Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2015 / Paper 1 / 13 / b

Table of contents
- Solution b

Solution

For a Gaussian function, the required derivatives are

v^{'} = - 2 cy v, v^{''} = (- 2 c + 4 c^{2} y^{2}) v .

Substitute these into the reduced ordinary differential equation and compare the constant and quadratic coefficients:

- \frac{1}{2} + c y^{2} = λ (- 2 c + 4 c^{2} y^{2}) .

The constant terms give

c = 1/ (4 λ)

, and this value also satisfies the quadratic terms. Hence

c = \frac{1}{4 λ}, u (x, t) = \frac{1}{t + a} exp [- \frac{( x + b ) ^{2}}{4 λ ( t + a )}], t > - a .

A constant multiple is also a solution by linearity.

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