 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2021/paper-1/8/b/solution

For the Frobenius solution, differentiate term by term:

y^{'} = \sum_{n \geq 0} (n + σ) a_{n} x^{n + σ - 1}, y^{''} = \sum_{n \geq 0} (n + σ) (n + σ - 1) a_{n} x^{n + σ - 2} .

The unshifted terms in the equation have coefficient

[4 (n + σ) (n + σ - 1) - 4 (n + σ) + 3] a_{n}

. The extra factors of

x

give the shifted contribution

[4 (n + σ) (n + σ - 1) + 1] a_{n} x^{n + σ + 1}

. Equating the coefficient of

x^{k + σ}

for

k \geq 1

[4 (k + σ)^{2} - 8 (k + σ) + 3] a_{k} + [4 (k + σ - 1) (k + σ - 2) + 1] a_{k - 1} = 0.

Factoring both coefficients is especially useful:

(2 k + 2 σ - 3) [(2 k + 2 σ - 1) a_{k} + (2 k + 2 σ - 3) a_{k - 1}] = 0.

The separate lowest-power equation is

(4 σ^{2} - 8 σ + 3) a_{0} = 0

. Do not cancel

2 k + 2 σ - 3

 New to topics? Read the docs here!