Solve the Recurrence relation an-3an-1-4an-2=4.3n where a0=1,a1=2

$a_n-3a_{n-1}-4a_{n-2}=4\cdot 3^n$

The associated linear homogeneous equation is 

$a_n-3a_{n-1}-4a_{n-2}=0$

The characteristic equation is

$r^2-3r-4=0$

$(r+1)(r-4)=0$

$r_1=-1, r_2=4$

The solution is

$a_n^{(h)}=\alpha_1(-1)^n+\alpha_2(4)^n$

Because $F(n)=4\cdot3^n,$ a reasonable trial solution is

$a_n^{(p)}=C\cdot3^n,$

where $C$ is a constant.

Substitute

$C\cdot3^n-3C\cdot3^{n-1}-4C\cdot3^{n-2}=4\cdot3^n$

$9C-9C-4C=36$

$C=-9$

Hence the particulr solution is

$a_n^{(p)}=-9\cdot3^n$

All solutions are of the form

$a_n=\alpha_1(-1)^n+\alpha_2(4)^n-9\cdot3^n$

where $\alpha_1$ and $\alpha_2$ are a constants.

$a_0=1, a_1=2$

$a_0=\alpha_1(-1)^0+\alpha_2(4)^0-9\cdot3^0=1$

$a_1=\alpha_1(-1)^1+\alpha_2(4)^1-9\cdot3^1=2$

$\alpha_1+\alpha_2=10$

$-\alpha_1+4\alpha_2=29$

$\alpha_1=2.2$

$\alpha_2=7.8$

$a_n=2.2(-1)^n+7.8(4)^n-9\cdot3^n$