,APM2611 Assignment 4 (COMPLETE ANSWERS) 2025 - DUE 24
September 2025 ; 100% trusted ,comprehensive and complete
reliable solution with clear explanation.
Question 1
Use the power series method to solve the initial value problem:
y00 − xy0 + 4y = 2, y(0) = 0, y0(0) =
Problem: Solve the initial value problem
y′′−x y′+4y=2,y(0)=0,y′(0)=1y'' - x\,y' + 4y = 2,\quad y(0) = 0,\quad
y'(0) = 1y′′−xy′+4y=2,y(0)=0,y′(0)=1
using the power series method.
Method (Power Series Approach)
1. Assume a power series form about x=0x=0x=0:
, 𝑦(𝑥) = ∑𝑛 = 0∞𝑎𝑛𝑥𝑛 ⇒ 𝑦′(𝑥) = ∑𝑛 = 1∞𝑛𝑎𝑛𝑥𝑛 − 1, 𝑦′′(𝑥)
= ∑𝑛 = 2∞𝑛(𝑛 − 1)𝑎𝑛𝑥𝑛 − 2. 𝑦(𝑥) = \𝑠𝑢𝑚_{𝑛
= 0}^{\𝑖𝑛𝑓𝑡𝑦} 𝑎_𝑛 𝑥^𝑛 \𝑞𝑢𝑎𝑑\𝑅𝑖𝑔ℎ𝑡𝑎𝑟𝑟𝑜𝑤
\𝑞𝑢𝑎𝑑 𝑦′(𝑥) = \𝑠𝑢𝑚_{𝑛
= 1}^{\𝑖𝑛𝑓𝑡𝑦} 𝑛 𝑎_𝑛 𝑥^{𝑛 − 1},\𝑞𝑢𝑎𝑑 𝑦′′(𝑥)
= \𝑠𝑢𝑚_{𝑛
= 2}^{\𝑖𝑛𝑓𝑡𝑦} 𝑛(𝑛 − 1) 𝑎_𝑛 𝑥^{𝑛 − 2}. 𝑦(𝑥) = 𝑛
= 0∑∞𝑎𝑛𝑥𝑛 ⇒ 𝑦′(𝑥) = 𝑛 = 1∑∞𝑛𝑎𝑛𝑥𝑛 − 1, 𝑦′′(𝑥)
= 𝑛 = 2∑∞𝑛(𝑛 − 1)𝑎𝑛𝑥𝑛 − 2.
2. Substitute into the differential equation:
∑𝑛 = 2∞𝑛(𝑛 − 1)𝑎𝑛𝑥𝑛 − 2 − 𝑥∑𝑛 = 1∞𝑛𝑎𝑛𝑥𝑛 − 1 + 4∑𝑛
= 0∞𝑎𝑛𝑥𝑛 = 2.\𝑠𝑢𝑚_{𝑛
= 2}^{\𝑖𝑛𝑓𝑡𝑦} 𝑛(𝑛 − 1) 𝑎_𝑛 𝑥^{𝑛 − 2} − 𝑥\𝑠𝑢𝑚_{𝑛
= 1}^{\𝑖𝑛𝑓𝑡𝑦} 𝑛 𝑎_𝑛 𝑥^{𝑛 − 1} + 4\𝑠𝑢𝑚_{𝑛
= 0}^{\𝑖𝑛𝑓𝑡𝑦} 𝑎_𝑛 𝑥^𝑛 = 2. 𝑛
= 2∑∞𝑛(𝑛 − 1)𝑎𝑛𝑥𝑛 − 2 − 𝑥𝑛
= 1∑∞𝑛𝑎𝑛𝑥𝑛 − 1 + 4𝑛 = 0∑∞𝑎𝑛𝑥𝑛 = 2.
3. Rewrite with same index power (let k=n−2k=n-2k=n−2 in the
first sum, and simplify the second):
∑𝑘 = 0∞(𝑘 + 2)(𝑘 + 1)𝑎𝑘 + 2𝑥𝑘 − ∑𝑛 = 1∞𝑛𝑎𝑛𝑥𝑛 + 4∑𝑛
= 0∞𝑎𝑛𝑥𝑛 = 2.\𝑠𝑢𝑚_{𝑘
= 0}^{\𝑖𝑛𝑓𝑡𝑦} (𝑘 + 2)(𝑘 + 1)𝑎_{𝑘 + 2} 𝑥^𝑘
− \𝑠𝑢𝑚_{𝑛 = 1}^{\𝑖𝑛𝑓𝑡𝑦} 𝑛 𝑎_𝑛 𝑥^𝑛 + 4\𝑠𝑢𝑚_{𝑛
= 0}^{\𝑖𝑛𝑓𝑡𝑦} 𝑎_𝑛 𝑥^𝑛 = 2. 𝑘
= 0∑∞(𝑘 + 2)(𝑘 + 1)𝑎𝑘 + 2𝑥𝑘 − 𝑛
= 1∑∞𝑛𝑎𝑛𝑥𝑛 + 4𝑛 = 0∑∞𝑎𝑛𝑥𝑛 = 2.
Write everything as a single summation in powers of xnx^nxn.
4. Match coefficients for each power xnx^nxn:
September 2025 ; 100% trusted ,comprehensive and complete
reliable solution with clear explanation.
Question 1
Use the power series method to solve the initial value problem:
y00 − xy0 + 4y = 2, y(0) = 0, y0(0) =
Problem: Solve the initial value problem
y′′−x y′+4y=2,y(0)=0,y′(0)=1y'' - x\,y' + 4y = 2,\quad y(0) = 0,\quad
y'(0) = 1y′′−xy′+4y=2,y(0)=0,y′(0)=1
using the power series method.
Method (Power Series Approach)
1. Assume a power series form about x=0x=0x=0:
, 𝑦(𝑥) = ∑𝑛 = 0∞𝑎𝑛𝑥𝑛 ⇒ 𝑦′(𝑥) = ∑𝑛 = 1∞𝑛𝑎𝑛𝑥𝑛 − 1, 𝑦′′(𝑥)
= ∑𝑛 = 2∞𝑛(𝑛 − 1)𝑎𝑛𝑥𝑛 − 2. 𝑦(𝑥) = \𝑠𝑢𝑚_{𝑛
= 0}^{\𝑖𝑛𝑓𝑡𝑦} 𝑎_𝑛 𝑥^𝑛 \𝑞𝑢𝑎𝑑\𝑅𝑖𝑔ℎ𝑡𝑎𝑟𝑟𝑜𝑤
\𝑞𝑢𝑎𝑑 𝑦′(𝑥) = \𝑠𝑢𝑚_{𝑛
= 1}^{\𝑖𝑛𝑓𝑡𝑦} 𝑛 𝑎_𝑛 𝑥^{𝑛 − 1},\𝑞𝑢𝑎𝑑 𝑦′′(𝑥)
= \𝑠𝑢𝑚_{𝑛
= 2}^{\𝑖𝑛𝑓𝑡𝑦} 𝑛(𝑛 − 1) 𝑎_𝑛 𝑥^{𝑛 − 2}. 𝑦(𝑥) = 𝑛
= 0∑∞𝑎𝑛𝑥𝑛 ⇒ 𝑦′(𝑥) = 𝑛 = 1∑∞𝑛𝑎𝑛𝑥𝑛 − 1, 𝑦′′(𝑥)
= 𝑛 = 2∑∞𝑛(𝑛 − 1)𝑎𝑛𝑥𝑛 − 2.
2. Substitute into the differential equation:
∑𝑛 = 2∞𝑛(𝑛 − 1)𝑎𝑛𝑥𝑛 − 2 − 𝑥∑𝑛 = 1∞𝑛𝑎𝑛𝑥𝑛 − 1 + 4∑𝑛
= 0∞𝑎𝑛𝑥𝑛 = 2.\𝑠𝑢𝑚_{𝑛
= 2}^{\𝑖𝑛𝑓𝑡𝑦} 𝑛(𝑛 − 1) 𝑎_𝑛 𝑥^{𝑛 − 2} − 𝑥\𝑠𝑢𝑚_{𝑛
= 1}^{\𝑖𝑛𝑓𝑡𝑦} 𝑛 𝑎_𝑛 𝑥^{𝑛 − 1} + 4\𝑠𝑢𝑚_{𝑛
= 0}^{\𝑖𝑛𝑓𝑡𝑦} 𝑎_𝑛 𝑥^𝑛 = 2. 𝑛
= 2∑∞𝑛(𝑛 − 1)𝑎𝑛𝑥𝑛 − 2 − 𝑥𝑛
= 1∑∞𝑛𝑎𝑛𝑥𝑛 − 1 + 4𝑛 = 0∑∞𝑎𝑛𝑥𝑛 = 2.
3. Rewrite with same index power (let k=n−2k=n-2k=n−2 in the
first sum, and simplify the second):
∑𝑘 = 0∞(𝑘 + 2)(𝑘 + 1)𝑎𝑘 + 2𝑥𝑘 − ∑𝑛 = 1∞𝑛𝑎𝑛𝑥𝑛 + 4∑𝑛
= 0∞𝑎𝑛𝑥𝑛 = 2.\𝑠𝑢𝑚_{𝑘
= 0}^{\𝑖𝑛𝑓𝑡𝑦} (𝑘 + 2)(𝑘 + 1)𝑎_{𝑘 + 2} 𝑥^𝑘
− \𝑠𝑢𝑚_{𝑛 = 1}^{\𝑖𝑛𝑓𝑡𝑦} 𝑛 𝑎_𝑛 𝑥^𝑛 + 4\𝑠𝑢𝑚_{𝑛
= 0}^{\𝑖𝑛𝑓𝑡𝑦} 𝑎_𝑛 𝑥^𝑛 = 2. 𝑘
= 0∑∞(𝑘 + 2)(𝑘 + 1)𝑎𝑘 + 2𝑥𝑘 − 𝑛
= 1∑∞𝑛𝑎𝑛𝑥𝑛 + 4𝑛 = 0∑∞𝑎𝑛𝑥𝑛 = 2.
Write everything as a single summation in powers of xnx^nxn.
4. Match coefficients for each power xnx^nxn:
