Stable and unstable manifolds of a linear system

$$x'=x+2y$$ $$y'=x$$ You can also sum both Differential equations: $$(x+y)'=2(x+y)$$ And integrate: $$(ln (x+y))'=2$$ $$ln (x+y)=2t+C$$ $$x+y=C_1e^{2t}$$ Then solve the DE: $$y'=x=C_1e^{2t}-y$$ $$y'+y=C_1e^{2t}$$

You can also go the slow way, from $dot x-x=2y$ it follows that $$ ddot x-dot x=2dot y=2x $$ which you can now solve as a scalar equation to get $$ x=c_1e^{2t}+c_2e^{-t} $$ and consequently $$ y=frac12(dot x-x)=frac12(c_1(2-1)e^{2t}+c_2(-1-1)e^{-t}) $$ so that $$ pmatrix{x(t)\y(t)}=c_1pmatrix{1\frac12}e^{2t}+c_2pmatrix{1\-1}e^{-t} $$

In the case of distinct real eigenvalues, the general solution to the linear system is given by

$$ x(t) = c_1 e^{lambda_1 t} v_1 + c_2 e^{lambda_2 t} v_2,$$

where $v_1, v_2$ are eigenvalues with respect to eigenvectors $lambda_1, lambda_2$ respectively, and $c_1, c_2$ are any real constants.

In our situation $lambda_1 =-1 <0$ and $lambda_2 = 2 >0$. Let $sin mathbb R^2$. Then there are unique $c_1, c_2$ so that $c_1 v_1 + c_2 v_2 = s$. Then $phi_t(s) = x(t)$ (with that unique $c_1, c_2$). Using $e^{lambda_1 t}to 0$ and $e^{lambda_2 t} to infty$ as $tto infty$,

$$ x(t) = c_1 e^{lambda_1 t} v_1 + c_2 e^{lambda_2 t} v_2 to vec 0$$

if and only if $c_2 = 0$. Thus $s$ lies in the stable manifold if and only if $s = c_1v_1$ for some $c_1$. That is, precisely when $s$ is an eigenvector of $A$ with eigenvalues $lambda_1 = -1$. Similarly one shows that the unstable manifolds are given by the second eigenspaces.