Solve the recurrence relation :
an= 2an−1 − 2an−2
subject to the initial conditions
a0 = 1, a1 = 3.
That sounds right.
The next step is to put these solutions into polar form. Once you have done this, the general solution to the recurrance relation will be of the form
$\displaystyle \displaystyle a_n = r^n(\alpha \cos n\theta +\beta \sin n\theta)$
using $\displaystyle a_0=1,a_1=3$ to solve for $\displaystyle \alpha$ and $\displaystyle \beta$