 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2005/ii/paper-2/31c/b/solution

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

The scaling of the equation suggests

q = t^{- 1/3} v (z)

z = x t^{- 1/3}

, for

t > 0

. Substitution gives

v^{(3)} + 6 v^{2} v^{'} - \frac{1}{3} (v + z v^{'}) = 0.

Integrating in

z

v^{''} = \frac{z}{3} v - 2 v^{3} + C, q (x, t) = t^{- 1/3} v (x t^{- 1/3}),

where

C

is an arbitrary integration constant. If the standard normalization

w^{''} = 2 w^{3} + s w + α

is desired, set

z = 3^{1/3} s

and

v (z) = i 3^{- 1/3} w (s)

. Then

α = - 3 i C

and

q (x, t) = i (3 t)^{- 1/3} w (x / (3 t)^{1/3}) .

The direct scaled equation avoids introducing imaginary variables when a real

v

is required.

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