An ode system with initial condition Solve the following system subject to the given initial data
$$x’(t)= x(t)-4y(t), x(0)=1 $$
$$y’(t)=x(t)+y(t), y(0)=1$$

So far i have gotten to the step where
$$x''(t)=x'(t)-4y'(t)$$
$$x''(t)=-3x(t)-12y(t)$$
I am not sure on how to continue after this
 A: Here's one way: 


*

*Rewrite the second equation, solving it for $x$ in terms of $y$ and $y'$. 

*Differentiate, giving you an equation for $x'$ in terms of $y'$ and $y''$. 

*Substitute what you've got into the first equation: now you have an equation with just $y$ and its derivatives, and I assume you will know how to solve that. 
There's another way, which involves writing the two equations as a single matrix-vector equation --- does that sound like something you are supposed to know about?
A: Differentiating $y'$ we have
$$   y''=x'+y'=x-4y+y'=(y'-y)-4y+y'=2y'-5y                                      $$
We obtain  equation 
$$y''-2y'+5y=0$$
The characteristic equation is $\lambda^2-2\lambda+5=0$
This equation has complex roots. If you do not know get real solutions you can see this question
Another way to solve a system is :
The solution is $e^{tA}x_0$ where 
$x_0 = \begin{bmatrix}
1\\
1 \\
\end{bmatrix}$
and 
$A=\begin{bmatrix}
1 & -4 \\
1 & 1
\end{bmatrix}$
If you don't know calculate $e^{tA}$, here is a simple way
