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

{\bf Question}

Find the solution of the following differential equations with the
given initial condition.

\begin{enumerate}

\item $\ds \dy=8x^3e^{-2y} \qquad y(1)=0\qquad (*)$

\item $\ds \dy=y\sin x \qquad y(\pi)=-3$

\item $\ds \dy=x^2(1+y) \qquad y(0)=3\qquad (*)$

\end{enumerate}




\vspace{0.25in}

{\bf Answer}

\begin{itemize}
\item[a)]
$\ds\frac{dy}{dx}=8x^3e^{-2y} \Rightarrow \int e^{2y}dy=\int8x^3dx
\Rightarrow \frac{1}{2}e^{2y}=2x^4+c$

now, when $x=1, y=0$ so

$\ds\frac{1}{2}e^{2(0)}=2(1)^4+c \Rightarrow \frac{1}{2}=2+c
\Rightarrow c=\frac{-3}{2}$

hence $\frac{1}{2}e^{2y}=2x^4-\frac{3}{2} \Rightarrow
y=\frac{1}{2}\ln(4x^3-3)$

\item[b)]
$\ds\frac{dy}{dx}=y\sin x \Rightarrow \int\frac{1}{y}dy=\int\sin
xdx \Rightarrow \ln y=-\cos x+c$

$\Rightarrow y=ke^{-\cos x}$

when $\ds x=\pi, y=-3$ so choose $k$ so that

$\ds-3=ke^{-\cos(\pi)} \Rightarrow -3e^{-1}=k \Rightarrow
y=-3e^{(-\cos x-1)}$

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

$\ds1+y=e^{\frac{1}{3}x^3+c} \Rightarrow y=ke^{\frac{1}{3}x^3}-1$

when $\ds x=0, y=3 \Rightarrow 3=ke^{\frac{1}{3}(0)^3}-1
\Rightarrow k=4$ hence $\ds y=4e^{\frac{1}{3}x^3}-1$

\end{itemize}


\end{document}