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Now, applying exponent rules (multiply exponent inside parentheses by the one outside parentheses), we get:
This is equivalent to
So:
The characteristic solution follows from solving the characteristic equation,
so that
A guess for the particular solution may be
, but this is already contained within the characteristic solution. We require a set of linearly independent solutions, so we can look to
which has second derivative
Substituting into the ODE, you have
Therefore the particular solution is
Note that you could have made a more precise guess of
but, of course, any solution of the form
is already accounted for within 
.
so that
A guess for the particular solution may be
which has second derivative
Substituting into the ODE, you have
Therefore the particular solution is
Note that you could have made a more precise guess of
but, of course, any solution of the form