4.5.2 Second Order Differential Equations (Non-Homogeneous Equation)

MOOC MAT438/ UiTM

Step 1 Find the general solution, yg
(solve homogeneous equation)

Step 2 Find the trial solution, yp Write the trial solution, yp based on R(x)
Find yp’
(solve Non-homogeneous equation)
Find yp’’
Step 3 Find the solution, y
Substitute yp , yp’ and yp’’ into the given
question (to find the values of a, b, c…)

Substitute the values of a, b, c… (obtained
above) into yp

If the initial condition is given Step 4 Applying the given initial condition
MOOC MAT438/ UiTM (to find the values of C1 and C2)

Solve the second order differential equation
′′+ ′ − 2 = 1 − , ℎ = 0, = 1, ′ = 0

Step 1 Find the general solution, yg Write the general solution, (based on the types of
(solve homogeneous equation) roots)
the general solution,
m1 m0
= 1 1 + 2 2
m2 m 1
= 1 + 2 −2
′′+ ′ − 2 = 0

2 + − 2 = 0

− 1 + 2 = 0

1 = 1, 2 = −2
1 ≠ 2 Distinct real roots

MOOC MAT438/ UiTM

Solve the second order differential equation
′′+ ′ − 2 = 1 − , ℎ = 0, = 1, ′ = 0

Step 2 Find the trial solution, yp From the given question,
′′+ ′ − 2 = 1 −
(solve Non-homogeneous equation)
′′ + ′ − 2 = 1 −

= Linear 0 + − 2 + = 1 −
− 2 − 2 = 1 −
Therefore, the trial solution

= + −2 + − 2 = − + 1

′ = Substitute , ′ and ′′ By comparing the coefficient of :
′′ = 0 into the given question
(to find a and b) = 1

∶ −2 = −1 2

constant : − 2 = 1 Substitute the
values of a and b
1 − 2 = 1
2 into

− 1 = 2 = − 1
2
MOOC MAT438/ UiTM 4

Solve the second order differential equation
′′+ ′ − 2 = 1 − , ℎ = 0, = 1, ′ = 0 the initial conditions is given !

If the initial condition is given (then continue with…)

Step 2 Find the trial solution, yp Step 4 Applying the given initial condition
(solve Non-homogeneous equation) (to find the values of C1 and C2)

= 1 , = − 1 Substitute into = + Find y’ (differentiate solution, y)
2 4

→ = 1 − 1 From = 1 + 2 −2 + 1 − 1
4
2 24

Step 3 Find the solution, y → ′ = 1 − 2 2 −2 + 1
2

= +

= 1 + 2 −2 + 1 − 1

24

(this is the final answer, if the initial
conditions is not given)

MOOC MAT438/ UiTM

Solve the second order differential equation
′′+ ′ − 2 = 1 − , ℎ = 0, = 1, ′ = 0

1 = 1 + 2 − 1
4

Step 4 Applying the given initial condition 4 = 4 1 + 4 2 − 1
(to find the values of C1 and C2)

= 1 + 2 −2 + 1 − 1 4 1 + 4 2 = 5 
2 4

′ = 1 − 2 2 −2 + 1 Step 6 Substitute the initial condition = 0,
2 ′ = 0 into ′ (naming it as equation )

Step 5 Substitute the initial condition = 0, 0 = 1 0 1− 2 2 0 1 1
= 1 into (naming it as equation ) 2
+

1 = 1 0 1 2 0 1+ 1 0 0 1 0 = 1 − 2 2 + 1
2 2
+ −
4

1 = 1 + 2 − 1 0 = 2 1 − 4 2 + 1
4 4 2 = 2 1 + 1


MOOC MAT438/ UiTM

Solve the second order differential equation
′′+ ′ − 2 = 1 − , ℎ = 0, = 1, ′ = 0

6 1 = 4

Step 5 Substitute the initial condition = 0, 1 = 2 (into )
= 1 into (naming it as equation ) 3

4 1 + 4 2 = 5  from  : 4 2 = 2 2 +1
3

Step 6 Substitute the initial condition = 0, 4 2 = 4 + 1
′ = 0 into ′ (naming it as equation ) 3

4 2 = 2 1 + 1  4 2 = 7
3

Step 7 Solve equation  and  simultaneously 2 = 7
12
(to find C1 and C2)

Substitute  into  : 4 1 + 2 1 + 1 = 5

MOOC MAT438/ UiTM

Solve the second order differential equation
′′+ ′ − 2 = 1 − , ℎ = 0, = 1, ′ = 0

Step 3 Find the solution, y Step 8 Write the particular solution (substitute back

the values of C1 and C2 into the general solution )

= 1 + 2 −2 + 1 − 1 = 2 + 7 −2 + 1 − 1 #
3 12 24
24

Step 7 Solve equation  and  simultaneously

(to find C1 and C2)

1 = 2 , 2 = 7
3 12

MOOC MAT438/ UiTM

Solve the second order differential equation
′′+ ′ − 2 = 1 − , ℎ = 0, = 1, ′ = 0

is my answer ✓ 2 7 1 1
correct ?? = 3 + 12 −2 + 2 − 4
Or… wrong,
′ = 2 − 7 −2 + 1 Substitute into the LHS question,
maybe ??? hmmm 36 2 and show that the LHS = RHS

MOOC MAT438/ UiTM ′′ = 2 + 7 −2

33

′′+ ′ − 2 = 12 − + 7 −2 + 2 − 7 −2 + 1 − 2 2 + 7 −2 + 1 − 1
36 2 3 12 2 4
33

= 2 0+ 7 −2 0

33

+ 2 − 7 −2 + 1
3 6 2

− 4 − 7 −2 − + 1
36 2

′′+ ′ − 2 = 1 − Shown !

MOOC MAT438/ UiTM

Solve the second order differential equation
′′−4 ′ + 4 = 2

Step 1 Find the general solution, yg Write the general solution, (based on the types of
(solve homogeneous equation) roots)
the general solution,
m1 m0
= 1 + 2
m2 m 1
= 1 2 + 2 2
′′−4 ′ + 4 = 0

2 − 4 + 4 = 0

− 2 − 2 = 0

1 = 2, = 2 2 = 2

1 = 2 Repeated real roots

MOOC MAT438/ UiTM

Solve the second order differential equation
′′−4 ′ + 4 = 2

Step 2 Find the trial solution, yp From the given question,
′′−4 ′ + 4 = 2
(solve Non-homogeneous equation)
′′ − 4 ′ + 4 = 2

= Cosine −4 2 − 4 2 − 4 −2 2 + 2 2
Therefore, the trial solution + 4 2 + 2 = 2

= 2 + 2 Substitute , ′ and ′′ −4 2 − 4 2 = 2
′ = −2 2 + 2 2 into the given question −8 2 + 8 2
′ ′ = −4 2 − 4 2 (to find a and b) +4 2 + 4 2

−8 2 + 8 2 = 1 2 + 0 2

MOOC MAT438/ UiTM

Solve the second order differential equation
′′−4 ′ + 4 = 2

Step 2 Find the trial solution, yp Step 3 Find the solution, y

(solve Non-homogeneous equation)

= +

−8 2 + 8 2 = 1 2 + 0 2 1
8
= 1 2 + 2 2 − 2 #

By comparing the coefficient of :

2 ∶ −8 = 1 = − 1 Substitute into this is the final answer,
2 ∶ 8 = 0 = 2 + 2 because the initial
8
conditions is not given !
= 0

→ = 0 2 − 1 2
8

→ = − 1 2
8

MOOC MAT438/ UiTM

Solve the second order differential equation = 2 = 2
′′−4 ′ + 4 = 2
′ = 2 ′ = 2 2

2 2 = ′ + ′


= 2 2 + 2 2 2

✓ 1
= 2 2 = 2 = 1 2 + 2 2 − 8 2

′ = 2 2 ′ = 2 2 1 Substitute into the LHS
4 question, and show that the
2 2 2 = ′ + ′ ′ = 2 1 2 + 2 2 + 2 2 2 − 2
LHS = RHS

= 2 2 2 + 4 2 2 ′′ = 4 1 2 + 2 2 2 + 2 2 2 + 4 2 2 + 1 2
2

00 0 0

′′−4 ′ + 4 = 4 1 2 + 2 2 2 + 2 2 2 + 4 2 2 + 1 2
2

−8 1 2 − 4 2 2 − 8 2 2 + 2 − 1 2
+4 1 2 + 4 2 2 2

′′−4 ′ + 4 = 2 Shown !

MOOC MAT438/ UiTM

MOOC MAT438/ UiTM

Solve the second order differential equation
′′+4 = 2 + 1 − 3

Step 1 Find the general solution, yg Write the general solution, (based on the types of
(solve homogeneous equation) roots)
the general solution,
m1 m0
= 1 + 2
m2 m 1 = 0 1 1 2 + 2 2

′′+ 0 ′ + 4 = 0 = 1 2 + 2 2

2 + 4 = 0

2 = −4

= ± −4 = ± 4 −1 = ±2

complex roots = ±2 = 0 ± 2
= ±
MOOC MAT438/ UiTM = 0 = 2

Solve the second order differential equation
′′+4 = 2 + 1 − 3

Step 2 Find the trial solution, yp From the given question,
′′+4 = 2 + 1 − 3
(solve Non-homogeneous equation)
′′ + 4 = 2 + 1 − 3
= Quadratic + Ecoxnpsotnaenntti−alExponential 2 + 9 3 + 4 2 + + + 3 = 2 + 1 − 3

Therefore, the trial solution 2 + 9 3 + 4 2 + 4 + 4 + 4 3 = 2 + 1 − 3

= 2 + + + 3 Substitute , ′ and ′′ 4 2 + 4 + 2 + 4 + 13 3 = 2 + 0 + 1 − 3
′ = 2 + + 3 3 into the given question
′ ′ = 2 + 9 3 (to find a, b and c)

MOOC MAT438/ UiTM

Solve the second order differential equation
′′+4 = 2 + 1 − 3

By comparing the coefficient of :

Step 2 Find the trial solution, yp 3 ∶ 13 = −1 = − 1

(solve Non-homogeneous equation) 13

4 2 + 4 + 2 + 4 + 13 3 = 2 + 0 + 1 − 3 By substituting the values of a, b, c and d into :
4 2 + 4 + 2 + 4 + 13 3 = 1 2 + 0 + 1 − 1 3
→ = 2 + + + 3

→ = 1 2 + 0 + 1 − 1 3
4 8 13

By comparing the coefficient of : Step 3 Find the solution, y

2 ∶ 4 = 1 = 1 this is the final answer,
because the initial
4
conditions is not given !
= +
∶ 4 = 0 = 0
1 1 1
Constant : 2 + 4 = 1 1 = 1 2 + 2 2 + 4 2 + 8 − 13 3 #
2 1 + 4 = 1 4 = 2
1 + 4 = 1
4 1
2 = 8
MOOC MAT438/ UiTM

Solve the second order differential equation
′′+4 = 2 + 1 − 3

is my answer ✓ = + + 1 2 + 1 − 1 3
correct ?? 1 2 2 2 4 8 13
Or… wrong,
′ = −2 1 2 + 2 2 2 + 1 − 3 3 Substitute into the LHS
maybe ??? hmmm 2 13 question, and show that the

1 9 LHS = RHS
2 13
′′ = −4 1 2 − 4 2 2 + − 3

0 0 1 − 13

−4 1 2 − 4 2 2 13

′′+4 = + 1 − 9 3

2 13

+ 4 1 2 + 4 2 2 + 2 + 1 − 4 3
2 13

′′+4 = 2 + 1 − 3 Shown !

FSKM/ UiTM Pahang © Amirah Hana

MOOC MAT438/ UiTM