 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2017/paper-2/9/b/solution

The same chain rule gives

u_{xx} = u^{''} (ξ)

and

u_{y} = u^{'} (ξ) / y

. Substituting these and part (a) into the partial differential equation gives

u_{xx} - \frac{1}{2} u_{y} - y u_{yy} = u^{''} - \frac{u ^{'}}{2 y} - y (\frac{u ^{''}}{y} - \frac{u ^{'}}{2 y ^{3/2}}) = 0 .

Thus every twice-differentiable profile

u (ξ)

gives a solution on

y > 0

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