How to find such $t$ that $2x_t = x_{t+10}$? I am given a difference equation $x_{t+2} - x_{t+1} + \frac{x_{t}}{4} = \frac{2^t +t^2}{4}$. I need to know whether it is possible to double $x$ in ten period of time? If I solve the equation (I hope I didn't make a mistake), I get $C_1(\frac{1}{2})^t + C_2(\frac{1}{2})^{t}t + \frac{1}{9}2^t +t^2 -8t +20$. I guess I need it show that $2x_t = x_{t+10}$, however it is difficult to compare straightforwardly. 
 A: It will never be true that $x_{t+10}=2x_t$ for all $t$.  
One approach to finding one example is to note that the term $\frac192^t$ will grow much more rapidly than doubling in ten periods so we need $C_1$ to be large enough to offset that.  If we want $x_{10}=2x_0$, we can always increase $x_{10}$ without changing $x_0$ by increasing $C_2$.  I just started with $C_1=1000, C_2=0$ and found $x_{10}\approx 162.754$.  Using Goal Seek I find $C_2\approx 188134$ is a solution.  Similarly $C_1=100, C_2 \approx 3904$ is another solution.  If you don't want the doubling to start at $x_0$ we can have $x_{12}=2x_2$ with $C_1=500, C_2\approx 249.074$ 
Added:  you can get the doubling for any value of $t$ even with $C_2=0$.  Let $u_t=\frac{1}{9}2^t +t^2 -8t +20$.  Then if you want the doubling from $s$ to $s+10$ we have 
$$x_s=C_1\frac 1{2^s}+u_s\\x_{s+10}=C_1\frac 1{2^{s+10}}+u_{s+10}$$
If we want $x_{s+10}=2x_s$ we just say
$$C_1\frac 1{2^{s+10}}+u_{s+10}=2\left(C_1\frac 1{2^s}+u_s\right)\\
C_1=\frac{u_{s+10}-2u_s}{2^{1-s}- 2^{-s-10}}$$
