Question

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

Answer 1

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

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.

$a_0=1, a_1=2$

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

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

$$\alpha_1+\alpha_2=10$$

$$-\alpha_1+4\alpha_2=29$$

$$\alpha_1=2.2$$

$$\alpha_2=7.8$$

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