Example 2.1.4.
Consider the equation
\begin{equation*}
x'' + 2x' + x = 0.
\end{equation*}
The characteristic equation is
\begin{equation*}
\lambda^{2} + 2\lambda + 1 = (\lambda + 1)^{2} = 0.
\end{equation*}
Thus, there is a repeated real root \(\lambda = -1\) and we have a solution \(x_1(t) = e^{-t}\text{.}\)
We seek a second solution that is not a multiple of \(x_1(t) = e^{-t}\text{.}\) We will “guess” that a solution to \(x'' + 2x' + x = 0\) of the form \(v(t) x_1(t)\) exists for some unknown function \(v(t)\) and attempt to solve for \(v(t)\text{.}\) Indeed, if
\begin{equation*}
x(t) = v(t) x_1(t) = v(t) e^{-t},
\end{equation*}
then
\begin{equation*}
x'(t) = v(t) x_1'(t) + v'(t) x_1(t) = -v(t) e^{-t} + v'(t) e^{-t}
\end{equation*}
and
\begin{align*}
x''(t) & = v''(t) x_1(t) + 2v'(t) x_1'(t) + v(t) x_1''(t)\\
& = v''(t) e^{-t} - 2 v'(t) e^{-t} + v(t) e^{-t}.
\end{align*}
Consequently,
\begin{align*}
x'' + 2 x' + x & = [v'' e^{-t} - 2 v' e^{-t} + v e^{-t}] + 2 [-v e^{-t} + v' e^{-t}] + [v e^{-t}]\\
& = e^{-t} v''\\
& = 0,
\end{align*}
and \(v'' = 0\text{.}\) Therefore, \(v = c_1 t + c_2\text{.}\) Letting \(c_1 = 1\) and \(c_2 = 0\) gives \(v(t) = t\text{,}\) and a second solution to our equation is given by
\begin{equation*}
x(t) = t e^{-t}\text{.}
\end{equation*}
Note that we could just as well have chosen, say, \(c_{1} = 0\) and \(c_{2}=1\text{,}\) or \(c_{1} = 1\) and \(c_{2}=1\text{.}\) The former, however, results in \(x(t)=e^{-t}\text{,}\) which is the solution we already knew. The latter choice gives \(x(t) = te^{-t} + e^{-t}\text{.}\) While this solution is distinct from the one we started with, it still contains information from the original solution in the second term. The assignment \(c_{1}=1\text{,}\) \(c_{2}=0\) is the simplest choice of constants which gives another solution that is, in a sense, independent from the first.