### Example 26.

\(\ \ \)Solve the initial value problem
\begin{equation*}
y'' - y' - 12y = 0, \quad y(0) = 4, \quad y'(0) = -2\text{.}
\end{equation*}

## Solution.

Since we already know the general solution to (5) is given by

\begin{equation*}
y = c_1e^{-3t} + c_2e^{4t}\text{.}
\end{equation*}

If you are interested in the verification steps, see the solution below.

## Solution. Verification as a solution

First find \(y, y', \text{ and } y''\text{:}\)

\begin{align*}
y =\amp\ c_1e^{-3t} + c_2e^{4t} \\
y' =\amp\ -3c_1e^{-3t} + 4c_2e^{4t} \\
y'' =\amp\ 9c_1e^{-3t} + 16c_2e^{4t}
\end{align*}

and substitute these into \(y'' - y' - 12y\) and simplify:

\begin{align*}
y'' - y' - 12y =\amp\ \left(9c_1e^{-3t} + 16c_2e^{4t}\right) \\
\amp\ \hspace{1em} - \left( -3c_1e^{-3t} + 4c_2e^{4t} \right) -12\left( c_1e^{-3t} + c_2e^{4t} \right)\\
=\amp\ \ul{\ul{9c_1e^{-3t}}} + \ul{16c_2e^{4t}} \\
\amp\ \hspace{1em} + \ul{\ul{3c_1e^{-3t}}} - \ul{4c_2e^{4t}} - \ul{\ul{12c_1e^{-3t}}} - \ul{12c_2e^{4t}}\\
=\amp\ 0
\end{align*}

Which shows \(0 = 0\) and so \(y = c_1e^{-3t} + c_2e^{4t}\) is a solution.

We would like determine the spcific values of \(c_1\) and \(c_2\) that correspond to the initial conditions \(y(0) = 4\) and \(y'(0) = -2\text{.}\)

Since one initial condition involves the derivative, we will need to compute this before applying this condition.

\begin{equation*}
y' = -3c_1e^{-3t} + 4c_2e^{4t}.
\end{equation*}

Now let’s apply the initial conditions:

\begin{align*}
y(0) =\amp\ 4 \amp\Rightarrow \hspace{0.2cm} 4
=\amp\ c_1e^{-3\cdot 0} + c_2e^{4\cdot 0}\\
\amp \amp 4
=\amp\ c_1 + c_2 \\
\amp \amp \amp \\
y'(0) =\amp\ -5 \amp\Rightarrow \hspace{0.2cm} -5
=\amp\ -3c_1e^{-3\cdot} + 4c_2e^{4\cdot} \\
\amp \amp -5 =\amp\ -3c_1 + 4c_2
\end{align*}

We now have two equations and two unknowns,

\begin{align*}
c_1 + c_2 =\amp\ 4 \\
-3c_1 + 4c_2 =\amp\ -5
\end{align*}

There are multiple ways to solve this system of equations, and the reader is encouraged to verify that \(c_1 = 3\) and \(c_2 = 1\text{.}\) Here is one

## Solution. Solving for \(c_1\) and \(c_2\)

We will solve for \(c_2\) in the first equation and then substitute into the second.

\begin{align*}
4 =\amp\ c_1 + c_2 \\
\Rightarrow c_2 =\amp\ 4 - c_1 \\
\amp \\
-5 =\amp\ -3c_1 + 4c_2 \\
\Rightarrow -5 =\amp\ -3c_1 + 4(4 - c_1) \\
-5 =\amp\ -3c_1 + 16 - 4c_1 \\
-21 =\amp\ -7c_1 \\
3 =\amp\ c_1 \\
\amp \\
c_2 =\amp\ 4 - c_1 \\
=\amp\ 4 - 3 \\
=\amp\ 1
\end{align*}

Thus, the solution we that satisfies the DE and both of the provided initial conditions is

\begin{equation*}
y = 3e^{-3t} + e^{4t}.
\end{equation*}