Simple pendulum and angular position

Okay, I just took the final for my first DE class! I really liked it, now I can actually know how to derive some of these equations that they just hand to you in physics and chemistry.
So anyway, there is the solution to the angular position(from vertical)of the simple pendulum; the one example we had from class(physics) was theta(t)=Theta-max[e^(-bt/2m)], but i think that the cos(omega*t+phase change) part was ommited?
So I am really interested in deriving this! So I know the motion of the pendulum is given by the diffy Q I(alpha)+b(omega)+g(theta)/L=0 where sin theta has been replace by theta because of small angle approximation, and omega and alpha are the 1st and 2nd derivatives of theta(t). So then what?. Do I solve the aux eq r^2I +rb+g/L or what? Am I on the right track?
This is for my own curiosity only, Uhh please don't tell me to "go study my book a little better" heh heh. Hell even my physics prof. would flounder trying to explain, and dudes got a doctoral degree! Like I said, we are not expected to derive these, but enquiring minds want to know! Thanks

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KillaMarcilla
oh god, I can't interpret your formulae in my current state

Anywayz, I think I know how to solve the problem you're talking about

If you say that the sum of forces is: Spring - Gravity - Damping you get: (saying that damping is directly proportional to speed) (X is displacement)
Net_Force = Spring_Constant*X - Weight - Damping_Constant*dX/dT
Net_Force = Mass*Acceleration
Acceleration = second derivative of X with respect to time, dX^2/d^2T
And that gives you:

Mass*dX^2/d^2T + Damping_Constant*dX/dT - Spring_Constant*X + Weight = 0

Which appears to be a linear second-order differential equation that you could solve, I'm sure

Unless you're talking about the pendulum deal where you don't say that sin(theta) =~ theta.. My Mathematics level is only a 16, unfortunately, so I can't solve that

I've heard that it can be modeled with some sort of fancy mathematical shape and fancy nonlinear differential equations-solving, but I'm afraid I can take you no farther than that