Finding the particular solution to the following non-homogeneous recurrence relation

Question

$$x_n = 2x_{n-1}+2^n$$ $$x_1 =5$$

Finding the homogenous solution is easy enough but when attempting to solve the particular solution I arrive at:

$$C_22^n = 2C_22^{n-1} + 2^n $$ $$2C_2 = 2C_2 + 2 $$ $$ 2=0$$

At this point, I'm not sure how to proceed.

Answer 1

$x_n = 2x_{n-1}+2^n \\ x_n=An2^{n} \\ x_{n-1}=A(n-1)2^{n-1} \\ x_n=2x_{x-1}+2^n \text{ gives us } \\ A n 2^{n}=2A(n-1)2^{n-1}+2^n \\ An2^{n}=A(n-1)2^{n}+2^n \\ An=A(n-1)+1 \\ An=An-A+1 \\ -A+1=0 \\ A=1$

Answer 2

Assume $x_0=0$. From: $$ x_n-2x_{n-1} = 2^n \tag{1} $$ it follows that: $$ x_{n-1}-2x_{n-2} = 2^{n-1}, $$ $$ x_{n-2}-2x_{n-3} = 2^{n-2}, $$ $$ \ldots \tag{2} $$ hence: $$\begin{eqnarray*} x_n &=& (x_n-2x_{n-1})+2(x_{n-1}-2x_{n-2})+2^2(x_{n-2}-2x_{n-3})+\ldots\\ &=& \sum_{j=1}^{n}2^{j-1}(x_{n-j+1}-2 x_{n-j})=\sum_{j=1}^{n}2^{j}\cdot 2^{n-j+1}=n 2^{n+1}.\tag{3}\end{eqnarray*}$$ If $x_1=5$, the solution to $x_{n}=2 x_{n-1}+2^n $ is given by: $$ x_n = 2^{n-1} + n 2^{n+1} = (1+4n)\, 2^{n-1}.\tag{4}$$