 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2025/paper-1/8/c/solution

In the shifted variable

t = y - 1

x = \sum_{j \geq 0} a_{j} t^{j + σ}

into

t (t + 2) x^{''} + x^{'} + x = 0

. Equating the coefficient of

t^{j + σ}

gives

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

j = 0, 1, \dots

, is

a_{j + 1} = - \frac{( j + σ ) ( j + σ - 1 ) + 1}{( j + 1 + σ ) [ 2 ( j + σ ) + 1 ]} a_{j} .

σ = 0

, this becomes

a_{j + 1} = - \frac{j ^{2} - j + 1}{( j + 1 ) ( 2 j + 1 )} a_{j}, a_{1} = - a_{0} .

σ = 1/2

b_{j}

; then

b_{j + 1} = - \frac{j ^{2} + 3/4}{( j + 3/2 ) ( 2 j + 2 )} b_{j}, b_{1} = - \frac{1}{4} b_{0} .

With

a_{0} = b_{0} = 1

, the first terms of the two Frobenius solutions are

x_{1} = 1 - t + \frac{1}{6} t^{2} + O (t^{3}), x_{2} = t^{1/2} (1 - \frac{1}{4} t + \frac{7}{160} t^{2} + O (t^{3})) .

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