Prove that the Green´s Function satisfy the Differential Equation Prove that the Green´s Function: 
$$G(t,t´)=\theta(t-t´) \dfrac{\sin{[(t-t´)(\omega_0^2-\gamma^2)^{1/2}}]}{(\omega_0^2-\gamma^2)^{1/2}}\cdot e^{-\gamma(t-t´)}  \tag{1}$$ satisfy the Differential Equation: $$\left(\dfrac{d^2}{dt^2}+2\gamma\dfrac{d}{dt}+\omega_0^2\right)G(t,t´)=\delta(t-t´) \tag{2}$$


Starting with the equation:
$$\ddot{x}(t)+2\gamma\dot{x}(t)+\omega_0^2x(t)=F(t) \tag{3}$$
And using the Fourier transform, we get the given Greens Function:
$$G(t,t´)=\theta(t-t´) \dfrac{\sin{[(t-t´)(\omega_0^2-\gamma^2)^{1/2}}]}{(\omega_0^2-\gamma^2)^{1/2}}\cdot e^{-\gamma(t-t´)} \tag{4}$$
But i dont know how i can show that this Green function satisfy the given equation.
 A: *

*We know the homogeneous second-order ODE with constant coefficients 
$$\left(\dfrac{d^2}{dt^2}+2\gamma\dfrac{d}{dt}+\omega_0^2\right)x(t) ~=~0 \tag{A}$$
has complete solution
$$x(t)~=~\sum_{\pm}C_{\pm}\exp\left\{\lambda_{\pm}t\right\},\tag{B}$$
where 
$$\lambda_{\pm}~=~-\gamma\pm \sqrt{\gamma^2-\omega_0^2}\tag{C} $$
solves the characteristic equation
$$  \lambda^2+2\gamma\lambda+\omega_0^2~=~0.\tag{D}$$

*Adjust the two constants $C_{\pm}$ so that 
$$ x(0)~=~0 \quad\wedge\quad x^{\prime}(0)~=~1. \tag{E}$$
In other words, 
$$x(t)~\stackrel{(B)+(C)+(E)}{=}~\frac{\sin\left(t\sqrt{\omega_0^2-\gamma^2}\right)}{\sqrt{\omega_0^2-\gamma^2}} e^{-\gamma t}. \tag{F} $$

*Define the retarded Green's function as 
$$G(t)~:=~\theta(t) x(t).\tag{G}$$

*Show with above properties (E) that
$$G^{\prime}(t)~\stackrel{(E)+(G)}{=}~\theta(t) x^{\prime}(t)\tag{H}$$
and 
$$G^{\prime\prime}(t)~\stackrel{(E)+(H)}{=}~\theta(t) x^{\prime\prime}(t)+\delta(t)\tag{I}$$ 
as distributions.

*Finally show that the retarded Green's function satisfy OP's second equation
$$\left(\dfrac{d^2}{dt^2}+2\gamma\dfrac{d}{dt}+\omega_0^2\right)G(t) ~\stackrel{(A)+(G)+(H)+(I)}{=}~\delta(t). \tag{J} $$
