 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/iii/solution

For

0 < t < 1

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

The two terms inside the bracket are respectively

\frac{t}{2} - \frac{1}{4} + \frac{1}{4} e^{- 2 t}, \frac{t}{2} + \frac{1}{4} - \frac{3}{4} e^{2 t - 2} .

Adding them yields

x (t) = - \frac{t}{4} - \frac{1}{16} e^{- 2 t} + \frac{3}{16} e^{2 t - 2}, 0 < t < 1.

The affine term gives

x^{''} - 4 x = t

, while both exponential terms are homogeneous solutions. The values at the joins are

x (0) = (3 e^{- 2} - 1) /16

and

x (1) = - (1 + e^{- 2}) /16

, agreeing with parts (i) and (ii). Their first derivatives also match:

x^{'} (0) = (3 e^{- 2} - 1) /8

and

x^{'} (1) = (1 + e^{- 2}) /8

. Thus the pieces introduce no unwanted Dirac delta distributions.

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