\documentclass[a4paper,12pt]{article}
\usepackage{epsfig}
\newcommand{\un}{\underline}
\newcommand{\ds}{\displaystyle}
\begin{document}
\parindent=0pt

\begin{center}
\textbf{Ordinary Differential Equations}

\textit{\textbf{Classification}}
\end{center}

\textbf{Question}

Show that $y=-e$ is a solution of $y''-y=e$.

Find a solution $y$ to satisfy $y(1)=0$ and $y'(1)=1$.


\textbf{Answer}

If $y=y_1(x)=-e$ then this will give $y_1'=0$ and $y_1''=0$.
Thus $$y_1''-y_1 = 0 + e=e.$$

$y_2=Ae^x + Be^{-x}$ is a solution of $y''-y=0$ and so
$$y=y_1(x)+y_2(x) = -e + Ae^x + Be^{-x}$$
is also a solution.

The solution will satisfy
\begin{eqnarray*}
0 & = & y(1) = Ae + \frac{B}{e} -e\\
1 & = & y'(1)= Ae-\frac{B}{e}
\end{eqnarray*}
if $A$ and $B$ take the values
\begin{eqnarray*}
A & = & (e+1)/(2e) \\
B & = & e(e-1)/2
\end{eqnarray*}
So the solution is
$$y=-e + \frac{1}{2}(e+1)e^{x-1}+\frac{1}{2}(e-1)e^{1-x}$$

\end{document}
