Why is $r^2-9=0$ the characteristic equation for $y'-9y=0$ I'm trying to learn how to solve second order differential equations and I don't understand something here:
http://tutorial.math.lamar.edu/Classes/DE/SecondOrderConcepts.aspx
Question says find two solutions to $y'-9y=0$
It then says the characteristic equation is:
$r^2-9=0 \Rightarrow +-3$ and I understand why $-3$ and $3$ are the solutions, but not the equation.  So before I can understand the two solutions I just need help understanding the characteristic equation.  
I'm specifically confused at how $ar^2+br+c=0$ is related to $r^2-9=0$?  I would think it would be $r-9=0$ since $y'$ is the first derivative.
Sorry if this is a silly question!

At first, observe that $y(x) \equiv 0$ solves the equation. Then the usual ansatz for such equations for $y \not \equiv 0$ is $y(x) = Ae^{\lambda x}$ with $\Bbb C \ni A \neq 0$ and $\lambda \in \Bbb C$, since this gives you constant coefficients when you derive. Assuming, that you mean the equation $y''-9y=0$ (since you said second order ODE), you have to insert the ansatz into the equation:
$$A \lambda^2 e^{\lambda x} - 9Ae^{\lambda x} = 0$$
Since $Ae^{\lambda x} \neq 0$ for all $x$, we may divide and obtain
$$\lambda^2 - 9 = 0,$$
which is the characteristic equation for this ODE.
