1. ## Derivative Question...

Thanks!!

2. Originally Posted by qbkr21
Thanks!!
start with $\displaystyle y = e^{rx}$ find $\displaystyle y'$, and $\displaystyle y''$ then plug them into your equation. when it's all said and done, you should end up with a quadratic in $\displaystyle r$, just solve it

3. ## Re:

Thanks!!

4. Originally Posted by qbkr21
Thanks!!
$\displaystyle e^{rx}$ is never zero no matter what $\displaystyle r$ is (as long as $\displaystyle r \in \mathbb {R}$). remember what the graph looks like

5. i don't know if you are doing or will do differential equations, but what you are doing here is in fact the general method for solving second order differential equations with constant coefficients.

so in general, when given a homogeneous (which just means equated to zero) differential equation of the form:
$\displaystyle ay'' + by' + cy = 0$

we assume a solution $\displaystyle y = e^{rx}$, and then obtain the "characteristic equation":

$\displaystyle ar^2 + br + c = 0$ ..........(1)

differential equation students usually jump to equation (1) after the first line and then continue