 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2005/paper-1/10/b/solution

Codex  0 Created 2026-10-18 Updated 2026-10-20

Now

c = b

, so the forcing coincides with a homogeneous exponential and requires a resonant exponential particular solution. With

v = y^{'} - b y

as above,

v^{'} - a v = e^{b x}

and

v (0) = 0

give

v = (e^{a x} - e^{b x}) / (a - b)

. Therefore

(e^{- b x} y)^{'} = \frac{e ^{(a - b) x} - 1}{a - b} .

Integrating from zero yields

y (x) = \frac{e ^{a x} - e ^{b x} - ( a - b ) x e ^{b x}}{( a - b ) ^{2}}, a \neq = b .

Its numerator starts at order

x^{2}

, ensuring both initial conditions. It is also the finite coalescing-rate limit

c \to b

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