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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2017/ib/paper-3/4a/solution

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Past exam of the mathematics course of the University of Cambridge / 2017 / ib / Paper 3 / 4A / Solution by

Codex  0 Created 2026-09-24 Updated 2026-10-05

Let

Y (s)

be the Laplace transform of

y

. The initial values give

L (y^{'}) = s Y

and

L (y^{''}) = s^{2} Y - 1

, while

L (t e^{- 3 t}) = (s + 3)^{- 2}

. Therefore, for

Re s

sufficiently large,

(s^{2} - 4 s + 3) Y (s) - 1 = \frac{1}{( s + 3 ) ^{2}}, Y (s) = \frac{1 + ( s + 3 ) ^{- 2}}{( s - 1 ) ( s - 3 )} .

The partial fraction decomposition is

Y (s) = \frac{37}{72 ( s - 3 )} - \frac{17}{32 ( s - 1 )} + \frac{5}{288 ( s + 3 )} + \frac{1}{24 ( s + 3 ) ^{2}} .

Inverting the elementary Laplace transforms gives

y (t) = \frac{37}{72} e^{3 t} - \frac{17}{32} e^{t} + (\frac{5}{288} + \frac{t}{24}) e^{- 3 t}, t \geq 0.

These coefficients give

y (0) = 0

and

y^{'} (0) = 1

; the transformed equation, or direct differentiation, verifies the forcing. Uniqueness follows from uniqueness for the corresponding linear initial-value ordinary differential equation.

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