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

{\bf Question}

Show that the following are exact and find the solution in each
case.

\begin{enumerate}

\item $\ds  (xy^2+y)dx +(x^2y+x)dy=0$

\item $\ds (e^x \sin y+2x)dx+(e^x \cos y+2y)dy=0$

\item $\ds e^{xy}(1+xy)dx + x^2 e^{xy}dy=0\qquad (*)$

\item $\ds \left(2x+1-\frac{y^2}{x^2}\right)dx+\frac{2y}{x}dy=0$

\end{enumerate}




\vspace{0.25in}

{\bf Answer}

\begin{itemize}
\item[a)]
$\ds p=xy^2+y,\,\,\,\, q=x^2y+x,\,\,\,\, \frac{\pl P}{\pl
y}=2xy+1,\,\,\,\, \frac{\pl q}{\pl x}=2xy+1,\,\,$  exact.

$\ds\frac{\pl F}{\pl x}=xy^2+y \Rightarrow
F=\frac{1}{2}x^2y^2+xy+f(y) \Rightarrow \frac{\pl F}{\pl
y}=x^2y^2+x+\frac{df}{dy}$

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

so the solution is $\ds\frac{1}{2}x^2y^2+xy=A.$

\item[b)]
$\ds p=e^x\sin y+2x,\,\,\,\, q=e^x\cos y+2y,$

$\ds\frac{\pl P}{\pl y}=e^x\cos y,\,\,\,\, \frac{\pl Q}{\pl
x}=e^x\cos y, \,\,\,\,$ exact.

$\ds\frac{\pl F}{\pl x}=e^x\sin y+2x \Rightarrow F=e^x\sin
y+x^2+f(y)$

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

$\ds\frac{\pl F}{\pl y}=e^x\cos y+2y \Rightarrow \frac{df}{dy}=2y
\Rightarrow f(y)=y^2+c$

so the solution $\ds e^x\sin y+x^2+y^2=A$

\item[c)]
$\ds p=e^{xy}(1+xy),\,\,\,\, q=x^2e^{xy}$

$\ds\frac{\pl p}{\pl y}=xe^{xy}(1+xy)+e^{xy}x,\,\,\,\, \frac{\pl
q}{\pl x}=2xe^{xy}+x^2ye^{xy},$ hence it is exact.

$\ds\frac{\pl F}{\pl x}=e^{xy}(1+xy),\,\,\,\, \frac{\pl F}{\pl
y}=x^2e^{xy} \Rightarrow F=xe^{xy}+g(x)$

$\ds \Rightarrow \frac{\pl F}{\pl
x}=e^{xy}+xye^{xy}+\frac{dy}{dx}$ hence need $\ds\frac{dy}{dx}=0
\Rightarrow g(x)=c$

So the solution is $\ds xe^{xy}=A \Rightarrow
y=\frac{1}{x}\ln\left(\frac{A}{x}\right).$

\item[d)]
$\ds p=2x+1-\frac{y^2}{x^2},\,\,\,\, q=\frac{2y}{x},\,\,\,\,
\frac{\pl p}{\pl y}=-\frac{2y}{x^2},\,\,\,\, \frac{\pl q}{\pl
x}=-\frac{2y}{x^2},$ exact.

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

$\ds\Rightarrow \frac{\pl F}{\pl
x}=-\frac{y^2}{x^2}+\frac{dy}{dx}$ hence need
$\ds\frac{dy}{dx}=2x+1 \Rightarrow g=x^2+x+c$

so the solution is $\ds\frac{y^2}{x}+x^2+x=A$

\end{itemize}


\end{document}
