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

{\bf Question}

Find the orthogonal trajectories to the following one-parameter
families of curves ($c$ is the parameter that is constant on each
curve). In each case sketch some of the curves and their
orthogonal trajectories.
\begin{enumerate}
\item
$$ y^2=4cx $$
\item
$$ y=ce^{-x} \qquad (*) $$

\item
$$ x= ce^{-y^2}  \qquad (*) $$

\end{enumerate}




\vspace{0.25in}

{\bf Answer}

\begin{itemize}
\item[a)]
$\ds y^2=4cx \Rightarrow$ these are parabolas.

\begin{center}
\epsfig{file=114-4-1.eps, width=50mm}
\end{center}


ODE of curves $\ds\left\{\begin{array}{l}c=\frac{y^2}{4x}\\
2y\frac{dy}{dx}=4c\end{array}\right.$

$\ds\Rightarrow 2y\frac{dy}{dx}=\frac{4y^2}{4x} \Rightarrow
\frac{dy}{dx}=\frac{y}{2x}$

ODE for orthogonal trajectory is $\ds
-\frac{1}{\left(\frac{dy}{dx}\right)}=\frac{y}{2x} \Rightarrow
\frac{dy}{dx}=-\frac{2x}{y}$

$\ds\Rightarrow \int ydy=-\int2xdx \Rightarrow
\frac{1}{2}y^2=-x^2+A \Rightarrow x^2+\frac{1}{2}y^2=A$

these are ellipses centred at (0,0).

\begin{center}
\epsfig{file=114-4-2.eps, width=60mm}

(ellipses represent orthogonal trajectories)
\end{center}


\item[b)]
$\ds y=ce^{-x}$

\begin{center}
\epsfig{file=114-4-3.eps, width=60mm}
\end{center}

ODE of curves $\ds\left\{\begin{array}{l}c=e^xy\\
\frac{dy}{dx}=-ce^{-x}\end{array}\right.$

$\ds \Rightarrow \frac{dy}{dx}=-e^xye^{-x} \Rightarrow
\frac{dy}{dx}=-y$

Orthogonal trajectories satisfy $\ds
-\frac{1}{\left(\frac{dy}{dx}\right)}=-y \Rightarrow
\frac{dy}{dx}=\frac{1}{y}$

$\ds \Rightarrow \int ydy=\int dx \Rightarrow \frac{1}{2}y^2=x+d$
these are parabolas.

\begin{center}
\epsfig{file=114-4-4.eps, width=60mm}

(parabolas represent orthogonal trajectories)
\end{center}


\item[c)]
$\ds x=ce^{-y^2}$

\begin{center}
\epsfig{file=114-4-5.eps, width=50mm}
\end{center}


ODE of curves $\ds\left\{\begin{array}{l}c=xe^{-y^2}\\
1=-2cy\frac{dy}{dx}e^{-y^2}\end{array}\right.$

$\ds \Rightarrow 1=-2xe^{y^2}y\frac{dy}{dx}e^{-y^2} \Rightarrow
1=-2xy\frac{dy}{dx} \Rightarrow \frac{dy}{dx}=\frac{-1}{2xy}$

ODE for orthogonal trajectory $\ds
\frac{-1}{\frac{dy}{dx}}=\frac{-1}{2xy} \Rightarrow
\frac{dy}{dx}=2xy$

$\int\frac{1}{y}dy=\int2xdx \Rightarrow \ln y=x^2+A \Rightarrow
y=Be^{x^2}$

\begin{center}
\epsfig{file=114-4-6.eps, width=50mm}

(top and bottom curves represent orthogonal trajectories)
\end{center}

\end{itemize}


\end{document}