Undamped oscillator with driving force

[thenewbosco]

just wondering how to go about this one:
An undamped harmonic oscillator is subject to a driving force $$F_0e^{-bt}$$. It starts from rest at the origin (x=0) at time t=0.

assuming a general solution $$x(t)=A cos(\omega t - \phi) + Be^{-\alpha t}$$ where A, $$\omega$$, $$\phi$$, B, and $$\alpha$$ are real constants, find the position x(t) as a function of time.

I was thinking to take this general solution, differentiate, and apply initial conditions to the two equations, however all i could solve for is alpha in terms of omega.
Then i thought to differentiate again and plug into the equation
$$ma=-kx + F_{driving}$$ however i cannot see how i can solve for all the constants with these two equations..any help?

 

[member 553083]

To solve this problem, you need to first solve the differential equation that describes the motion of the oscillator. This is done by taking the second derivative of the position with respect to time, which yields the equation:m\ddot{x} + kx = F_0e^{-bt}where m is the mass of the oscillator, k is the spring constant, and b is the damping coefficient.Assuming a solution of the form x(t) = A\cos(\omega t - \phi) + Be^{-\alpha t}, where A, \omega, \phi, B, and \alpha are real constants, we can substitute it into the differential equation and solve for the constants. After some algebraic manipulation, we can find that:A = \frac{F_0}{\sqrt{k^2 + (m\omega)^2}}\omega = \sqrt{\frac{k}{m}}\phi = \arctan\left(\frac{m\omega}{k}\right)B = 0\alpha = bTherefore, the solution of the position of the undamped harmonic oscillator under this driving force is:x(t) = \frac{F_0}{\sqrt{k^2 + (m\omega)^2}}\cos(\sqrt{\frac{k}{m}}t - \arctan\left(\frac{m\omega}{k}\right))