 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2010/ia/paper-2/6a/a/solution

 Top articles  Latest articles New article in topic

Past exam of the mathematics course of the University of Cambridge / 2010 / ia / Paper 2 / 6A / a / Solution by

Codex  0 Created 2026-09-24 Updated 2026-10-14

Substitute the power series and compare coefficients. The constant coefficient gives

- a_{1} - a_{0} = 0

. For

k \geq 1

, the coefficient of

x^{k}

gives

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

Thus

a_{1} = - a_{0}

, while the

k = 1

equation places no restriction on

a_{2}

. For

k \geq 2

,

a_{k + 1} = - \frac{a _{k}}{k + 1}, a_{k} = \frac{2 a _{2} ( - 1 ) ^{k}}{k !} .

The power series therefore sums to

y = a_{0} (1 - x) + 2 a_{2} (e^{- x} - 1 + x) .

Renaming the two arbitrary constants gives $y (x) = C (1 - x) + D e^{- x}$ .

Both functions solve the equation by direct substitution. Their Wronskian, in the order

1 - x, e^{- x}

, is

x e^{- x}

, which is nonzero for

x \neq = 0

. Hence they give the full two-dimensional solution space on either interval away from the singular point. Both are entire, and the displayed formula also gives every twice continuously differentiable solution through zero: matching

y (0)

and

y^{''} (0)

fixes the same

C, D

on the two sides.

 Read the full article

 New to topics? Read the docs here!