APM2611 Assignment 4 2025 (Answer Guide) - Due 24 September 2025

APM2611 Assignment 4 2025 (Answer Guide) - Due 24 September 2025 APM2611 Assignment 4 - Question 1 Question 1 Use the power-series method to solve the initial-value problem: y'' - x y' + 4y = 2, y(0)=0, y'(0)=1. Solution 1. Assume a power series Assume y(x) = Σ (from n=0 to ∞) aₙ xⁿ. Then y'(x) = Σ (from n=0 to ∞) (n+1)aₙ₊₁ xⁿ, y''(x) = Σ (from n=0 to ∞) (n+2)(n+1)aₙ₊₂ xⁿ. Also, -x y'(x) = -Σ (from n=0 to ∞) n aₙ xⁿ. Substituting into the differential equation gives: Σ (n=0 to ∞)[(n+2)(n+1)aₙ₊₂ + (4-n)aₙ] xⁿ = 2. 2. Recurrence relations For n=0: 2a₂ + 4a₀ = 2. For n ≥ 1: (n+2)(n+1)aₙ₊₂ + (4-n)aₙ = 0 ⇒ aₙ₊₂ = -(4-n)/( (n+2)(n+1) ) * aₙ. 3. Apply initial conditions From y(0)=0 ⇒ a₀=0. From y'(0)=1 ⇒ a₁=1. For n=0: 2a₂ + 0 = 2 ⇒ a₂ = 1. 4. Compute coefficients Using recurrence: n=1: a₃ = -(4-1)/(3·2) * a₁ = -3/6 * 1 = -1/2. n=2: a₄ = -(4-2)/(4·3) * a₂ = -2/12 * 1 = -1/6. n=3: a₅ = -(4-3)/(5·4) * a₃ = -1/20 * (-1/2) = 1/40. n=4: a₆ = -(4-4)/(6·5) * a₄ = 0. n=5: a₇ = -(4-5)/(7·6) * a₅ = -(-1/42) * (1/40) = 1/1680. n=6: a₈ = -(4-6)/(8·7) * a₆ = 0. n=7: a₉ = -(4-7)/(9·8) * a₇ = -(-3/72)(1/1680) = 1/40320. 5. Series solution y(x) = a₀ + a₁x + a₂x² + a₃x³ + a₄x⁴ + a₅x⁵ + a₆x⁶ + a₇x⁷ + a₈x⁸ + a₉x⁹ + … Substitute known coefficients: y(x) = x + x² - (1/2)x³ - (1/6)x⁴ + (1/40)x⁵ + (1/1680)x⁷ + (1/40320)x⁹ + … 6. Remark The constructed power series satisfies the differential equation and the initial conditions. The recurrence relation aₙ₊₂ = -(4-n)/((n+2)(n+1)) · aₙ determines all higher coefficients. This is the required power series solution.