Solve the differential equation

\({x^2}\frac{{{\text{d}}y}}{{{\text{d}}x}} = {y^2} + 3xy + 2{x^2}\)

given that y = −1 when x =1. Give your answer in the form \(y = f(x)\) .

put y = vx so that \(\frac{{{\text{d}}y}}{{{\text{d}}x}} = v + x\frac{{{\text{d}}v}}{{{\text{d}}x}}\)     M1

substituting,     M1

\(v + x\frac{{{\text{d}}v}}{{{\text{d}}x}} = \frac{{{v^2}{x^2} + 3v{x^2} + 2{x^2}}}{{{x^2}}}{\text{ }}( = {v^2} + 3v + 2)\)     (A1)

\(x\frac{{{\text{d}}v}}{{{\text{d}}x}} = {v^2} + 2v + 2\)     A1

\(\int {\frac{{{\text{d}}v}}{{{v^2} + 2v + 2}} = \int {\frac{{{\text{d}}x}}{x}} } \)     M1

\(\int {\frac{{{\text{d}}v}}{{{{(v + 1)}^2} + 1}} = \int {\frac{{{\text{d}}x}}{x}} } \)     (A1)

\(\arctan (v + 1) = \ln x + c\)     A1

Note: Condone absence of c at this stage.

\(\arctan (\frac{y}{x} + 1) = \ln x + c\)     M1

When x = 1, y = −1     M1

c = 0     A1

\(\frac{y}{x} + 1 = \tan \ln x\)

\(y = x(\tan \ln x - 1)\)     A1

[11 marks]

Most candidates recognised this differential equation as one in which the substitution y = vx would be helpful and many reached the stage of separating the variables. However, the integration of \(\frac{1}{{{v^2} + 2v + 2}}\) proved beyond many candidates who failed to realise that completing the square would lead to an arctan integral. This highlights the importance of students having a full understanding of the core calculus if they are studying this option.