 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2020/paper-1/13/b/iv/solution

y_{c} + 2 y_{p, 2} + y_{p, 1}

y = e^{- 3 x /2} (A cos \frac{3 x}{2} + B sin \frac{3 x}{2}) + \frac{2 x}{3} - \frac{2}{3 3} + \frac{1}{3} e^{- 3 x} .

A = 2/ (33) - 1/3

. At

x = π

, the cosine vanishes and the sine equals

- 1

. Comparing with the second boundary condition then gives

- B e^{- 3 π /2} + \frac{2 π}{3} = 0, B = \frac{2 π}{3} e^{3 π /2} .

The unique solution satisfying both boundary conditions is therefore

y = e^{- 3 x /2} [(\frac{2}{3 3} - \frac{1}{3}) cos \frac{3 x}{2} + \frac{2 π}{3} e^{3 π /2} sin \frac{3 x}{2}] + \frac{2 x}{3} - \frac{2}{3 3} + \frac{1}{3} e^{- 3 x} .

Uniqueness follows because a homogeneous solution zero at both endpoints would have

A = 0

from the first and

B = 0

from the second.

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