 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/a/solution

T = t + a > 0

y = (x + b) T^{- 1/2}

x

t

are

y_{x} = T^{- 1/2}, y_{t} = - \frac{y}{2 T} .

u = T^{- 1/2} v (y)

u_{t} = - \frac{1}{2} T^{- 3/2} v + T^{- 1/2} v^{'} (- \frac{y}{2 T}) = - \frac{1}{2} T^{- 3/2} (v + y v^{'}),

u_{x} = T^{- 1} v^{'}, u_{xx} = T^{- 3/2} v^{''} .

Substituting in the heat equation and cancelling the positive common factor

T^{- 3/2}

gives

- \frac{1}{2} (v + y v^{'}) = λ v^{''} .

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