APM2611
EXAM PACK
2025
, APM2611
Assignment 04
Due 24 September 2025
,Question 1. Power Series Method to Solve an Initial Value Problem
Solve the initial value problem using the power series method:
y ′′ − xy ′ + 4y = 2, y(0) = 0, y ′ (0) = 1.
Step 1: Assume a Power Series Solution
Assume:
∞
X
y(x) = an xn .
n=0
Then,
∞
X ∞
X
′ n−1 ′′
y (x) = nan x , y (x) = n(n − 1)an xn−2 .
n=1 n=2
Step 2: Substitute into the Differential Equation
Substituting into the given equation:
∞
X ∞
X ∞
X
n−2 n−1
n(n − 1)an x −x nan x +4 an xn = 2.
n=2 n=1 n=0
Rewriting:
∞
X ∞
X ∞
X
k k
(k + 2)(k + 1)ak+2 x − kak x + 4ak xk = 2.
k=0 k=0 k=0
Combine terms:
∞
X
[(k + 2)(k + 1)ak+2 + (4 − k)ak ] xk = 2.
k=0
Equating coefficients:
For k = 0:
1
2(1)(0)a2 + 4a0 = 2 ⇒ a0 = .
2
For k ≥ 1:
(k − 4)ak
(k + 2)(k + 1)ak+2 + (4 − k)ak = 0 ⇒ ak+2 = .
(k + 2)(k + 1)
Step 3: Apply Initial Conditions
Given:
y(0) = a0 = 0, y ′ (0) = a1 = 1.
1
Since the previous equation gave a0 = 2
, there is inconsistency. This indicates a
1
, particular solution is needed. Assume the particular solution is a constant yp = 12 , since
it satisfies:
yp′′ = 0, −xyp′ = 0, 4yp = 2.
Homogeneous solution uses:
a0 = 0, a1 = 1.
Use recurrence:
−4a0
a2 = = 0,
2·1
−3a1 1
a3 = =− ,
3·2 2
−2a2
a4 = = 0,
4·3
−1a3 1
a5 = = .
5·4 40
Therefore, the homogeneous series is:
1 1
yh (x) = x − x3 + x5 + · · · .
2 40
The full solution:
1 1 1
y(x) = yh (x) + yp = x − x3 + x5 + · · · + .
2 40 2
2
EXAM PACK
2025
, APM2611
Assignment 04
Due 24 September 2025
,Question 1. Power Series Method to Solve an Initial Value Problem
Solve the initial value problem using the power series method:
y ′′ − xy ′ + 4y = 2, y(0) = 0, y ′ (0) = 1.
Step 1: Assume a Power Series Solution
Assume:
∞
X
y(x) = an xn .
n=0
Then,
∞
X ∞
X
′ n−1 ′′
y (x) = nan x , y (x) = n(n − 1)an xn−2 .
n=1 n=2
Step 2: Substitute into the Differential Equation
Substituting into the given equation:
∞
X ∞
X ∞
X
n−2 n−1
n(n − 1)an x −x nan x +4 an xn = 2.
n=2 n=1 n=0
Rewriting:
∞
X ∞
X ∞
X
k k
(k + 2)(k + 1)ak+2 x − kak x + 4ak xk = 2.
k=0 k=0 k=0
Combine terms:
∞
X
[(k + 2)(k + 1)ak+2 + (4 − k)ak ] xk = 2.
k=0
Equating coefficients:
For k = 0:
1
2(1)(0)a2 + 4a0 = 2 ⇒ a0 = .
2
For k ≥ 1:
(k − 4)ak
(k + 2)(k + 1)ak+2 + (4 − k)ak = 0 ⇒ ak+2 = .
(k + 2)(k + 1)
Step 3: Apply Initial Conditions
Given:
y(0) = a0 = 0, y ′ (0) = a1 = 1.
1
Since the previous equation gave a0 = 2
, there is inconsistency. This indicates a
1
, particular solution is needed. Assume the particular solution is a constant yp = 12 , since
it satisfies:
yp′′ = 0, −xyp′ = 0, 4yp = 2.
Homogeneous solution uses:
a0 = 0, a1 = 1.
Use recurrence:
−4a0
a2 = = 0,
2·1
−3a1 1
a3 = =− ,
3·2 2
−2a2
a4 = = 0,
4·3
−1a3 1
a5 = = .
5·4 40
Therefore, the homogeneous series is:
1 1
yh (x) = x − x3 + x5 + · · · .
2 40
The full solution:
1 1 1
y(x) = yh (x) + yp = x − x3 + x5 + · · · + .
2 40 2
2
