All you need to do is take the general solution and sub it into the d.e. EG
Substituting into the d.e.
Now collecting like terms
Now since and are roots of the terms on the LHS are both zero therefore the d.e. is satisfied.
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.