Solve the differential equation

$${x^2}\frac{{{\text{d}}y}}{{{\text{d}}x}} = {y^2} + xy + 4{x^2},$$

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

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

the equation becomes $v + x\frac{{{\text{d}}v}}{{{\text{d}}x}} = {v^2} + v + 4$     A1

$\int {\frac{{{\text{d}}v}}{{{v^2} + 4}} = \int {\frac{{{\text{d}}x}}{x}} }$     A1

$\frac{1}{2}\arctan \left( {\frac{v}{2}} \right) = \ln x + C$     A1A1

substituting$(x,{\text{ }}v) = (1,{\text{ }}2)$

$C = \frac{\pi }{8}$     M1A1

the solution is

$\arctan \left( {\frac{y}{{2x}}} \right) = 2\ln x + \frac{\pi }{4}$     A1

$y = 2x\tan \left( {2\ln x + \frac{\pi }{4}} \right)$     A1

[9 marks]

Most candidates recognised this differential equation as one in which the substitution $y = vx$ would be helpful and many carried the method through to a successful conclusion. The most common error seen was an incorrect integration of $\frac{1}{{4 + {v^2}}}$ with partial fractions and/or a logarithmic evaluation seen. Some candidates failed to include an arbitrary constant which led to a loss of marks later on.