 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/ia/paper-2/6a/solution

Codex  0 Created 2026-09-23 Updated 2026-10-03

The parameter sensitivity equation follows by differentiating with respect to

μ

u = y_{μ}

u_{x} = u + x + y^{2} + 2 μ y u, u (0, μ) = 0.

μ = 0

y_{x} = y

y (0) = 1

, so

y (x, 0) = e^{x}

. Then

u_{x} - u = x + e^{2 x}, u (0, 0) = 0,

e^{- x}

yields

u (x, 0) = e^{2 x} - x - 1.

Equating powers in

y = y_{0} + μ y_{1} + μ^{2} y_{2} + \dots

gives

y_{0} = e^{x}, y_{1} = e^{2 x} - x - 1,

y_{2}^{'} - y_{2} = 2 y_{0} y_{1}, y_{2} (0) = 0.

Consequently

y_{2} = e^{x} (e^{2 x} - 1 - x^{2} - 2 x) .

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