Mr Fantastic,

I have been looking over your post today and i am a little lost to how you factorized the equation.

The deritive that you stated

25e^(-5t)(cos(10t) + -2sin(20t))

How did you come to this every time i do this i end up with some thing different so i must be factorizing wrong.

Mr F says: I took out the common factor of 25 e^(-5t).
I would like to know where the 2 and 20 come from seeing that cos(10t) = -10sin(10t)

Mr F says: The 20 comes from the fact that I made a typo that unfortunately got carried through when I did a copy and paste. I've edited the reply in red.
and the next part of the equation where did the 25e^(-5t) go to as you ended up with

Mr F says: $\displaystyle {\color{red}25e^{-5t} \left( \cos (10t) + 2 \sin ( {\color{red}1}0t) \right) = 0}$ implies that either $\displaystyle {\color{red}25e^{-5t} = 0}$ or $\displaystyle {\color{red}\cos (10t) + 2 \sin (10t) = 0}$. I thought it would be obvious that $\displaystyle {\color{red}25e^{-5t} = 0}$ has no real solution so I just gave $\displaystyle {\color{red}\cos (10t) + 2 \sin (10t) = 0}$.
0 = Cos(10t) + 2sin(20t) = tan(10t) = -1/2

When i do it i end up with

0 = Cos(10t) + 2sin(10t) = 2tan(10t)

Mr F says: I'm sorry but that's totally wrong. Cos(10t) + 2sin(10t) is not the same as 2 tan(10t). Cos(10t) + 2sin(10t) = 0 => Cos(10t) = -2sin(10t) => 1 = -2 sin(10t)/cos(10t) = -2 tan(10t) => -1/2 = tan(10t).
?Where have i lost my factorizing?

I have attached the graph that i am working from to find the max positive rebound and when it occurs.