 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2015/paper-2/20/a/solution

ν u_{s}^{''} = - G sin (nπ x)

. Since the second derivative of the sine is

- (nπ)^{2}

times itself,

u_{s} (x) = \frac{G}{ν n ^{2} π ^{2}} sin (nπ x) + A x + B .

The zero boundary values at

x = 0, 1

A = B = 0

. Therefore

u_{s} (x) = \frac{G}{ν n ^{2} π ^{2}} sin (nπ x) .

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