I solved this problem but am not super confident with my methods, if someone could take a look at it and tell me if it looks okay, that would be great.
Solve the following non-homogeneous recurrence relation
$a_{n}=2a_{n-1}+(-1)^{n}$
$a_{0}=2$
Homogeneous part
$h_{n}=2h_{n-1}$
$x=2$
$h_{n}=α(2)^{n}$
Find the constant
$a_{n}=c(-1)^{n}$
$a_{n-1}=c(-1)^{n-1}$
$c(-1)^{n}=c(-1)^{n-1}+(-1)^{n}$
$c(-1)^{1}=c(-1)^{0}+(-1)^{1}$
$-c=c-1$
$2c=1$
$c=\frac{1}{2}$
Solve for α using initial condition
$a_{n}=α(2)^{n}+\frac{1}{2}(-1)^{n}$
$2=α(2)^{0}+\frac{1}{2}(-1)^{0}$
$2=2α+\frac{1}{2}$
$\frac{3}{2}=2α$
$α=\frac{3}{4}$
$a_{n}=\frac{3}{4}(2)^{n}+\frac{1}{2}(-1)^{n}$
Thanks in advance for any help!
Let us look first for a particular solution, of the form $k(-1)^n$. Substituting, we find that $k(-1)^n =2k(-1)^{n-1}+(-1)^n$. Divide by $(-1)^n$. We get $k=-2k+1$, and therefore $k=\frac{1}{3}$.
A solution of the homogeneous equation is $2^n$, so the general solution of the homogeneous equation is $A\cdot 2^n$, and therefore the general solution of the inhomogeneous equation is $$a_n=A\cdot 2^n +\frac{1}{3}(-1)^n.$$ Finally, find $A$ so that the initial condition is satisfied.