The real and imaginary parts of a complex number $x + {\text{i}}y$ are related by the differential equation $(x + y)\frac{{{\text{d}}y}}{{{\text{d}}x}} + (x - y) = 0$.

By solving the differential equation, given that $y = \sqrt 3$ when x =1, show that the relationship between the modulus r and the argument $\theta$ of the complex number is $r = 2{{\text{e}}^{\frac{\pi }{3} - \theta }}$.

$(x + y)\frac{{{\text{d}}y}}{{{\text{d}}x}} + (x - y) = 0$

$\Rightarrow \frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{y - x}}{{x + y}}$

let $y = vx$     M1

$\frac{{{\text{d}}y}}{{{\text{d}}x}} = v + x\frac{{{\text{d}}v}}{{{\text{d}}x}}$     A1

$v + x\frac{{{\text{d}}v}}{{{\text{d}}x}} = \frac{{vx - x}}{{x + vx}}$     (A1)

$x\frac{{{\text{d}}v}}{{{\text{d}}x}} = \frac{{v - 1}}{{v + 1}} - v = \frac{{v - 1 - {v^2} - v}}{{v + 1}} = \frac{{ - 1 - {v^2}}}{{1 + v}}$     A1

$\int {\frac{{v + 1}}{{1 + {v^2}}}{\text{d}}v = - \int {\frac{1}{x}{\text{d}}x} }$     M1

$\int {\frac{v}{{1 + {v^2}}}{\text{d}}v + \int {\frac{1}{{1 + {v^2}}}{\text{d}}v = - \int {\frac{1}{x}{\text{d}}x} } }$     M1

$\Rightarrow \frac{1}{2}\ln \left| {1 + {v^2}} \right| + \arctan v = - \ln \left| x \right| + k$     A1A1

Notes: Award A1 for $\frac{1}{2}\ln \left| {1 + {v^2}} \right|$, A1 for the other two terms.

Do not penalize missing k or missing modulus signs at this stage.

$\Rightarrow \frac{1}{2}\ln \left| {1 + \frac{{{y^2}}}{{{x^2}}}} \right| + \arctan \frac{y}{x} = - \ln \left| x \right| + k$     M1

$\Rightarrow \frac{1}{2}\ln 4 + \arctan \sqrt 3 = - \ln 1 + k$     (M1)

$\Rightarrow k = \ln 2 + \frac{\pi }{3}$     A1

$\Rightarrow \frac{1}{2}\ln \left| {1 + \frac{{{y^2}}}{{{x^2}}}} \right| + \arctan \frac{y}{x} = - \ln \left| x \right| + \ln 2 + \frac{\pi }{3}$

attempt to combine logarithms     M1

$\Rightarrow \frac{1}{2}\ln \left| {\frac{{{y^2} + {x^2}}}{{{x^2}}}} \right| + \frac{1}{2}\ln \left| {{x^2}} \right| = \ln 2 + \frac{\pi }{3} - \arctan \frac{y}{x}$

$\Rightarrow \frac{1}{2}\ln \left| {{y^2} + {x^2}} \right| = \ln 2 + \frac{\pi }{3} - \arctan \frac{y}{x}$     (A1)

$\Rightarrow \sqrt {{y^2} + {x^2}}  = {{\text{e}}^{\ln 2 + \frac{\pi }{3}\arctan \frac{y}{x}}}$     (A1)

$\Rightarrow \sqrt {{y^2} + {x^2}}  = {{\text{e}}^{\ln 2}} \times {{\text{e}}^{\frac{\pi }{3} - \arctan \frac{y}{x}}}$     A1

$\Rightarrow r = 2{{\text{e}}^{\frac{\pi }{3} - \theta }}$     AG

[15 marks]

Most candidates realised that this was a homogeneous differential equation and that the substitution $y = vx$ was the way forward. Many of these candidates reached as far as separating the variables correctly but integrating $\frac{{v + 1}}{{{v^2} + 1}}$ proved to be too difficult for many candidates – most failed to realise that the expression had to be split into two separate integrals. Some candidates successfully evaluated the arbitrary constant but the combination of logs and the subsequent algebra necessary to obtain the final result proved to be beyond the majority of candidates.