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

\textit{\textbf{Classification}}
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\textbf{Question}

Find 2 values of $r$ for which $y=e^{rx}$ is a solution of
$y''-y'-2y=0$.

Find a solution to satisfy $y(0)=1$ and $y'(0)=2$.


\textbf{Answer}

$y=e^{rx}$ is a solution to $y''-y'-2y=0$ if
$r^2e^{rx}-re^{rx}-2e^{rx}=0$, i.e. $r^2-2-2=0$.

The roots of this quadratic equation are $r=2$ and $r=-1$.

The DE os linear and homogeneous, so any function of the form
$$y=Ae^{2x}+Be^{-x}$$
is a solution.

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


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