Differentiate the expression \({x^2}\tan y\) with respect to \(x\), where \(y\) is a function of \(x\).

Hence solve the differential equation \({x^2}\frac{{{\text{d}}y}}{{{\text{d}}x}} + x\sin 2y = {x^3}{\cos ^2}y\) given that \(y = 0\) when \(x = 1\). Give your answer in the form \(y = f(x)\).

\(\frac{{\text{d}}}{{{\text{d}}x}}({x^2}\tan y) = 2x\tan y + {x^2}{\sec ^2}y\frac{{{\text{d}}y}}{{{\text{d}}x}}\)     M1A1A1

\({x^2}\frac{{{\text{d}}y}}{{{\text{d}}x}} + 2x\sin y\cos y = {x^3}{\cos ^2}y\)     A1

\( \Rightarrow {x^2}{\sec ^2}y\frac{{{\text{d}}y}}{{{\text{d}}x}} + 2x\tan y = {x^3}\)     M1A1

\( \Rightarrow \frac{{\text{d}}}{{{\text{d}}x}}({x^2}\tan y) = {x^3}\)     A1

\({x^2}\tan y = \frac{{{x^4}}}{4} + c\)     A1

Note: Condone the omission of \(c\) in the line above

when \(x = 1,{\text{ }}y = 0 \Rightarrow c =  - \frac{1}{4}\)     M1

\(\tan y = \frac{{{x^2}}}{4} - \frac{1}{{4{x^2}}}\)

\(y = \arctan \left( {\frac{{{x^2}}}{4} - \frac{1}{{4{x^2}}}} \right)\)     A1

Many candidates made the connection between (a) and (b) and went on to solve the differential equation correctly. Candidates who failed to make the connection usually tried to write the equation in the form required for the use of an integrating factor but this led nowhere.

Many candidates made the connection between (a) and (b) and went on to solve the differential equation correctly. Candidates who failed to make the connection usually tried to write the equation in the form required for the use of an integrating factor but this led nowhere.