Solve the differential equation

\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{y}{x} + \frac{{{y^2}}}{{{x^2}}}\) (where x > 0 )

given that y = 2 when x = 1 . Give your answer in the form \(y = f(x)\) .

put y = vx so that \(\frac{{{\text{d}}y}}{{{\text{d}}x}} = v + x\frac{{{\text{d}}v}}{{{\text{d}}x}}\)     M1A1

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

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

separating variables, \(\int {\frac{{{\text{d}}x}}{x} = \int {\frac{{{\text{d}}v}}{{{v^2}}}} } \)     M1A1

hence \(\ln x = - {v^{ - 1}} + C\)     A1A1

substituting for v, \(\ln x = \frac{{ - x}}{y} + C\)     M1

Note: Do not penalise absence of C at the above stages.

substituting the boundary conditions,

\(0 = - \frac{1}{2} + C\)     M1

\(C = \frac{1}{2}\)     A1

the solution is \(\ln x = \frac{{ - x}}{y} + \frac{1}{2}\)     (A1)

leading to \(y = \frac{{2x}}{{1 - 2\ln x}}\) (or equivalent form)     A1

Note: Candidates are not required to note that \(x \ne \sqrt {\text{e}} \) .

[13 marks]

Many candidates were able to make a reasonable attempt at this question with many perfect solutions seen.