decoupling and integrating differential equations I am having trouble with the process of decoupling. If I have
$$\frac{dx}{dt}=-x+y$$
$$\frac{dy}{dt}=-x-y$$
I am trying to figure out how to solve for $x(t)$ and $y(t)$ by decoupling the system so that I only have one variable but I can't seem to get anywhere
 A: Assume we can write $x'$ as $Dx$ in which $D$ stands for the differntiation. So your system would be as follows:
$$
\left\{
        \begin{array}{ll}
            Dx=-x+y \\
            Dy=-x-y 
        \end{array}
    \right.
\longrightarrow \left\{
        \begin{array}{ll}
            (D+1)x+y=0 \\
            x+(D+1)y=0 
        \end{array}
    \right.$$ Now try to find find $x$ and $y$ from above by eliminating one of them first. For example, by multiplying the second equation by $-(D+1)$ and making a summation we have: $$y-(D+1)^2y=0$$ which is $$y-(y''+2y'+y)=0$$ or $$y''+2y'=0$$ This is a very simple second-order OE. Do the same way for finding a proper expression with respect to $x$ and...
A: Compute $d^2 x/dt^2$, giving
$$\frac{d^2 x}{dt^2} = - \frac{dx}{dt} + \frac{dy}{dt}$$
Use the second equation to replace $\frac{dy}{dt}$ with an expression in x and y, and use the first equation to replace $y$ with an expression in $dx/dt$ and $x$. The result is a second order equation in x.
Edit: For clarity, this gives
$$x'' = -x' + y' = -x' + (-x - y) = -x' -x - y = -x' - x - (x' + x)$$
$$x'' = -2x' - 2x$$
A: Note that you can write your system like so:
$$
   \frac{\mathrm{d}}{\mathrm{d}t}\begin{pmatrix}x\\y \end{pmatrix}
   +\begin{pmatrix}1 & -1 \\ 1 & 1\end{pmatrix}\begin{pmatrix}x\\y \end{pmatrix}
   =0.
$$
To effectively decouple the equations, you want the system to looke like
$$
   \frac{\mathrm{d}}{\mathrm{d}t}\begin{pmatrix}u\\v \end{pmatrix}
   +\begin{pmatrix}a & 0 \\ 0 & b\end{pmatrix}\begin{pmatrix}u\\v \end{pmatrix}
   =0,
$$
so you need to diagonalize the matrix $\begin{pmatrix}1 & -1 \\ 1 & 1\end{pmatrix}$, find the eigenvalues $a$ and $b$, and the corresponding eigenvectors $u$ and $v$. In this case, we find $a=1+i$, $b=1-i$, $u=(1,-i)$ and $v=(1,i)$. The final system is thus
\begin{align}
    \frac{\mathrm{d}}{\mathrm{d}t}u+(1+i)u=0 \\
    \frac{\mathrm{d}}{\mathrm{d}t}v+(1-i)v=0
\end{align}
with $u=x-iy$ and $v=x+iy$.
This is a somewhat more general method, although probably slightly not as effective for this particular problem, then the one T. Bongers suggested.
