Stable and unstable manifolds of a linear system Consider the linear system of the form $\dot{x}=Ax$, with $x \in R^2$.
(a) Solve the system and find stable and unstable manifolds in the case of
$A=\begin{pmatrix}
1 & 2 \\
1 & 0
\end{pmatrix}$
(b) Sketch the phase portrait of the above system.
To begin with, I found eigenvalues to be 2 and -1. Then I calculated the unstable subspace to be $E^u=\text{span} \left\{\begin{pmatrix} 2\\ 1\end{pmatrix}\right\}$ and
stable subspace to be $E^s=\text{span} \left\{\begin{pmatrix} 1\\ -1\end{pmatrix}\right\}$.
Now the system of equations can be written as $\dot{x}=x+2y$ and $\dot{y}=x$.
To find stable and unstable manifolds, I need to solve the differential equation which is where I am having difficulty.
Taking the first equation $(\dot{x}-x=2y)$, we can write it as $(-xe^{-t})'=xe^{-t}-\dot{x}e^{-t}=(x-\dot{x})e^{-t}=2ye^{-t}$.
Now I don't know how to proceed after that. Can the second equation be used to solve the first one? How do we find stable and unstable manifolds using the above calculations and find their phase portraits?
 A: 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 $s\in \mathbb R^2$. 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 $t\to \infty$,
$$ x(t) = c_1 e^{\lambda_1 t} v_1 + c_2 e^{\lambda_2 t} v_2 \to \vec 0$$
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.
A: You can also go the slow way, from $\dot x-x=2y$ it follows that
$$
\ddot x-\dot x=2\dot y=2x
$$
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
$$
\pmatrix{x(t)\\y(t)}=c_1\pmatrix{1\\\frac12}e^{2t}+c_2\pmatrix{1\\-1}e^{-t}
$$
A: $$x'=x+2y$$
$$y'=x$$
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:
$$y'=x=C_1e^{2t}-y$$
$$y'+y=C_1e^{2t}$$
