 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2022/paper-1/3/a/solution

y = x^{r}

x > 0

r (r - 1) + α r + \frac{( α - 1 ) ^{2}}{4} = (r + \frac{α - 1}{2})^{2} = 0.

Thus

r = (1 - α) /2

is a repeated indicial root. The two independent solutions are

x^{r}

and

x^{r} lo g x

y_{h} (x) = x^{(1 - α) /2} (C_{1} + C_{2} lo g x), x > 0.

Their Wronskian, with

W = y_{1} y_{2}^{'} - y_{1}^{'} y_{2}

, is

x^{2 r - 1} = x^{- α} \neq = 0

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