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\textbf{Ordinary Differential Equations}

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

If one solution of $y''-k^2y = 0$ is $y_1=e^{kx}$, guess and verify
a solution $y_2$ that is not a multiple of $y_1$.

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


\textbf{Answer}

As $y_1=e^{kx}$ is a solution. A sensible guess for $y_2$ is
$y_2=e^{-kx}$.

Since
$$y_2''-k^2y_2 = k^2e^{-kx} - k^2e^{-kx} =0$$
then $y_2$ is confirmed as a solution.

The DE is linear and homogeneous, so any function of the form
$$y = Ay_1+By_2 = Ae^{kx} + Be^{-kx}$$
is also a solution.

To satisfy the given conditions:
\begin{eqnarray*}
0 & = & y(1) = Ae^k + Be^{-k}\\
2 & = & y'(1) = Ake^k - Bke^{-k}
\end{eqnarray*}
provided that
\begin{eqnarray*}
A & = & e^{-k}/k\\
B & = & -e^k/k
\end{eqnarray*}
So the solution is
$$y = \frac{1}{k}e^{k(x-1)} - \frac{1}{k}e^{-k(x-1)}$$


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