APM2611 EXAM PACK ACCURATE SOLUTIONS 2025

APM2611
EXAM PACK
2025

, APM2611
Assignment 04
Due 24 September 2025

,Question 1: Power Series Method
Problem: Solve the initial value problem


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


using the power series method.

Step 1: Assume a power series solution.

∞
X
y(x) = an x n
n=0


∞
X ∞
X
′ n−1 ′′
y (x) = nan x , y (x) = n(n − 1)an xn−2
n=1 n=2



Step 2: Substitute into the differential equation.




y ′′ − xy ′ + 4y = 2
∞
X ∞
X ∞
X
n−2 n−1
n(n − 1)an x −x nan x +4 an x n = 2
n=2 n=1 n=0




Step 3: Align powers of x.
Let k = n − 2 in the first sum:

∞
X ∞
X ∞
X
k n
(k + 2)(k + 1)ak+2 x − nan x + 4an xn = 2
k=0 n=1 n=0



Shift index of the second sum to start at n = 0 by defining 0 · a0 = 0. Then:

∞
X
[(k + 2)(k + 1)ak+2 + 4ak − kak ] xk = 2
k=0


∞
X
[(k + 2)(k + 1)ak+2 + (4 − k)ak ] xk = 2
k=0



Step 4: Equate coefficients.


1

, For k = 0:
(2)(1)a2 + (4 − 0)a0 = 2 ⇒ 2a2 + 4a0 = 2

For k ≥ 1:

(k − 4)ak
(k + 2)(k + 1)ak+2 + (4 − k)ak = 0 ⇒ ak+2 =
(k + 2)(k + 1)


Step 5: Apply initial conditions.


y(0) = a0 = 0, y ′ (0) = a1 = 1


Step 6: Compute coefficients.




a0 = 0

a1 = 1
(0 − 4)a0
a2 = =0
2·1
(1 − 4)a1 1
a3 = =−
3·2 2
(2 − 4)a2
a4 = =0
4·3
(3 − 4)a3 1
a5 = =
5·4 40


Step 7: Homogeneous solution.

1 1
yh (x) = x − x3 + x5 + · · ·
2 40


Step 8: Particular solution.
Guess yp = C. Substitute into the equation:

1
yp′′ − xyp′ + 4yp = 0 + 0 + 4C = 2 ⇒ C =
2




2