Differential equations - transforming from one to another

Homework Statement

$$m\frac{d^{2}x}{dt^{2}}+c\frac{dx}{dt}+kx=F_{0}cos\omega t\\ LC\frac{d^{2}V_{c}}{dt^{2}}+RC\frac{dV_{c}}{dt}+V_{c}=V_{0}cos\omega t$$

These two equations are for two different physical phenomena. The first is for a mass moving along an x axis. The mass is affected by a springforce k, a dampener with constant c and an outer force F(t), here set to $$F_0cost(\omega t)$$.
Explanation of the second equation follows. The gist is that they are mathematically the same (I'm translating from danish, let me know if you need the rest).

a) show that both differential equations can be brought to the form:

1:
$$\frac{dx^{2}}{dt^{2}}+2\alpha \frac{dx}{dt}+\omega_{0}^{2}x=fcos\omega t$$

Where $$\alpha \geq 0$$ and we assume $$\omega > 0$$

Homework Equations

1:
$$m\frac{d^{2}x}{dt^{2}}+c\frac{dx}{dt}+kx=F_{0}cos\omega t, LC\frac{d^{2}V_{c}}{dt^{2}}+RC\frac{dV_{c}}{dt}+V_{c}=V_{0}cos\omega t$$

2:
$$\frac{dx^{2}}{dt^{2}}+2\alpha \frac{dx}{dt}+\omega_{0}^{2}x=fcos\omega t$$

The Attempt at a Solution

So I'm thinking that I somehow need to use:
$$\omega_{0}^{2} = c^2 + 4mk$$

and possible take the integral of the whole thing. But I am pretty lost.

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Use, $$x=Ae^{kt}+Be^{-kt}$$ to find the general solution and to find the particular integral $$x=C\cos\omega t+D\sin\omega t$$.