 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/d/solution

For the larger indicial root

σ = 3/2

, the recurrence in part (b) becomes

2 k [(2 k + 2) a_{k} + 2 k a_{k - 1}] = 0, k \geq 1.

No factor being cancelled is zero in this range, so

a_{k} = - \frac{k}{k + 1} a_{k - 1} = \frac{( - 1 ) ^{k}}{k + 1} a_{0} .

Thus the resulting Frobenius solution is

y (x) = a_{0} x^{3/2} k = 0 \sum \infty \frac{( - x ) ^{k}}{k + 1} = a_{0} x^{3/2} (1 - \frac{x}{2} + \frac{x ^{2}}{3} - \dots) .

There is exactly one free parameter,

a_{0}

, setting the overall scale. The analytic series factor has radius of convergence one. For a real-valued expression take

x > 0

; a complex version requires a local square-root branch.

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