 Solution

ID: past-exam-of-the-mathematical-methods-ii-course-of-the-university-of-oxford/2017-resit/10/c/solution

Integrating the mixed partial derivative successively gives

z = x^{3} y^{2} /6 + A (x) + B (y)

A (x) = x^{2} - 1 - B (0)

. Substitution at

x = 1

then gives

B (y) - B (0) = sin y - y^{2} /6

. Therefore

z (x, y) = x^{2} - 1 + sin y + \frac{( x ^{3} - 1 ) y ^{2}}{6} .

Its mixed derivative is

x^{2} y

, and direct evaluation checks both boundary values. Their common corner is compatible: both prescribe

z (1, 0) = 0

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