\documentclass[12pt]{article}
\newcommand{\ds}{\displaystyle}
\newcommand{\pl}{\partial}
\newcommand{\dy}{\frac{dy}{dx}}
\parindent=0pt
\begin{document}

{\bf Question}

Show that

\[
(x^2+y^2-x)dx-ydy=0
\]

is not exact, but that it becomes exact if multiplied by $\ds
(x^2+y^2)^{-1}$. Hence obtain the solution of the differential
equation $\ds y\dy=x^2+y^2-x$.  $\qquad (*)$



\vspace{0.25in}

{\bf Answer}

$\ds p=x^2+y^2-x, \,\,\,\, q=-y \Rightarrow \frac{\pl p}{\pl
y}=2y, \,\,\,\, \frac{\pl q}{\pl x}=0 \Rightarrow$ Not exact.

Consider $\ds (1-\frac{x}{x^2+y^2}dx-\frac{y}{x^2+y^2}dy=0,
\Rightarrow p=1-\frac{x}{x^2+y^2},$

$\ds q=-\frac{y}{x^2+y^2}, \,\,\,\, \frac{\pl p}{\pl
y}=\frac{2xy}{(x^2+y^2)^2}, \,\,\,\, \frac{\pl q}{\pl
x}=\frac{2xy}{(x^2+y^2)^2} \Rightarrow$ exact.

$\ds\frac{\pl F}{\pl x}=1-\frac{x}{x^2+y^2} \Rightarrow
F(x,y)=x-\frac{1}{2}\ln(x^2+y^2)+f(y)$

$\ds\Rightarrow \frac{\pl F}{\pl
y}=\frac{-y}{x^2+y^2}+\frac{df}{dy}$

$\ds\frac{\pl F}{\pl y}=-\frac{y}{x^2+y^2}$  hence
$\ds\frac{df}{dy}=0, \,\,\, f(y)=c$

so the solution is $\ds x-\frac{1}{2}\ln(x^2+y^2)=A$


\end{document}