥
1 𝑒 2𝑥
𝑥

𝑣=

𝑒𝑥
𝑥𝑒
1 + 𝑥2
𝑑𝑥
1 𝑒 2𝑥
𝑥

𝑢=−
𝑢=−

𝑣=

𝑥
𝑑𝑥
1 + 𝑥2

Using substitution method ,
𝑥 2
...


𝑦 ′′ − 4𝑦 ′ + 3𝑦 = 𝑒 𝑥

2
...


𝑦 ′′ + 2𝑦 ′ + 5𝑦 = 𝑒 −𝑥 sin 2𝑥

4
...


𝑦 ′′ − 2𝑦 ′ + 2𝑦 = 𝑒 𝑥 1 + sin 𝑥 ;

1

𝑦 0 = 0 and 𝑦

𝜋
2

=0

34

Applications of second order
linear differential equation

 Variety of applications in engineering fields such as:

 Electrical Circuits
 Spring/Mass Systems
 In electrical circuits, RLC circuits ( contains Resistor, inductor, capacitor),
voltage across that resistor, inductor and capacitor are given by
𝑖𝑅,

𝐿

𝑑𝑖
𝑑𝑡

and

1
𝐶

𝑞

respectively
...
The initial charge and current are
both zero
...


36

𝑞 ′′ + 40𝑞 ′ + 625𝑞 = 100 cos 10𝑡

Use undetermined coefficient method to solve [1]
Step 1: find 𝑦ℎ

𝑞 ′′ + 40𝑞′ + 625𝑞 = 0
↓
𝑚2 + 40𝑚 + 625 = 0
−40 ± 402 − 4(625)
𝑚=
2
−40 ± 30𝑖
𝑚=
2
𝑚 = −20 ± 15𝑖
⇒ 𝑞ℎ = 𝑒 −20𝑡 𝐴∗ cos 15𝑡 − 𝐵∗ sin 15𝑡

 Step 2: find the 𝑞 𝑝
𝐸 𝑡 = 100 cos 10𝑡 ⇒ 𝑞 𝑝 𝑡 = 𝐴 cos 10𝑡 + 𝐵 sin 10𝑡
37
𝑞 ′𝑝 = −10𝐴 sin 10𝑡 + 10𝐵 cos 10𝑡
𝑞 ′′ = −100𝐴 cos 10𝑡 − 100𝐵 sin 10𝑡
𝑝
Substitute 𝑞 𝑝 and its derivatives in
𝑞 ′′ + 40𝑞 ′ + 625𝑞 = 100 cos 10𝑡

Gets
−100𝐴 cos 10𝑡 − 100𝐵 sin 10𝑡 + 40 −10𝐴 sin 10𝑡 + 10𝐵 cos 10𝑡
+ 625 𝐴 cos 10𝑡 + 𝐵 sin 10𝑡 = 100 cos 10𝑡
−100𝐴 + 400𝐵 + 625𝐴 cos 10𝑡
+ −100𝐵 − 400𝐴 + 625𝐵 sin 10𝑡 = 100 cos 10𝑡
⇒ −525𝐴 + 400𝐵 = 100
⇒ −400𝐴 + 525𝐵 = 0
84
64
,B =

...

697
697

This yields 𝐴 =
Hence,𝑞 𝑝 𝑡

38
The general solution of the problem is
𝑞 = 𝑞ℎ + 𝑞 𝑝

84
64
𝑞= 𝑒
𝐴 cos 15𝑡 − 𝐵 sin 15𝑡 +
cos 10𝑡 +
sin 10𝑡
697
697
From question, the initial charge and current are both zero
means, the initial conditions is
𝑡 = 0, 𝑞 = 0; 𝑞′ = 0
84
464
∗ =−
∗ =−
⇒ 𝐴
; 𝐵
697
2091
−20𝑡

𝑞= 𝑒

−20𝑡

∗

∗

84
464
84
64
−
cos 15𝑡 −
sin 15𝑡 +
cos 10𝑡 +
sin 10𝑡
697
2091
697
697

39

THE END



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