 Solution

ID: 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/solution

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}) .

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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