 Solution

ID: past-exam-of-the-mathematical-methods-i-course-of-the-university-of-oxford/2014/5/solution

Rearrange as

x^{''} - 4 x^{'} + 3 x = 10 t

(r - 1) (r - 3)

e^{t}, e^{3 t}

. A particular polynomial solution

A t + B

obeys

3 A = 10

and

3 B - 4 A = 0

. Hence

x (t) = C_{1} e^{t} + C_{2} e^{3 t} + \frac{10}{3} t + \frac{40}{9} .

This is the full two-constant solution of the second-order linear differential equation.

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