You could search the net for spring dashpot.
Oh sorry, those are determined by initial conditions. You should have initial conditions that say something like and With these initial conditions you can solve for both and They are basically constants of integration that define the family of solutions.
Well, would be something given. Say you know the spring is extended a distance of at the initial time then we know and therefore
Because there are two unknowns we needed the second initial condition Say, then we can solve for using
Thus,
then substituting this into
we have
and
Okay, the way to solve this differential equation is by finding the roots to the equation
like you wanted to. This can be done using the quadratice formula, i.e.,
Unfortunately, you do not no if or is a repeated root and each one of these cases leads to a different general solution. Therefore, you would have to solve the differential equation using each of these cases and then add all the solutions together.
If are repeated real roots then the solution is of the form
If the solution is of the form
Finally, if the solution is of the form
The constants are all determined by the initial conditions.
EDIT: I am sorry this keeps getting more and more complicated, but your questions are so general they have to have a general answer.
I guess I don't understand what you are asking. If we assume that this leads you to conclude the only time this is zero is when Therefore, the we must solve for the roots of the characteristic equatioin Again, I am not sure what you are having issues with, and so I am sorry I am not of much help.