Question

how do you get from y=Ae^-4x + Be^-6x
to differential eqution f''(x) +10f(x) +24y=0

Answer 1

A and B are 2 arbritary constants

Answer 2

I think you part from the differential equation. Then, you use the replacement,
\[f(x)=C e^{rx}\]
So,
\[f''(x)+10f'(x)+24f(x)=0\Rightarrow r^2+10r+24=0\Rightarrow r_1=-6, r_2=-4\]
Then, the general solution is,
\[f(x)=Ae^{-4x}+Be^{-6x}\]