 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2018/paper-2/20/a/ii/solution

u (x, 0) = a x^{2} + d = 0

for all

x

, so

a = d = 0

. Since

u_{y} = b x + 2 cy

b = 0

2 a - 2 c = 1

, fixing

c = - 1/2

. Thus

a = 0, b = 0, c = - \frac{1}{2}, d = 0, u (x, y) = - \frac{y ^{2}}{2} .

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