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.