\documentclass[a4paper,12pt]{article}
\setlength\oddsidemargin{0pt} \setlength\evensidemargin{0pt}
\setlength\topmargin{0pt}
\begin{document}
\parindent=0pt
{\bf Exam Question

Topic: Chain Rule}

The function $z(x,y)$ satisfies the equation $$x^2\frac{\partial
z}{\partial x}-y^2\frac{\partial z}{\partial y}=0.$$ Transform the
equation using the change of variables given by
$$u=\frac{1}{x}+\frac{1}{y},\ \ v=y,$$ and deduce that
$\displaystyle z=f\left(\frac{x+y}{xy}\right)$ where $f$ is an
arbitrary function.

\vspace{0.5in}

{\bf Solution}

\begin{eqnarray*}
\frac{\partial z}{\partial x}&=&\frac{\partial z}{\partial
u}.\frac{\partial u}{\partial x}+\frac{\partial z}{\partial
v}.\frac{\partial v}{\partial x}=\frac{\partial z}{\partial
u}\left(-\frac{1}{x^2}\right)+\frac{\partial z}{\partial
v}.0=-\frac{1}{x^2}\frac{\partial z}{\partial u}\\ \frac{\partial
z}{\partial y}&=&\frac{\partial z}{\partial u}.\frac{\partial
u}{\partial y}+\frac{\partial z}{\partial v}.\frac{\partial
v}{\partial y}=\frac{\partial z}{\partial
u}\left(-\frac{1}{y^2}\right)+\frac{\partial z}{\partial v}.1.
\end{eqnarray*}

So if $$x^2\frac{\partial z}{\partial x}-y^2\frac{\partial
z}{\partial y}=0$$ it follows that $$-\frac{\partial z}{\partial
u}+\frac{\partial z}{\partial u}-y^2\frac{\partial z}{\partial
v}=0\ \ \mathrm{so}\ \ \frac{\partial z}{\partial v}=0,$$ i.e. $z$
is a function of $u$ only, and so $\displaystyle
z=f(u)=f\left(\frac{x+y}{xy}\right).$


\end{document}
