Homework | Find the general solution to the recurrence relation A question I have been stuck on for quite a while is the following
Find the general solution to the recurrence  relation
$$a_n = ba_{n-1} - b^2a_{n-2}$$
Where $b \gt 0$ is a constant.
I don't understand how the general solution can be found with $b$ and $b^2$ in the relation.
Any help or advice would be greatly appreciated.
EDIT
Using $a_n = t^n$ I found the quadratic equation $t^2 - bt + b^2$
Which then comes to:
$$\frac{b \pm \sqrt{-4b^2 + b}}{2}$$
Therefore I have complex roots as $-4b^2 + b$ will be a negative number.
How do I continue from this point?
EDIT
Using $a_n = b^nc_n$ I came to $c_n = c_{n-1} - c_{n-2}$. Substituting $t^n$ for $c_n$ I get the quadratic $t^2 - t + t$. Which solves to:
$$
\frac{1 \pm i\sqrt{3}}{2}
$$
$\Rightarrow D = \frac{1}{2}\sqrt{1 + 2\sqrt{3}}$ and $tan\theta = \frac{1}{\sqrt{3}}$
$\Rightarrow a_n = \left(\frac{\sqrt{1 + 2\sqrt{3}}}{2}\right)(Acos(n\theta) + Bsin(n\theta))$

If I'm correct you made a small error in your quadratic equation. Substituting $a_n=t^n$ gives:
$$t^n=bt^{n-1}+b^2t^{n-2}\quad\Longrightarrow\quad t^2-bt-b^2=0.$$
This equation has two roots: $t_1=\varphi b$ and $t_2=\psi b$, where:
$$\varphi=\frac{1+\sqrt{5}}{2}, \qquad\psi=\frac{1-\sqrt{5}}{2}.$$
The solution is now given by a combination of these two solutions:
$$a_n = \alpha t_1 + \beta t_2 = \alpha(\varphi b)^n + \beta(\psi b)^n = \left(\alpha\varphi^n+\beta\psi^n\right)b^n.$$
Finally, $\alpha$ and $\beta$ depend on the choices for $a_0$ and $a_1$. Choose $\alpha$ and $\beta$ such that the solution gives the correct values for $a_0$ and $a_1$, so:
$$\alpha + \beta = a_0, \qquad \left(\alpha\varphi+\beta\psi\right)b=a_1.$$
Solving this set of equation leads to:
$$\alpha=\frac{a_1-a_0\psi b}{(\varphi-\psi)b}, \qquad \beta=\frac{a_0\varphi b-a_1}{(\varphi-\psi)b}.$$
The expression for $a_n$ together with the expressions for $\varphi,\psi,\alpha$ and $\beta$ constitute the solution.
