for distinct roots (k1, k2) of the equation k^2 + 2Bk + w^2 show that x(t) = Ae^(k1t) + Be^(k2t) is a solution of the following differential equation: (d^2)x/dt^2 + 2B(dx/dt) + (w^2)x = 0
What you need to do here is substitute in the boundary conditions to determine A and B so...
and
You can solve these simultaneous as 2 equations with 2 unknowns (A and B)
This problem is a bit tricky since throughout the B in the original equation is different from the B in the general solution.
Regarding the general behavior it depends on the sign of and . If we assume the B in the original equation is real and positive and the w is real and less than B then the roots are
which are both definitely negative and in that case the exponential terms go to zero as t goes to infinity.
Hope this helps.
OK so if W > B and B > 0 then you have the roots equal to
and
where
Then your solution becomes
So when t goes to infinity the exponential term sends the function to zero. The other terms just cause the function to wiggle around but the overall trend is .
If B < 0 then the function x goes to infinity.