How to solve simple systems of differential equations Say we are given a system of differential equations 
$$
\left[ \begin{array}{c} x' \\ y' \end{array} \right] = A\begin{bmatrix} x \\ y \end{bmatrix} 
       $$ 
Where $A$ is a $2\times 2$ matrix. 
How can I in general solve the system, and secondly sketch a solution $\left(x(t), y(t) \right)$, in the $(x,y)$-plane?
For example, let's say 
$$\left[ \begin{array}{c} x' \\ y' \end{array} \right] = \begin{bmatrix} 2 & -4 \\ -1 & 2 \end{bmatrix} \left[ \begin{array}{c} x \\ y \end{array} \right]$$
Secondly I would like to know how you can draw a phane plane? I can imagine something like setting $c_1 = 0$ or $c_2=0$, but I'm not sure how to proceed. 
 A: Quite generally,
$$
\left[ \begin{array}{c} x(t) \\ y(t) \end{array} \right] = \mathrm e^{tA}\begin{bmatrix} x(0) \\ y(0) \end{bmatrix},
$$
where, by definition,
$$
\mathrm e^{tA}=\sum_{n=0}^{+\infty}\frac{t^n}{n!}A^n.
$$
To compute $\mathrm e^{tA}$ in the case at hand, note that $A^2=4A$, hence
$$
\mathrm e^{tA}=I+\sum_{n=1}^{+\infty}\frac{t^n}{n!}4^{n-1}A=I+\frac{\mathrm e^{4t}-1}4A.
$$
Hence,
$$
x(t)=\frac{\mathrm e^{4t}+1}2x(0)+(1-\mathrm e^{4t})y(0),
$$
and
$$
y(t)=\frac{1-\mathrm e^{4t}}4x(0)+\frac{\mathrm e^{4t}+1}2y(0).
$$
A: $\newcommand{\vect}[1]{\mathbb{#1}}$Try finding out about diagonalisation of matrices.  (If you do not already know about this.)
The basic idea is that I can find a particular matrix P, and a diagonal matrix $\Lambda$; these combine in such a way that $A = P \Lambda P^{-1}$.  (These matrices relate to the eigenvectors and eigenvalues of your matrix $A$.)
The way we use diagonalisation is as follows.  Let me redefine the question slightly so that it is easier for me to explain: I shall use the differential equation
$$
\begin{bmatrix}x_1'\\x_2'\end{bmatrix} = A \begin{bmatrix}x_1\\x_2\end{bmatrix}.
$$
Let me call the vector of functions $\vect{x}$: then I can write our equation as
$$
\vect{x}' = A \vect{x}.
$$
Replacing $A$ with my expression $A = P\Lambda P^{-1}$, I have
$$
\vect{x}' = P \Lambda P^{-1} \vect{x},
$$
or
$$
P^{-1} \vect{x}' = \Lambda P^{-1} \vect{x}.
$$
But the matrix $P$ is just a constant matrix, so if I were to define $\vect{y} = P^{-1} \vect{x}$, then we would simply have
$$
\vect{y}' = \Lambda \vect{y}.
$$
Big deal, you might think: this is just like the old equation.  But remember that $\Lambda$ is a diagonal matrix.  So in fact this equation looks like
$$
\begin{bmatrix}y_1'\\y_2'\end{bmatrix} = \begin{bmatrix}\lambda_1 & 0 \\0 & \lambda_2\end{bmatrix} \begin{bmatrix}y_1\\y_2\end{bmatrix},
$$
which is far simpler to solve.  I can solve the differential equations for our $y_i$ functions, and then use the equation
$$
\vect{x} = P \vect{y}
$$
to find the solution to our original problem.
[EDIT: removed references to $P$ being orthogonal which are incorrect.]
A: If you don't want change variables then,
there is a simple way for calculate $e^A$(all cases).
Let me explain about. Let A be a matrix and $p(\lambda)=\lambda^2-traço(A)\lambda+det(A)$ the characteristic polynomial. We have 2 cases:
$1$) $p$ has two distinct roots
$2$) $p$ has one root with multiplicity 2
The case 2 is more simple:
In this case we have $p(\lambda)=(\lambda-a)^2$. By Cayley-Hamilton follow that $p(A)=(A-aI)^2=0$. Now develop $e^x$ in taylor series around the $a$
$$e^x=e^a+e^a(x-a)+e^a\frac{(x-a)^2}{2!}+...$$
Therefore $$e^A=e^aI+e^a(A-aI)$$
Note that $(A-aI)^2=0$ $\implies$ $(A-aI)^n=0$ for all $n>2$ 
Case $1$:
Let A be your example. The eigenvalues are $0$ and $4$. Now we choose a polynomial $f$ of degree $\le1$ such that   $e^0=f(0)$   and   $e^4=f(4)$( there is only one). In other words what we want is a function $f(x)=cx+d$ such that 
$$1=d$$
$$e^4=c4+d$$
Solving this system we have $c=\dfrac{e^4-1}{4}$ and $d=1$.
I say that 
$$e^A=f(A)=cA+dI=\dfrac{e^4-1}{4}A+I$$
In general if $\lambda_1$ and $\lambda_2$ are the distinct eigenvalue, and $f(x)=cx+d$ satisfies $f(\lambda_1)=e^{\lambda_1}$ and $f(\lambda_2)=e^{\lambda_2}$, then
$$e^A=f(A)$$
If you are interested so I can explain more (it is not hard to see why this is true)
Now I will solve your equation using above.
What we need is $e^{tA}$
The eigenvalues of $tA$ is $0$ and $4t$.
Then $e^{tA}=\dfrac{e^{4t}-1}{4t}A+I$  for $t$ distinct of $0$
