Example 2.1.7.
In Example 2.1.2, we used the characteristic equation to deduce that
\begin{equation*}
x_{1}(t) = e^{-5t}\quad \text{ and } \quad x_{2}(t) = e^{2t}
\end{equation*}
both solve the equation \(x'' + 3 x' - 10 x = 0\text{.}\) Taking an arbitrary linear combination of \(x_{1}\) and \(x_{2}\) then gives a two-parameter family of solutions
\begin{equation*}
c_{1}e^{-5t} + c_{2}e^{2t}.
\end{equation*}
(There is an easy way to verify that \(x_{1}\) and \(x_{2}\) are not constant multiples of one another: just confirm that their ratio \(\frac{e^{-5t}}{e^{2t}} = e^{-7t}\) is not constant!)
Now, let \(t_{0}\text{,}\) \(x_{0}\text{,}\) and \(v_{0}\) be arbitrary. Theorem 2.1.6 gives that there is a unique solution \(x(t)\) to the differential equation satisfying \(x(t_{0}) = x_{0}\) and \(x'(t_{0}) = v_{0}\text{.}\) If we are able to choose \(c_{1}\) and \(c_{2}\) so that the system of equations
\begin{align*}
c_{1}e^{-5t_{0}} + c_{2}e^{2t_{0}} &= x_{0}\\
-5c_{1}e^{-5t_{0}} + 2c_{2}e^{2t_{0}} &= v_{0}.
\end{align*}
is satisfied, then \(x(t) = c_{1}e^{-5t} + c_{2}e^{2t}.\)
Solving the first equation for \(c_{1}\) gives
\begin{equation*}
c_{1} = x_{0}e^{5t_{0}} - c_{2}e^{7t_{0}}.
\end{equation*}
Substituting this into the second equation and solving for \(c_{2}\) results in
\begin{equation*}
c_{2} = \frac{1}{7}\left( v_{0}e^{-2t_{0}} - 5x_{0} \right).
\end{equation*}
We have demonstrated assignments of \(c_{1}\) and \(c_{2}\) in terms of arbitary \(t_{0}\text{,}\) \(x_{0}\text{,}\) and \(v_{0}\text{.}\) So
\begin{equation*}
x(t) = c_{1}e^{-5t} + c_{2}e^{2t}
\end{equation*}
is the general solution to \(x'' + 3 x' - 10 x = 0\text{.}\) Here, we established this by explicitly solving a system of equations for \(c_{1}\) and \(c_{2}\text{.}\) Going forward, we will develop some theory which will allow us to reach the same conclusion with less explicit computation.