 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2023/paper-2/15/b/ii/solution

Here

α = 4

and

β = 4

(D - 2)^{2} u = 0

. Its repeated characteristic root gives

u = (A + Bt) e^{2 t}

A = 0

. The original equation

u^{'} = 3 u - v

gives

u^{'} (0) = - 1

, hence

B = - 1

Recover

v = 3 u - u^{'}

to obtain

u (t) = - t e^{2 t}, v (t) = (1 - t) e^{2 t} .

Direct differentiation verifies

v^{'} = u + v

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