Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2023 / Paper 1 / 8 / c

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- Solution c

Solution

Normalize the free leading coefficient to

a_{0} = 1

. The linear recurrence relation from part (b) gives

4 a_{2} + 4 a_{0} = 0, 16 a_{4} + 2 a_{0} = 0, 36 a_{6} + 12 a_{4} = 0.

Consequently the first four nonzero coefficients are

a_{0} = 1, a_{2} = - 1, a_{4} = - \frac{1}{8}, a_{6} = \frac{1}{24},

and the single pure Frobenius solution is

y_{1} (z) = \frac{1}{z} (1 - z^{2} - \frac{1}{8} z^{4} + \frac{1}{24} z^{6} + O (z^{8})) .

For clarity about the second independent solution, a choice with no added multiple of

y_{1}

is

y_{2} = y_{1} lo g z + \frac{5}{2} z + \frac{3}{8} z^{3} - \frac{11}{144} z^{5} + \frac{131}{9216} z^{7} + O (z^{9}) .

These companion coefficients follow by reduction of order with the normalized Wronskian

W = z^{- 3} (1 + z^{2})^{3/2} exp [3 z^{2} / (2 (1 + z^{2}))]

. The logarithm requires a local branch; it does not supply a second ordinary power series solution.

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