APM2611
Assignment 4
Detailed Solutions
DUE 24 September 2025
, APM2611
Assignment 4: Detailed Solutions
DUE 24 September 2025
Question 1: Power Series Solution
Solve:
y′′ − xy′ + 4y = 2, y(0) = 0, y′(0) = 1
Step 1: Assume a power series solution
Let:
Step 2: Plug into the DE
We substitute into the DE:
Step 3: Align all terms as power series
Shift indices:
∞
⇒ X [(n + 2)(n + 1)an+2 + (4 − n)an]xn = 2
n=0
Step 4: Match coefficients Right-hand side: 2 + 0x + 0x2 + ...
• For n = 0:
2a2 + 4a0 = 2
• For n ≥ 1:
Step 5: Use initial conditions
a0 = 0, a1 = 1 ⇒ 2a2 + 4(0) = 2 ⇒ a2 = 1
1
Assignment 4
Detailed Solutions
DUE 24 September 2025
, APM2611
Assignment 4: Detailed Solutions
DUE 24 September 2025
Question 1: Power Series Solution
Solve:
y′′ − xy′ + 4y = 2, y(0) = 0, y′(0) = 1
Step 1: Assume a power series solution
Let:
Step 2: Plug into the DE
We substitute into the DE:
Step 3: Align all terms as power series
Shift indices:
∞
⇒ X [(n + 2)(n + 1)an+2 + (4 − n)an]xn = 2
n=0
Step 4: Match coefficients Right-hand side: 2 + 0x + 0x2 + ...
• For n = 0:
2a2 + 4a0 = 2
• For n ≥ 1:
Step 5: Use initial conditions
a0 = 0, a1 = 1 ⇒ 2a2 + 4(0) = 2 ⇒ a2 = 1
1
