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?

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 $c_1.$
