so the characteristic solution is
As a guess for the particular solution, let's back up a bit. The reason the choice of
works for the characteristic solution is that, in the background, we're employing the substitution
, so that
is getting replaced with a new function
. Differentiating yields
Now the ODE in terms of
is linear with constant coefficients, since the coefficients
and
will cancel, resulting in the ODE
Of coursesin, the characteristic equation will be
, which leads to solutions
, as before.
Now that we have two linearly independent solutions, we can easily find more via variation of parameters. If
are the solutions to the characteristic equation of the ODE in terms of
, then we can find another of the form
where
where
is the Wronskian of the two characteristic solutions. We have
and recalling that
, we have
