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

{\bf Question}

Obtain the general solution of the following differential
equations:

\begin{enumerate}

\item  $\ds x\dy +y = 2e^{-x}\qquad (*)$

\item  $\ds (y-x^3)+(x+y^3)\dy=0$

\item  $\ds x\dy = 3y + x^4$

\item  $\ds \dy={{x^2-xy+y^2} \over {xy}}$

\item  $\ds \dy+y\cot x=2x\hbox{ cosec} x\qquad (*)$

\item  $\ds (3x^2y^2+\sin x)\dy +(2xy^3+y\cos x)=0$

\end{enumerate}




\vspace{0.25in}

{\bf Answer}

\begin{itemize} \item[a)]
$\ds\frac{dy}{dx}+\frac{1}{x}y=2\frac{1}{x}e^{-x} \Rightarrow
I(x)=\exp(\int\frac{1}{x}dx)=\exp(\ln x)=x$

$\ds\frac{d}{dx}(xy)=2e^{-x} \Rightarrow xy=-2e^{-x}+c \Rightarrow
y=\frac{c-2e^{-x}}{x}$

\item[b)]
$\ds p=y-x^3, \,\,\,\, q=x+y^3, \,\,\,\, \frac{\pl p}{\pl y}=1,
\,\, \frac{\pl q}{\pl x} \Rightarrow$ exact.

$\ds \frac{\pl F}{\pl x}=y-x^3 \Rightarrow
F=xy-\frac{1}{4}x^4+f(y) \Rightarrow \frac{\pl F}{\pl
y}=x+\frac{df}{dy}$

$\ds \frac{\pl F}{\pl y}=x+y^3 \,\,\,$ hence need $\ds\frac{\pl
f}{\pl y}=y^3 \,\,\,$ so $\ds f=\frac{1}{4}y^4+c$.

There is a solution if $\ds
F(x,y)=xy-\frac{1}{4}x^4+\frac{1}{4}y^4=A$

\item[c)]
$\ds\frac{dy}{dx}-\frac{3}{x}y=x^3 \Rightarrow
I(x)=\exp(\int-\frac{3}{x}dx)=\exp(-3\ln x)=x^{-3}$

$\ds\frac{d}{dx}\left(\frac{1}{x^3}y\right)=1 \Rightarrow
\frac{1}{x^3}y=x+c \Rightarrow y=x^4+cx^3$

\item[d)]
$\ds\frac{dy}{dx}=\frac{1-\frac{y}{x}+\left(\frac{y}{x}\right)^2}{\frac{y}{x}}$,
put $\ds y=xv \Rightarrow x\frac{dv}{dx}+v=\frac{1-v+v^2}{v}$

$\ds x\frac{dv}{dx}=\frac{1-v+v^2}{v}-v=\frac{1-v}{v} \Rightarrow
\int\frac{v}{1-v}dv=\int\frac{1}{x}dx$

$\ds\int-1+\frac{1}{1-v}dv=\ln|x|+c \Rightarrow
-v-\ln|1-v|=\ln|x|+c$

$\ds e^{-v}\frac{1}{1-v}=Ax \Rightarrow
e^{-\frac{y}{x}}\frac{1}{x-y}=A$

\item[e)]
$\ds\frac{dy}{dx}+\frac{\cos x}{\sin x}y=2x\frac{1}{\sin x}
\Rightarrow I(x)=exp\left(\int\frac{\cos x}{\sin x}dx\right)$

$\ds I(x)=\exp(\ln(\sin x))=\sin x \Rightarrow \frac{d}{dx}((\sin
x)y)=2x$

$\ds\Rightarrow (\sin x)y=x^2+c \Rightarrow y=\frac{x^2+c}{\sin
x}$

\item[f)]
$\ds p=2xy^3+y\cos x, \,\,\,\, q=3x^2y^2+\sin x$

$\ds\frac{\pl P}{\pl y}=6xy^2+\cos x, \,\,\,\, \frac{\pl q}{\pl
x}=6xy^2+\cos x \Rightarrow$ exact.

$\ds\frac{\pl F}{\pl x}=2xy^3+y\cos x \Rightarrow F=x^2y^3+y\sin
x+f(y)$

$\ds \Rightarrow \frac{\pl F}{\pl y}=3x^2y^2+\sin x+\frac{df}{dy}$

$\ds\frac{\pl F}{\pl y}=3x^2y^2+\sin x$ hence need
$\ds\frac{df}{dy}=0 \Rightarrow f=c$

solution is $\ds F(x,y)=x^3y^3+y\sin x=A$
\end{itemize}


\end{document}
