APM2611 Assignment 4 (COMPLETE ANSWERS) 2025 - DUE 24 September 2025 ; 100% correct solutions and explanations.

, APM2611 Assignment 4 (COMPLETE ANSWERS) 2025
- DUE 24 September 2025 ; 100% correct solutions and
explanations.
Question 1

𝒑𝒐𝒘𝒆𝒓 𝒔𝒆𝒓𝒊𝒆𝒔 𝒎𝒆𝒕𝒉𝒐𝒅.

𝑊𝑒 𝑤𝑎𝑛𝑡 𝑡𝑜 𝑠𝑜𝑙𝑣𝑒:

𝑦′′ − 𝑥𝑦′ + 4𝑦 = 2, 𝑦(0) = 0, 𝑦′(0) = 1. 𝑦′′ − 𝑥 𝑦′ + 4𝑦 = 2,\𝑞𝑢𝑎𝑑 𝑦(0)
= 0,\𝑞𝑢𝑎𝑑 𝑦′(0) = 1. 𝑦′′ − 𝑥𝑦′ + 4𝑦 = 2, 𝑦(0) = 0, 𝑦′(0) = 1.


𝑺𝒕𝒆𝒑 𝟏: 𝑨𝒔𝒔𝒖𝒎𝒆 𝒂 𝒑𝒐𝒘𝒆𝒓 𝒔𝒆𝒓𝒊𝒆𝒔 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏

𝐿𝑒𝑡

𝑦(𝑥) = ∑𝑛 = 0∞𝑎𝑛𝑥𝑛. 𝑦(𝑥) = \𝑠𝑢𝑚_{𝑛 = 0}^\𝑖𝑛𝑓𝑡𝑦 𝑎_𝑛 𝑥^𝑛. 𝑦(𝑥) = 𝑛 = 0∑∞𝑎𝑛𝑥𝑛.

𝑇ℎ𝑒𝑛

𝑦′(𝑥) = ∑𝑛 = 1∞𝑛𝑎𝑛𝑥𝑛 − 1, 𝑦′′(𝑥) = ∑𝑛 = 2∞𝑛(𝑛 − 1)𝑎𝑛𝑥𝑛 − 2. 𝑦′(𝑥) = \𝑠𝑢𝑚_{𝑛
= 1}^\𝑖𝑛𝑓𝑡𝑦 𝑛 𝑎_𝑛 𝑥^{𝑛 − 1},\𝑞𝑞𝑢𝑎𝑑 𝑦′′(𝑥) = \𝑠𝑢𝑚_{𝑛
= 2}^\𝑖𝑛𝑓𝑡𝑦 𝑛(𝑛 − 1)𝑎_𝑛 𝑥^{𝑛 − 2}. 𝑦′(𝑥) = 𝑛 = 1∑∞𝑛𝑎𝑛𝑥𝑛 − 1, 𝑦′′(𝑥) = 𝑛
= 2∑∞𝑛(𝑛 − 1)𝑎𝑛𝑥𝑛 − 2.


𝑺𝒕𝒆𝒑 𝟐: 𝑺𝒖𝒃𝒔𝒕𝒊𝒕𝒖𝒕𝒆 𝒊𝒏𝒕𝒐 𝒕𝒉𝒆 𝑫𝑬

𝑇ℎ𝑒 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑡𝑖𝑎𝑙 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑖𝑠

𝑦′′ − 𝑥𝑦′ + 4𝑦 = 2. 𝑦′′ − 𝑥 𝑦′ + 4𝑦 = 2. 𝑦′′ − 𝑥𝑦′ + 4𝑦 = 2.

 𝐹𝑜𝑟 𝑦′′𝑦′′𝑦′′:

𝑦′′ = ∑𝑛 = 2∞𝑛(𝑛 − 1)𝑎𝑛𝑥𝑛 − 2. 𝑦′′ = \𝑠𝑢𝑚_{𝑛 = 2}^\𝑖𝑛𝑓𝑡𝑦 𝑛(𝑛 − 1)𝑎_𝑛 𝑥^{𝑛 − 2}. 𝑦′′
= 𝑛 = 2∑∞𝑛(𝑛 − 1)𝑎𝑛𝑥𝑛 − 2.

𝑆ℎ𝑖𝑓𝑡 𝑖𝑛𝑑𝑒𝑥 𝑚 = 𝑛 − 2𝑚 = 𝑛 − 2𝑚 = 𝑛 − 2:

𝑦′′ = ∑𝑚 = 0∞(𝑚 + 2)(𝑚 + 1)𝑎𝑚 + 2𝑥𝑚. 𝑦′′ = \𝑠𝑢𝑚_{𝑚
= 0}^\𝑖𝑛𝑓𝑡𝑦 (𝑚 + 2)(𝑚 + 1) 𝑎_{𝑚 + 2} 𝑥^𝑚. 𝑦′′ = 𝑚
= 0∑∞(𝑚 + 2)(𝑚 + 1)𝑎𝑚 + 2𝑥𝑚.