 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2012/paper-1/13/c/solution

e^{k x}

(e^{k x} y)^{'} = a e^{k x} sin (m x)

. The required antiderivative is

\int e^{k x} sin (m x) d x = \frac{e ^{k x}}{k ^{2} + m ^{2}} [k sin (m x) - m cos (m x)] .

After dividing by

e^{k x}

, the general solution is the oscillatory term plus

C e^{- k x}

C = 1 + am / (k^{2} + m^{2})

, so

y (x) = (1 + \frac{am}{k ^{2} + m ^{2}}) e^{- k x} + \frac{a [ k sin ( m x ) - m cos ( m x )]}{k ^{2} + m ^{2}} .

The real, nonzero parameters ensure

k^{2} + m^{2} > 0

; direct substitution verifies both the equation and

y (0) = 1

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