Getting equation from differential equations I have:
$\dfrac {dx} {dt}$=$-x+y$
$\dfrac {dy}{dt}$=$-x-y$
and I am trying to find $x(t)$ and $y(t)$ given that $x(0)=0$ and $y(0)=1$
I know to do this I need to decouple the equations so that I only have to deal with one variable but the decoupling is what I am having trouble with
Do I set them equal to each other and then just move like terms to separate sides getting two different equations and then integrate?
 A: Differentiate both w.r.t. t again, we end up with
$$
d^2x/dt^2 = -dx/dt+dy/dt
$$
$$
d^2y/dt^2 = -dx/dt-dy/dt
$$
Then add both of them. Remember that $y=dx/dt + x$
We get
$$
d^2(x+y)/dt^2 = -2dx/dt
$$
Substitute the value of $y = dx/dt +x $
You get,
$$
d^3x/dt^3 + 2d^2x/dt^2 + 2dx/dt = 0
$$
Mathematica(or setting $dx/dt = u$) will give the answer as
$$
x(t) = 1/2 exp(-t) ((c_2-c_1) sin(t)-(c_1+c_2) cos(t))+c_3
$$
A: If I understand your question correctly, you are seeking for a possiblity to separate the equations as common with some of the systems, by shifting the system parallel to the existing $x$ and $y$ coordinates. In simple words you want to find a way by addition or subtraction or multiplication or division. A fair answer is that this is not possible with this system, you can not just scale it. You need to rotate it.
You would see this from the solution of the equation with
$$x(t) = e^{-t} \sin t$$
$$y(t) = e^{-t} \cos t$$
you see there the trigonometric term and $\sin$ and $\cos$ depicting that $y$ and $x$ are shifted at a rotation of $\pi/2$ against each other.
But a priori you dont have of course the solution and the only way to know about the demand for a rotation is due to the fact that you feel by yourself intuitively the system has a resistence and you can not just stretch it. The positive and negative signs just make it impossible to manage a separation by only scaling - following your words.
There is no other way to solve the system by separation except via transformation in the eigenspace accepting a rotation.
You will see a way to caculation of the eigenvalues and solution of the system here>>>
It wont make sense to copy/paste all.
The solution path for such equations via transformation is standard straight forward and cares about all stretching/scaling and rotation required. You can find it also in other books/literature.
Hope this helps out of all confusion.
A: Suppose you change your variables from $(x,y)$ to $(u,v)$ according to the rule
$$\begin{cases}x=a u+bv \\y=cu+dv \end{cases}. $$
This can be written in a matrix form $\mathbf x=P \mathbf u$ where $P=\begin{pmatrix} a & b \\ c & d \end{pmatrix}$.
The ODE transforms accordingly, from
$$\frac{ \mathrm d \mathbf x}{\mathrm d t}=A \mathbf x $$ to
$$P\frac{ \mathrm d \mathbf u}{\mathrm d t}=AP \mathbf u,   $$ or
$$\frac{ \mathrm d \mathbf u}{\mathrm d t}=P^{-1}AP \mathbf u.  $$
If you choose the "right" matrix $P$ the equations for $u,v$ become uncoupled and you can solve them. Then it takes a simple multiplication by $P^{-1}$ to get back to $x,y$. 
