 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2021/paper-2/15/c/i/solution

The dependent variable itself does not occur, so reduce order by putting

p = y^{'}

. Then

p^{'} + 2 p = 3 x

e^{2 x}

(e^{2 x} p)^{'} = 3 x e^{2 x}, p = \frac{3}{2} x - \frac{3}{4} + C e^{- 2 x} .

Integrating once more and renaming the constants gives

y (x) = \frac{3}{4} x^{2} - \frac{3}{4} x + A + B e^{- 2 x} .

The two homogeneous terms are the constant solution and the exponential associated with the roots

0, - 2

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