 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2023/paper-1/3/b/ii/solution

For

t > 1

, every forcing point has

τ < t

. Hence the Green's function gives

x (t) = - \frac{1}{4} e^{- 2 t} \int_{0}^{1} τ e^{2 τ} d τ .

Since

\int τ e^{2 τ} d τ = (τ /2 - 1/4) e^{2 τ}

, the remaining integral is

(e^{2} + 1) /4

. Therefore

x (t) = - \frac{e ^{2} + 1}{16} e^{- 2 t}, t > 1.

This decays at positive infinity.

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