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?