I shall solve an inhomogeneous recurrence relation: $$x_n=2x_{n-1}+2^n,\quad x_0=2$$ My approach:
The homogeneous part: $$x_n=2x_{n-1}\implies x_n-2x_{n-1}=0$$ With $x_n=x^n$ approach: $$x^n-2x^{n-1}=0$$ The characteristic equation has the form: $$x-2=0\implies x_0=2$$ The recurrence looks like this: $$x_n=c*2^n$$ Find the coefficient c: $$2^0*c=x_0\implies c=x_0\implies c=2$$ So the corresponding homogeneous recursion has the form: $$x_n=2*2^n=2^{n+1}$$
The inhomogeneous part is of the form $x_n=a*2^n$: Plugging this into recursion gives the equation: $$a*2^n=2*(a*2^{n-1})+2^n\implies a2^n=a2^n+2^n$$ Divided by $2^n$ gives: $$a=a+1$$ And that's wrong. I've tried to solve the problem a couple of times, cause I thought I was doing some of the calculations wrong... Now I'm just lost and have no idea how to grasp this kind of thing.
Let $x_{np}$ a particular solution for the recurrence relation. Since the recurrence relations is of the form $g(n)=a^n$, then $x_{np}=qa^n$ except if $a$ is solution for the characteristic equation with multiplicity $s$, in which case $x_{np}=qn^sa^n$.
It was your error: 2 is solution for the caracteristic equation, which is $x-2=0$, with multiplicity $s=1$. Then, the particular solution is of the form $x_{np}=qn2^n$.
Now you can continue.