 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/ia/paper-2/6c/solution

Codex  0 Created 2026-09-24 Updated 2026-09-29

y = \sum_{n \geq 0} a_{n} x^{n}

and compare the coefficient of

x^{n}

. This gives

(n + 2) (n + 1) a_{n + 2} + (λ^{2} - n^{2}) a_{n} = 0,

a_{n + 2} = \frac{n ^{2} - λ ^{2}}{( n + 1 ) ( n + 2 )} a_{n} .

The two arbitrary coefficients

a_{0}, a_{1}

determine all solutions. The condition

y^{'} (0) = 0

a_{1} = 0

, leaving the even series. It terminates precisely when

λ^{2} = (2 k)^{2}

for some nonnegative integer

k

. Therefore the permitted real values are

λ = 0

λ = \pm 2 k

for

k \geq 1

; these are the even Chebyshev polynomials up to normalization.

The degrees below six occur for

λ = 0, \pm 2, \pm 4

. Taking

a_{0} = 1

, the corresponding polynomials are

1, 1 - 2 x^{2}, 1 - 8 x^{2} + 8 x^{4} .

For the forced equation, try

y = a_{0} + a_{2} x^{2} + a_{4} x^{4}

. Comparing the coefficients of

x^{4}, x^{2}, 1

gives

- 15 a_{4} = 8, 12 a_{4} - 3 a_{2} = 0, 2 a_{2} + a_{0} = - 3.

Thus one polynomial satisfying

y^{'} (0) = 0

y (x) = \frac{19 - 32 x ^{2} - 8 x ^{4}}{15} .

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