I am reading Greiner's "Quantum Electrodynamics". In example 1.5 he derives the Green's function for diffusion. I am stuck on a step in the derivation.

He has the defining differential equation as $$ {\bf \nabla}^\prime G-a^2{\partial G \over \partial t^\prime}=-4\pi\delta^3({\bf x}^\prime-{\bf x})\delta(t^\prime-t) \label{eq4} \tag{1} $$ where $a$ is a constant. He then defines \begin{align} \tau&=t^\prime-t\\ {\bf R}&={\bf x}^\prime-{\bf x}. \tag{2} \label{eq5} \end{align} The Fourier transform of the Green's function is given as $$ G({\bf x}^\prime,t^\prime,x,t)={1\over (2\pi)^3}\int {\rm d}^3 p e^{i{\bf p}\cdot({\bf x}^\prime-{\bf x})} g({\bf p},\tau)\, . \tag{3} \label{eq6} $$ We insert Eq. (\ref{eq6}) into the left hand side of Eq. (\ref{eq4}) to get $$ {\bf \nabla}^\prime G-a^2{\partial G \over \partial \tau}={1\over (2\pi)^3}\int {\rm d}^3 p e^{i{\bf p}\cdot{\bf R}} \left(-p^2 g-a^2{\partial g \over \partial \tau}\right)\, . \tag{4} \label{eq8} $$ Using, $$ \delta^3({\bf R})={1\over (2\pi)^3}\int {\rm d}^3 p e^{i{\bf p}\cdot{\bf R}} \tag{5} \label{eq7} $$ this results in a differential equation for $g$: $$ a^2{\partial g\over \partial \tau}+p^2 g=4\pi\delta(\tau)\,. \label{eq9} \tag{6} $$ Greiner then claims that this equation has the following solution: $$ g={4\pi \over a^2}e^{-(p^2\tau/a^2)}\Theta(\tau) \label{eq10} \tag{7} $$ To prove this he substitutes Eq. (\ref{eq10}) into Eq. (\ref{eq9}) and uses the relation $$ { {\rm d}\Theta \over {\rm d} \tau}=\delta(\tau) \label{eq11} \tag{8} $$ to get $$ 4\pi e^{-p^2\tau/a^2}{{\rm d} \Theta \over {\rm d} \tau}=4\pi\delta(\tau) \,. \label{eq12} \tag{9} $$

As far as I can see Eqs. (\ref{eq12}) and (\ref{eq11}) imply that $e^{-p^2\tau/a^2}=1$. But I don't think this is correct. My question is what am I missing or is this an error in the textbook?