\documentclass[a4paper,12pt]{article}
\newcommand{\ds}{\displaystyle}
\newcommand{\un}{\underline}
\newcommand{\undb}{\underbrace}
\newcommand{\pl}{\partial}
\parindent=0pt
\begin{document}

{\bf Question}

Find the first order partial differential equations for which the
following are general solutions and describe them as linear,
homogeneous,etc. as appropriate (in each case $f$ is an arbitrary
real function)

\begin{description}
\item[(a)]
$u=xf(x^2+y^2)$

\item[(b)]
$u=xy+(x-y)f(x+y)$

\item[(c)]
$u=x+f(uy)$

\item[(d)]
$u=f\left(\ds\frac{xy}{u}\right)$

\end{description}


\medskip

{\bf Answer}

\begin{description}
\item[(a)]
$\begin{array} {rcrcl} u_x & = & f(x^2+y^2) & + &
2x^2f'(x^2+y^2)\\ u_y & = & {} & {} & 2yxf'(x^2+y^2) \end{array}$

Therefore

$\ \ \ \ \ \ \ \ \ xyu_x=xyf+2x^3yf'\ \ \ +$

$\un{\ \ \ \ \ \ \ -x^2u_y=\ \ \ \ \ -2x^3yf'}$

$xyu_x-x^2u_y=xyf$

or $xyu_x-x^2u_y=yu$, linear and homogeneous.

\item[(b)]
$u_x=y+f(x+y)+9x-y)f'(x+y)$

$u_y=x-f(x+y)+(x-y)f'(x+y)$

Therefore $(x-y)(u_x-u_y)=2u-(x^2+y^2)$, linear
\un{non}-homogeneous

\item[(c)]
$\begin{array} {rcrcl} u_x & = & 1 & + & f'(uy)u_x \Rightarrow
u_x(1-f'(uy)=1\\ u_y & = & & & f'(uy)(yu_y+u) \end{array}$

$\begin{array} {rcrcl} uu_x & = & u & + & uu_xf'\\ -yu_y & = & & -
& f'(yu_y+u) \end{array}$

$\Rightarrow \un{uu_x-yu_y=u}$,\ quasi-linear

\item[(d)]
$xu_x-yu_y=0$ linear, homogeneous

\end{description}
\end{document}
