Solve the recurrence relation: $$a_n = 5a_{n-1} - 6a_{n-2} + 3$$ With initial conditions $a_1 = \frac{3}{2}$ and $a_2 = 3$
I tried solving it with roots but I get a third degree polynomial with irrational roots and I'm sure that method isn't working.
Can someone help?
We can write down the equation to be like this: $$a_n-5a_{n-1}+6a_{n-2}=3.$$ First, we solve the homogeneous equation that means: \begin{equation}\tag 1\label 1a_n-5a_{n-1}+6a_{n-2}=0.\end{equation} Assume $a_n=\alpha^n$, then from \eqref{1} we get the characteristic equation: $$\alpha^{n+2}-5\alpha^{n+1}-6\alpha^{n}=0.$$ Assume $\alpha^n\neq 0$, we get: $$\alpha^2-5\alpha+6=0\iff \alpha=2$$ or $$\alpha=3,$$ hence the solution of the homogeneous equation is $$a_n=c_12^n+c_23^n$$ for some constants $c_1$ and $c_2$.
Second, we solve the non homogeneous equation that means: $$a_n-5a_{n-1}+6a_{n-2}=3$$ Assume $a_n=k$, for a constant k then we get $$a_n-5a_{n-1}+6a_{n-2}=3\iff k-5k+6k=3\iff k=\frac{3}{2}$$ Then, the solution of non homogenous equation is: $$a_n=\frac{3}{2}$$ Hence, the total solution is: $$\boxed{a_n=c_13^n+c_22^n+\frac{3}{2}}$$
With initial conditions, you can find $c_1$ and $c_2$ Hope this help you!
