APM2611 Assignment 4 |COMPLETE ANSWERS & SOLUTIONS|- Due 24 September 2025

APM2611
Assignment 4

Due 24 September 2025

, Assignment 04

Due date: Wednesday, 24 September 2025

Question 1

Use the power series method to solve the initial value problem:

𝑦 ″ − 𝑥𝑦 ′ + 4𝑦 = 2, 𝑦(0) = 0, 𝑦 ′ (0) = 1

Answer

Assume the solution is of the form:

∞

𝑦(𝑥) = ∑ 𝑎 𝑛 𝑥𝑛
𝑛=0


Then,

∞ ∞
′ 𝑛−1
𝑦 (𝑥) = ∑ 𝑛 𝑎𝑛 𝑥 = ∑( 𝑛 + 1)𝑎 𝑛+1 𝑥 𝑛
𝑛=1 𝑛=0

∞ ∞

𝑦 ″ (𝑥) = ∑ 𝑛 (𝑛 − 1)𝑎 𝑛 𝑥 𝑛−2 = ∑( 𝑛 + 2)(𝑛 + 1)𝑎 𝑛+2 𝑥 𝑛
𝑛=2 𝑛=0


Substitute into the differential equation:

𝑦 ″ − 𝑥𝑦 ′ + 4𝑦 = 2

∞ ∞ ∞

∑( 𝑛 + 2)(𝑛 + 1)𝑎 𝑛+2 𝑥 𝑛 − 𝑥 ∑( 𝑛 + 1)𝑎 𝑛+1 𝑥 𝑛 + 4 ∑ 𝑎 𝑛 𝑥𝑛 = 2
𝑛=0 𝑛=0 𝑛=0


We simplify the middle term:

∞ ∞ ∞

𝑥 ∑( 𝑛 + 1)𝑎 𝑛+1 𝑥 𝑛 = ∑( 𝑛 + 1)𝑎 𝑛+1 𝑥 𝑛+1 = ∑ 𝑛 𝑎𝑛 𝑥 𝑛
𝑛=0 𝑛=0 𝑛=1


Now all terms become:

, ∞

∑ [(𝑛 + 2)(𝑛 + 1)𝑎 𝑛+2 − 𝑛𝑎𝑛 + 4𝑎𝑛 ] 𝑥 𝑛 = 2
𝑛=0

∞

∑ [(𝑛 + 2)(𝑛 + 1)𝑎 𝑛+2 + (4 − 𝑛)𝑎 𝑛 ] 𝑥 𝑛 = 2
𝑛=0


Right-hand side: 2 = ∑∞ 𝑛
𝑛=0 𝑐𝑛 𝑥 with 𝑐0 = 2, 𝑐𝑛 = 0 for 𝑛 ≥ 1


So,

(𝑛 + 2)(𝑛 + 1)𝑎 𝑛+2 + (4 − 𝑛)𝑎 𝑛 = 𝑐 𝑛

Recurrence relation:

(𝑛 + 2)(𝑛 + 1)𝑎 𝑛+2 = 𝑐 𝑛 − (4 − 𝑛)𝑎 𝑛

Initial values:

Given:

𝑦(0) = 𝑎 0 = 0, 𝑦 ′ (0) = 𝑎 1 = 1

Now compute coefficients:

n = 0:

(2)(1)𝑎2 + (4 − 0)𝑎0 = 2 ⇒ 2𝑎 2 + 4(0) = 2 ⇒ 𝑎 2 = 1

n = 1:

1
(3)(2)𝑎3 + (4 − 1)𝑎1 = 0 ⇒ 6𝑎 3 + 3(1) = 0 ⇒ 𝑎 3 = −
2

n = 2:

1
(4)(3)𝑎4 + (4 − 2)𝑎2 = 0 ⇒ 12𝑎 4 + 2(1) = 0 ⇒ 𝑎 4 = −
6

n = 3:

1 1
(5)(4)𝑎5 + (4 − 3)𝑎3 = 0 ⇒ 20𝑎 5 + 1(− ) = 0 ⇒ 𝑎 5 =
2 40