<s>[INST] Solve the recurrence relation $a_n=2a_{n-1}+15a_{n-2}+8$ for $n\geq2$, $a_0=0$, $a_1=1$ My task:
$a_n=2a_{n-1}+15a_{n-2}+8$ for $n\geq2$,  $a_0=0$, $a_1=1$
My solution
$x^{2}-2x-15$
$\Delta=64$
$x1=-3 $
$x2=5$
So I am gonna use following formula:
$a_n=ar^{n}+br^{n}$
$a_n=a*(-3)^{n}+b*5^{n}$
$+8$ is the problem, so I am looking for $c$ that $b_n:=a_n+c\implies b_n=2b_{n-1}+15b_{n-2}$
$$b_n=2(b_{n-1}-c)+15(b_{n-2}-c)+8+c=2b_{n-1}+15b_{n-2}+8-16c$$ I am setting $c=\frac{1}{2}$ so 
$$b_n=2b_{n-1}+15b_{n-2}\implies\exists a,\,b:\,b_n=a*(-3)^{n}+b*5^{n}.$$From $b_0=\frac{1}{2},\,b_1=-\frac{1}{2}$, after finding $a,\,b$. Then $a_n=b_n-\frac{1}{2}$.
$$a=\frac{3}{8}$$
$$b=\frac{1}{8}$$
$b_2=\frac{52}{8}$
$a_2=\frac{52}{8}-\frac{1}{2}=6$
Actual $a_2=10, a_3$=43
So $a_2$ from $b_n$ method is not equal to actual $a_n$.
It means I am doing something wrong here, could anyone point out the mistake?
 [/INST] Assuming $a(n) = \gamma^n$ and substituting into the homogeneous 
$$
\gamma^n-2\gamma^{n-1}-15\gamma^{n-2}=0\to \gamma^n\left(1-\frac{2}{\gamma}-\frac{15}{\gamma^2}\right)=0
$$
and solving for $\gamma$ we have
$$
a_h(n) = C_1(-3)^n + C_2 5^n
$$
and the particular dictates
$$
a_p(n)-2a_p(n-1)-15a_p(n-2) = 8
$$
so making $a_p(n) = C_0$ and substituting into the particular we have
$$
C_0-2C_0-15C_0 = 8\to C_0 = -\frac 12
$$
and finally
$$
a(n) = a_h(n)+a_p(n) = C_1(-3)^n + C_2 5^n-\frac 12
$$
NOTE
$$
a(0) = C_1+C_2-\frac 12 = 0\\
a(1) = C_1(-3)+C_25-\frac 12 =1
$$
and solving for $C_1, C_2$ gives
$$
a(n) = \frac 18\left(-4+(-3)^n+3 \cdot 5^n\right)
$$
</s>