\(\displaystyle a_0\cos(\omega t)\) is the forcing or input function \(\displaystyle y(t)\)

The initial conditions **are** zero, that is how you construct a transfer function.

CB

Ok, I get that part. What happens when the initial condition isn't zero anymore? lets just say it is 1. Do i have to take the Laplace of (ao cos wt) = s / (s^2 + w^2), multiple it by the transfer function where it will equal to

s / (s^2 + w^2) * 1 / (k + r s + r s^2) ?

or i still use the regular transfer function 1 / (k + r s + r s^2) and plug one into that?

Another question I have is, lets say I solve the second order differential equation and I come up with the general solution. If i take the Laplace transform of that is that similar to the transfer function? What is the differences?

Thanks,