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

{\bf Question}

For each of the following differential equations sketch:\\
\hspace*{3cm} i) Several isoclines\\ \hspace*{3cm} ii) The
direction field \\ \hspace*{3cm} iii) Several solution curves


\begin{enumerate}

\item $\ds \dy={x \over y}$

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

\item $\ds \dy=xy$

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

\end{enumerate}




\vspace{0.25in}

{\bf Answer}

\begin{enumerate}
\item

\textbf{Isoclines}: Curves where $\frac{x}{y}=C=$constant, $$\mathbf{\textrm{i.e. }x=cy}.$$
\begin{center}
\epsfig{file=114-1-1.eps, width=70mm}
\end{center}

\textbf{Direction field}: On $x=cy$, $\dy = c$

\begin{center}
\epsfig{file=114-1-2.eps, width=70mm}
\end{center}

\textbf{Solution curves}: (connect direction field lines)

\begin{center}
\epsfig{file=114-1-3.eps, width=70mm}
\end{center}

\item

\textbf{Isoclines} are $-\frac{x}{y}=C$, $\Rightarrow y=-\frac{1}{c}x$

\textbf{Direction field} on $y = -\frac{1}{c} x$, $\dy=c$ 

\begin{center}
\epsfig{file=114-1-4.eps, width=70mm}
\end{center}

\textbf{Solution curves}

\begin{center}
\epsfig{file=114-1-5.eps, width=50mm}
\end{center}

\item

\textbf{Isoclines} are $xy=c$, (hyperbola) $\Rightarrow y=-\frac{c}{x}$

\textbf{Direction field} on $y = \frac{c}{x}$, draw small line with
slope $c$. 

\begin{center}
\epsfig{file=114-1-6.eps, width=70mm}
\end{center}

\textbf{Solution curves}

\begin{center}
\epsfig{file=114-1-7.eps, width=70mm}
\end{center}

\item

\textbf{Isoclines} are $x^2+y=c$, (parabola) $\Rightarrow y=-x^2+c$

\textbf{Direction field} draw short lines of slope $c$ on the curves $y=-x^2+c$.

\begin{center}
\epsfig{file=114-1-8.eps, width=70mm}
\end{center}

\textbf{Solution curves}, join together direction fields

\begin{center}
\epsfig{file=114-1-9.eps, width=70mm}
\end{center}

\end{enumerate}

\end{document}