Critical points of a system of linear differential equations I have the following system of linear ODEs
$$ \begin{aligned} x' &= x + 2y \\ y' &= 2x + 4y \end{aligned} $$
Looking for critical points and their classification, I set $x'=0$ and $y'=0$, ending up with $x=-2y$. Does this mean that I have infinitely many critical points? Is there another way to approach this problem?
 A: The system is easy to solve just divide $y'$ by $x'$ to get
$$\frac{dy/dt}{dx/dt}=\frac{dy}{dx}=2\\
y=2x+c_1$$
using the chain rule.
We then solve for $x$
$$x'-3x=c_1\\
\left(xe^{-3t}\right)'=c_1e^{-3t}\\
x=-\frac{c_1}{3}+c_2e^{3t}$$
and for $y$
$$y=2x+c_1\\
y=\frac{c_1}{3}+2c_2e^{3t}.$$
We want
$$x'=y'=0\\
3c_2e^{3t}=6c_2e^{3t}=0\\
c_2=0.$$
But what exactly is $c_2$?
To solve for it, we need to choose a set of initial conditions that we will solve for, for example $x(0)=x_0$ and $y(0)=y_0$, to get rid of the exponential terms.
$$x_0=-\frac{c_1}{3}+c_2\\
y_0=\frac{c_1}{3}+2c_2$$
Add the $2$ equations to get
$$c_2=\frac{x_0+y_0}{3}.$$
So we essentially need
$$x_0+y_0=0.$$
That is
$$x(0)+y(0)=0.$$
Whenever that happens, we get a critical point.
So yes, we get an infinite number of solutions since we have no condition on $c_1.$
A: You are right, the system of ODEs has infinitely many critical points as $x'=0$ if and only if $y'=0$, so the equation $x=-2y$ indeed characterises all the critical points. To classify the critical points you need to analyse the eigenvalues of the Jacobian of the system of ODEs. Since the system of ODEs is linear, its Jacobian is the same at any point:
$$
  J = \begin{bmatrix} 1 & 2 \\ 2 & 4 \end{bmatrix}
$$
and so
$$
  det(J - \lambda I) 
= \begin{vmatrix} 1-\lambda & 2 \\ 2 & 4-\lambda \end{vmatrix} 
= \begin{vmatrix} 1-\lambda & 2 \\ 2\lambda & -\lambda \end{vmatrix} 
= \begin{vmatrix} 5-\lambda & 2 \\ 0 & -\lambda \end{vmatrix} 
= - \lambda (5-\lambda).
$$
It follows that the eigenvalues of the Jacobian $J$ are $0$ and $5$ at any of the critical points. This means that there is a solution can leave from any critical point going forward in time either in the direction $v=(v^x,v^y)$ or in the direction $-v$, where $v$ is an eigenvector of the Jocabian $J$ associated with the eigenvalue $5$. In contrast, no solution can leave from any of the critical points going backwards in time.
To find $v$ one needs to solve the equation
$$
  0 = (J-5I) v = \begin{bmatrix} -4 & 2 \\ 2 & -1 \end{bmatrix}
\begin{bmatrix} v^x \\ v^y \end{bmatrix},
$$
which is equivalent to $v^y = 2 v^x$, so we can choose for example $v = \begin{bmatrix} 1 \\ 2 \end{bmatrix}$.
