<html><head><meta name="color-scheme" content="light dark"></head><body><pre style="word-wrap: break-word; white-space: pre-wrap;">\documentclass[12pt]{article}
\usepackage{epsfig}
\newcommand{\ds}{\displaystyle}
\newcommand{\pl}{\partial}
\newcommand{\dy}{\frac{dy}{dx}}
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\begin{document}

{\bf Question}

Write each of the following second order differential equation as
a pair of first order equations for $x(t)$ and $v(t)$ \Big(where
$v=dx/dt$\Big). Hence obtain a differential equation for $v$ as a
function of $x$. Obtain the general solution of this equation and
hence roughly sketch the phase-plane.
\begin{enumerate}

\item $\displaystyle \frac{d^2x}{dt^2} - x = 0    $

\item $\displaystyle \frac{d^2x}{dt^2}- 2 \frac{dx}{dt}+ x = 0 $

\item $\displaystyle \frac{d^2x}{dt^2} - x^2 = 0   \qquad (*)$

\item $\displaystyle \frac{d^2x}{dt^2} + \frac{dx}{dt}  -
\frac{dx}{dt} e^x= 0    \qquad (*)
$

\end{enumerate}



\vspace{0.25in}

{\bf Answer}

\begin{itemize}
\item[a)]
$\ds \frac{dx}{dt}=v, \,\,\,\, \frac{dv}{dt}=x \Rightarrow
\frac{dx}{dt}\frac{dt}{dv}=\frac{v}{x} \Rightarrow
\frac{dx}{dv}=\frac{v}{x}$

This is separable $\ds \int xdx=\int vdv \Rightarrow
\frac{1}{2}x^2=\frac{1}{2}v^2+A $

$\ds \Rightarrow x^2-v^2=2A$, this is a hyperbola.

\begin{center}
\epsfig{file=114-10-1.eps, width=50mm}

phase-plane
\end{center}

\item[b)]
$\ds \frac{dx}{dt}=v \Rightarrow \frac{dv}{dt}-2v+x=0$

$\ds \frac{dv}{dt}=2v-x, \,\,\,\, \frac{dx}{dt}=v$

$\ds \Rightarrow \frac{dv}{dx}=\frac{2v-x}{v}=2-\frac{x}{v}$

This is homogeneous, so $\ds v=xV \Rightarrow
x\frac{dV}{dx}+V=2-\frac{1}{v}$

$\ds x\frac{dV}{dx}=-V+2-\frac{1}{V}$, this is separable.

$\ds \int\frac{-VdV}{V^2-2V+1}=\int\frac{1}{x}dx$

$\ds
\int\frac{-\frac{1}{2}(2V-2)}{V^2-2V+1}-\frac{1}{(V-1)^2dV}=\int\frac{1}{x}dx$

$\ds -\frac{1}{2}\ln\left(V^2-2V+1\right)+\frac{1}{V-1}=\ln x+A$

$\ds \frac{e^\frac{1}{V-1}}{\sqrt{V^2-2V+1}}=cx \Rightarrow
\frac{e^\frac{x}{v-x}}{\sqrt{\left(\frac{v}{x}\right)^2-\frac{2v}{x}+1}}=cx$

For the phase plane use isoclines / direction fields etc.


DIAGRAM


\item[c)]
$\ds \frac{dx}{dt}=v \Rightarrow \frac{dv}{dt}-x^2=0$

$\ds \frac{dv}{dt}=x^2, \,\,\,\, \frac{dx}{dt}=v \Rightarrow
\frac{dv}{dx}=\frac{x^2}{v}$, this is separable.

$\ds \int vdv=\int x^2dx$

$\ds \Rightarrow \frac{1}{2}v^2=\frac{1}{3}x^3+A$

Can draw phase plane using isoclines and direction fields.


DIAGRAM


\item[d)]
$\ds \frac{dx}{dt}=v \Rightarrow \frac{dv}{dt}+v-ve^x=0$

$\ds \frac{dv}{dt}=-v+ve^x, \,\,\,\, \frac{dx}{dt}=v \Rightarrow
\frac{dv}{dx}=-1+e^x$

$\ds \Rightarrow v=-x+e^x+B$

Can draw phase plane by graph or isoclines and direction field.


DIAGRAM


\end{itemize}


\end{document}
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