Past exam of the mathematics course of the University of Cambridge / 2022 / ia / Paper 2 / 5C / b

Table of contents
- Solution b

Solution

After division by

x

, the coefficient of

y^{'}

is

P (x) = 1/ x - 1

. The Abel identity for the Wronskian gives

W^{'} = - P (x) W = (1 - \frac{1}{x}) W,

so

W (x) = C \frac{e ^{x}}{x} .

For

λ = 1

, direct substitution confirms that

y_{1} = 1 - x

. Reduction of order gives a second solution proportional to

y_{2} = (1 - x) \int \frac{e ^{x}}{x ( 1 - x ) ^{2}} d x .

To extract the requested coefficients, put

y_{2} = (1 - x) lo g x + b_{1} x + b_{2} x^{2} + \dots .

For the differential operator

L [y] = x y^{''} + (1 - x) y^{'} + y,

one finds

L [(1 - x) lo g x] = x - 3.

The analytic correction

h = \sum_{n \geq 1} b_{n} x^{n}

must therefore satisfy

L [h] = 3 - x

. Its constant and linear coefficients give

b_{1} = 3, 4 b_{2} = - 1.

Hence

y_{2} = (1 - x) lo g x + 3 x - \frac{1}{4} x^{2} + \dots .

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