If $y = Ae^{2x} + Be^{3x}$, what is $\frac{d^2y}{dx^2} - 5\frac{dy}{dx} + 6y$?

Explanation

1. 01

   Each term $e^{mx}$ gives $(m^2 - 5m + 6)e^{mx}$.

2. 02

   $m^2 - 5m + 6 = (m - 2)(m - 3)$ is 0 for $m = 2$ and $m = 3$, so the expression is 0 for every $A$ and $B$.