Differential equation of $y = e^{rx}$ I am trying to find what values of r in $y = e^{rx}$ satsify $2y'' + y' - y = 0$
I thought I was being clever and knew how to do this so this is how I proceeded.
$$y' = re^{rx}$$
$$y'' = r^2 e^{rx}$$
$$2(r^2 e^{rx}) +re^{rx} -e^{rx} = 0 $$
I am not sure how to proceed from here, the biggest thing I am confused on is that I am working with a variable x, with no input conditions at all, and a variable r (the constant) so how do I do this?

Here's the best part: $e^{rx}$ is never zero. Thus, if we factor that out, it is simply a quadratic in $r$ that remains.
