I have a question concerning the Wikipedia's article on green function stating in the section "table of green fuctions", that a green function satisfying: $$(\gamma+\partial_t)G(t)=\delta(t)$$ has an explicit form of $$G(t)=\Theta(t)e^{-\gamma t}$$ where $\Theta(t)$ is the Heaviside step function.
My question concerns a proof of this fact, since the direct insertion yields $$\gamma\Theta(t)e^{-\gamma t}-\gamma\Theta(t)e^{-\gamma t}+\delta(t)e^{-\gamma t}=\delta(t)e^{-\gamma t}\neq\delta(t)$$ Using the fact, that $d_t \Theta(t)=\delta(t)$.
Is there an error in this procedure? What is the correct procedure to prove this fact?
Only to complement Artem's correct answer:
Better writing $(\gamma+\partial_t)G(t)=\delta$. Now let $\psi(t)= e^{-\gamma t}$. By definition of product of a distribution by a smooth function (in the case of delta in fact only required to be continuous), $$ \langle \psi\delta,\phi\rangle=\langle \delta ,\psi\phi\rangle=(\psi\phi)(0)=\phi(0)=\langle\delta,\phi\rangle. $$