APM2611 Assignment 4 due 24 September 2025

APM2611
ASSIGNMENT 4 2025

UNIQUE NO.
DUE DATE: 24 SEPTEMBER 2025

, Differential Equations


Q1. Power–series solution

Solve 𝑦 ″ − 𝑥 𝑦 ′ + 4𝑦 = 2 with 𝑦(0) = 0, 𝑦 ′ (0) = 1 .

Seek 𝑦(𝑥) = σ ∞ 𝑛
𝑛=0 𝑎𝑛 𝑥 . Then

∞ ∞

𝑦 ′ ෍ = 𝑛 𝑎𝑛 𝑥 𝑛−1 , 𝑦 ″ ෍ = 𝑛 (𝑛 − 1)𝑎𝑛 𝑥 𝑛−2 .
𝑛=1 𝑛=2


Plugging in and aligning powers of 𝑥:

∞ ∞ ∞

( ෍ 𝑛 + 2)(𝑛 + 1)𝑎 𝑛+2 𝑥 ෍ − 𝑛 𝑎𝑛 𝑥 ෍ 4 + 𝑎𝑛 𝑥 𝑛 = 2.
𝑛 𝑛

𝑛=0 𝑛=1 𝑛=0


Equate coefficients:

 𝑛 = 0: 2𝑎 2 + 4𝑎0 = 2 ⇒ 𝑎2 = 1 − 2𝑎 0 .
 𝑛 ≥ 1: (𝑛 + 2)(𝑛 + 1)𝑎 𝑛+2 + (4 − 𝑛)𝑎 𝑛 = 0 ⇒

𝑛−4
𝑎 𝑛+2 = 𝑎 (𝑛 ≥ 1) .
(𝑛 + 2)(𝑛 + 1) 𝑛

Initial conditions: 𝑎0 = 𝑦(0) = 0, 𝑎 1 = 𝑦 ′ (0) = 1 . Then

1−4 2−4 3−4
𝑎2 = 1 − 2(0) = 1, 𝑎 3 = 𝑎 = − 1Τ2 , 𝑎4 = 𝑎 = − 1Τ6 , 𝑎5 = 𝑎
3⋅2 1 4⋅3 2 5⋅4 3
= 1 Τ40 , 𝑎 6 = 0, 𝑎 7 = 1 Τ1680 , 𝑎 9 = 1 Τ40320 , …

Hence the power series about 𝑥 = 0 is

𝑥3 𝑥4 𝑥5
2
𝑥7 𝑥9
𝑦(𝑥) = 𝑥 + 𝑥 − − + + + +⋯ ,
2 6 40 1680 40320