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

{\bf Question}

Find the general solutions of the following differential
equations:

\begin{enumerate}

\item $\ds \dy={{y-1} \over {x+3}}\qquad (*)$

\item $\ds \dy={1 \over {(x-1)y^3}}$

\item $\ds \dy=3x^2(1+y^2)$

\end{enumerate}




\vspace{0.25in}

{\bf Answer}

\begin{itemize}
\item[a)]
$\ds\frac{dy}{dx}=\frac{y-1}{x+3} \Rightarrow
\int\frac{dy}{y-1}=\int\frac{dx}{x+3}$

\begin{eqnarray*}
\Rightarrow \ln(y-1)&=&\ln(x+3)+c\\ y-1&=&e^c(x+3)\\
y-1&=&k(x+3)\\  y&=&k(x+3)+1
\end{eqnarray*}
For any constant k.

\item[b)]
$\ds\frac{dy}{dx}={1}{(x-1)y^3} \Rightarrow \int
y^3dy=\int\frac{dx}{x-1}\Rightarrow \frac{1}{4}y^4=\ln(x-1)+c$

\begin{eqnarray*}
\Rightarrow y&=&\pm(4\ln(x-1)+c)^\frac{1}{4}\\
y&=&\pm\sqrt2(\ln(k(x-1)))^\frac{1}{4}
\end{eqnarray*}
For any constant k.

\item[c)]
$\ds\frac{dy}{dx}=3x^2(1+y^2) \Rightarrow
\int\frac{dy}{1+y^2}=\int3x^2dx \Rightarrow \tan^{-1}y=x^3+c$

$\ds y=\tan(x^3+c)$
\end{itemize}


\end{document}