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

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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