Checking validity of Green's Function I have the following ODE: 
$$\left[\frac{d^2}{dt^2}+2\gamma\frac{d}{dt}+\gamma^2\right]x(t) = f(t),$$ where $f(t)$ is an arbitrary driving force. I have already obtained the causal Green's function, $G(t,t')$, which is:
$$G(t,t') = \begin{cases}
0 &\text{, if }t\leq t'\\
(t-t')e^{-\gamma\,(t-t')} &\text{, if } t>t'
\end{cases}$$
Is there any way for me to check if the Green's function I calculated is correct?
I know that the Green's function should satisfy the ODE for which $f(t)$ is a delta function (an impulse centred at $t=t'$. I have tried substituting back my Green's function into the relevant ODE (modifying the RHS), but I'm not sure what to expect? I think I got $0$, but I'm not entirely confident of that answer, nor what to make of it.

EDIT: Fixed a small typo for the solution for $G(t,t')$.
Okay, here's what I got:
Let $\displaystyle G=H(t-t')e^{-\gamma(t-t')}$. I need to compute the derivatives for $t-t'>0$:
$$\begin{align}
\frac{dG}{dt} &= \frac{dH}{dt}e^{-\gamma(t-t')}-H(t-t')\gamma e^{-\gamma(t-t')} \\
&= \delta(t-t')e^{-\gamma(t-t')}-\gamma G(t-t') \\
\frac{d^2G}{dt^2}&= \frac{d\delta}{dt}e^{-\gamma(t-t')}-\gamma\delta(t-t')e^{-\gamma(t-t')}-\gamma\frac{dG}{dt} \\
&= \frac{d\delta}{dt}e^{-\gamma(t-t')}-2\gamma\delta(t-t')e^{-\gamma(t-t')}+\gamma^2G
\end{align}$$
Putting them altogether, I get:
$$\text{L.H.S.}=\frac{d\delta}{dt}e^{-\gamma(t-t')}$$
I don't see how this is equal to $\delta(t-t')$..
 A: HINT. Write your solution in the form 
$$G(t, t')=H(t-t')(t-t')e^{\gamma(t-t')}, $$ 
where $H$ is the Heaviside step function defined by the formula 
$$H(t)=\begin{cases} 1, & t> 0 \\ 0, &t\le0\end{cases}.$$
Plug your solution into the differential equation and check that it is verified with $f(t)=\delta(t-t')$. Use the following property of the Heaviside step function: 
$$\frac{dH}{dt}(t)=\delta(t).$$
SOLUTION. In general, for the second order constant-coefficient linear differential operator 
$$L=\frac{d^2}{dt^2} + a \frac{d}{dt} + b,$$ 
the formula $G(t, t')=H(t-t')u(t-t')$, where 
$$\begin{cases} Lu= 0 , & t>0 \\ u(0)=0 \\ u'(0)=1\end{cases}$$
produces the unique solution to the problem 
$$\begin{cases} L_t G(t, t') = \delta(t-t'), & t, t'>0 \\ G(0, t')=0 \\ \partial_t G(0, t')=0 \end{cases}$$
(Here $L_t$ means: apply the operator with respect to the $t$ variable). 
Proof: using Leibniz's formula and the fact that $Lu=0$ one obtains 
$$L_t(u(t-t')H(t-t'))=u(t-t')\delta'(t-t') + a u(t-t')\delta(t-t') +2u'(t-t')\delta(t-t').$$
To deal with the derivative of the delta function one uses the following formulas: 
$$f(t)\delta'(t) = f(0)\delta'(t) - f'(0)\delta(t),\qquad f(t)\delta(t)=f(0)\delta(t) $$ 
which are obtained by Leibniz's rule and integration by parts. It is here that one needs the assumption that the operator has constant coefficients. 
This is rather hasty, but I hope it helps nonetheless
