Use Power Series to solve system of differential equations Problem:
Hello, I wonder how you would use a Power Series to solve a system of differential equations.
Lets say I have the system
$$\begin{cases}(1)\text{ }\text{ }x_1'=2x_1+4x_2 \\
(2)\text{ }\text{ }x_2'=3x_1-2x_2.\end{cases}$$
How would you solve such a system using power series? I can't really see it (this particular system came straight out of my head, so I don't know how easy it is to solve using power series).

I was thinking to just let $x_1=\sum\limits_{n\geq 0}a_nt_1^n$ and then we would have $x_1'=\sum\limits_{n\geq 1}n\cdot a_nt_1^{n-1}$. I guess I would do the same for $x_2$ to get $x_2=\sum\limits_{n\geq 0}b_nt_2^n$ and $x_2'=\sum\limits_{n\geq 1}n\cdot b_nt_2^{n-1}$ and then substitute this into the system to get
$$$$\begin{cases}(1)\text{ }\text{ }\sum\limits_{n\geq 1}n\cdot a_nt_1^{n-1}=2\sum\limits_{n\geq 0}a_nt_1^n+4\sum\limits_{n\geq 0}b_nt_2^n \\
(2)\text{ }\text{ }\sum\limits_{n\geq 1}n\cdot b_nt_2^{n-1}=3\sum\limits_{n\geq 0}a_nt_1^n-2\sum\limits_{n\geq 0}b_nt_2^n.\end{cases}$$$$

Questions:
I don't really know where to go from here. I don't really know how I should treat a differential equation with two different power-series in it. So my thoughts were to try to eliminate $t_1$ (or $t_2$) from one of the equations and then solve that equations just as I would normally do and then substitute(?) that solution back to the other equation.
If this is correct, how do I do the elimination part? Should I first try to match the indexes with each other and then do the elimination?

Continue of problem:
I would, in this case, have
$$$$\begin{cases}\sum\limits_{n\geq 0}(n+1)\cdot a_{n+1}t_1^{n}=2\sum\limits_{n\geq 0}a_nt_1^n+4\sum\limits_{n\geq 0}b_nt_2^n \\
\sum\limits_{n\geq 0}(n+1)\cdot b_{n+1}t_2^{n}=3\sum\limits_{n\geq 0}a_nt_1^n-2\sum\limits_{n\geq 0}b_nt_2^n.\end{cases}$$$$
That is,
$$$$\begin{cases}\sum\limits_{n\geq 0}((n+1)a_{n+1}t_1^{n}-2a_n)t_1^n=4\sum\limits_{n\geq 0}b_nt_2^n \\
3\sum\limits_{n\geq 0}a_nt_1^n=\sum\limits_{n\geq 0}((n+1)\cdot b_{n+1}+2b_n)t_2^{n}.\end{cases}$$$$
AND NOW, I seriously don't know how to do the elimination. I want to subtract a multiple of (1) from (2) for example to get rid of one of the series. But I guess this wasn't any good idea. I also thought about trying to get it "back" into a second degree differential equation, but it wasn't that easy as I first thought.

Last Questions:
Why would I even solve a system of differential equations using power series, in this case I could just find the eigenvalues and eigenvectors and then I'm done. Can you solve a System like
$$X'=AX+B$$ using power series? Then I can understand why the professor mentioned that power series can be good to know for the exam when you want to solve a system of diff. equations.
I am pretty lost right now and would love to get some help. At least get the layout, the method, how to use powerseries to solve system of differential equations. Thanks :)
Edit:
Now I am really not sure, should $x_1$ and $x_2$ depend on the same variable $t$ or two different, say $t_1$ and $t_2$?
 A: From (1) we have $x_2 = \frac14 x_1' - \frac12 x_1$, and differentiating yields \begin{align}
x_1'' &= 2x_1' + 4x_2'\\
&= 2x_1'+4(3x_1-2x_2)\\
&= 2x_1' + 12x_1 - 2x_1'+4x_1,\\
\end{align}
so we may instead consider the second-order ODE
$$x_1''-16x_1=0. \tag 3$$
Suppose $$x_1(t) = \sum_{n=0}^\infty a_n t^n.$$ Then differentiating and substituting into (3) yields
$$\sum_{n=0}^\infty(n+1)(n+2)a_{n+2}t^n - \sum_{n=0}^\infty 16a_nt^n=0, $$
so that $(n+1)(n+2)a_{n+2}=16a_n$, or $a_{n+2}=\frac{16}{(n+1)(n+2)}a_n$. It follows by induction that
\begin{align}
a_{2(n+1)} &= \frac{16^{n+1}}{(2(n+1))!}a_0\\
a_{2n+3} &= \frac{16^{n+1}}{(2n+3)!}a_1
\end{align} for $n\geqslant 0$, and as $a_0=x_1(0)$, $a_1=x_1'(0)$, we have $$a_n = \frac{x_1(0)4^n+x_1'(0)(-4)^n}{n!}, $$ so that $$x_1(t)=\sum_{n=0}^\infty \frac{x_1(0)4^n+x_1'(0)(-4)^n}{n!}t^n=x_1(0)e^{4t}+x_1'(0)e^{-4t}. $$
It follows that 
\begin{align}
x_2(t) &= \frac14\frac{\mathsf d}{\mathsf dt}\left[x_1(0)e^{4t}+x_1'(0)e^{-4t}\right]-\frac12\left(x_1(0)e^{4t}+x_1'(0)e^{-4t}\right)\\
&=\left(1-\frac12 x_1(0)\right)e^{4t}-\left(1+\frac12 x_1'(0)\right)e^{-4t}.
\end{align}
