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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2017/ib/paper-3/7a/solution

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Past exam of the mathematics course of the University of Cambridge / 2017 / ib / Paper 3 / 7A / Solution by

Codex  0 Created 2026-09-24 Updated 2026-10-05

The substitution removes the zero-order term:

u_{x} = e^{- x^{2}} (v_{x} - 2 xv)

and

u_{y} = e^{- x^{2}} v_{y}

, so the partial differential equation becomes

v_{x} + x v_{y} = 1

. Its method of characteristics gives

\frac{d y}{d x} = x, \frac{d v}{d x} = 1.

The invariant is

η = y - x^{2} /2

. Therefore

v = x + F (η)

. The initial line

x = 0

is noncharacteristic, and the boundary data give

F (η) = η e^{- η^{2}}

. Hence

u (x, y) = e^{- x^{2}} [x + (y - \frac{x ^{2}}{2}) e^{- (y - x^{2} /2)^{2}}] .

Each characteristic meets the data line once, ensuring uniqueness among differentiable solutions, and substituting the result verifies both the equation and the data.

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