Find the general solution of the differential equation \(t\frac{{{\text{d}}y}}{{{\text{d}}t}} = \cos t - 2y\) , for t > 0 .

recognise equation as first order linear and attempt to find the IF     M1

\({\text{IF}} = {{\text{e}}^{\int {\frac{2}{t}{\text{d}}t} }} = {t^2}\)     A1

solution \(y{t^2} = \int {t\cos t{\text{d}}t} \)     M1A1

using integration by parts with the correct choice of u and v     (M1)

\(\int {t\cos t{\text{d}}t = t\sin t + \cos t( + C)} \)     A1

obtain \(y = \frac{{\sin t}}{t} + \frac{{\cos t + C}}{{{t^2}}}\)     A1

[7 marks]

Perhaps a small number of candidates were put off by the unusual choice of variables but in most instances it seemed that candidates who recognised the need for an integration factor could make a good attempt at this problem. Candidates who were not able to simplify the integrating factor from \({e^{2\ln t}}\) to \({t^2}\) rarely gained full marks. A significant number of candidates did not gain the final mark due to a lack of an arbitrary constant or not dividing the constant by the integration factor.