$y' - ay = \delta(t - T) , t > 0$ Solve for $y$:
$$y' - ay = \delta(t - T) , t > 0  \text { and } y(0) = y_0.$$
 A: For $0<t<T$, $y'-ay=0$.  So, $y(t)=y_0e^{at}$.
For $T<t$, $y'-ay=0$.  So, $y(t)=Ae^{at}$.
Applying the discontinuity at $T$, we see that $y(T^+)-y(T^-)=1$ so that 
$$(A-y_0)e^{aT}=1$$
Solving for $A$ we find that $A=y_0+e^{-aT}$.
Putting it all together 
$$y(t)=y_0e^{at}+\begin{cases}0&,0<t<T\\\\
e^{a(t-T)}&,T,t\end{cases}$$
This can be written more succinctly as 
$$y(t)=y_0e^{at}+e^{a(t-T)}H(t-T)$$
where $H(t)$ is the Heaviside (unit step) function.
A: I guess we can assume that $T>0$. We can solve the differential equation on $[0,T)$ and on $(T,+\infty)$ separately. On the first interval, it's easy to see that the solution is $y(t)=y_0e^{at}$ and on the second interval it's $y(t)=ce^{at}$ for some $c$. Now we just need to find $c$. And we can do it by integrating the differential equation around $T$:
Let $\varepsilon>0$ be small enough. Then we have that
$$\int_{T-\varepsilon}^{T+\varepsilon}\delta(t-T)\mathrm{d}t=1$$
$$\int_{T-\varepsilon}^{T+\varepsilon}ay=2 \varepsilon q(\varepsilon)$$
For some $q(\varepsilon) \in \mathbb{R}$, because $y$ is bounded.
$$\int_{T-\varepsilon}^{T+\varepsilon}y'=y(T+\varepsilon)-y(T-\varepsilon)=ce^{a(T+\varepsilon)}-y_0e^{a(T-\varepsilon)}$$
i.e. we get that
$$ce^{a(T+\varepsilon)}-y_0e^{a(T-\varepsilon)}+\varepsilon q(\varepsilon) = 1$$
Taking the limit $\varepsilon \to 0+0$ we get that
$$ce^{aT}-y_0e^{aT}= 1$$
i.e.
$$c= 1e^{-aT}+y_0$$
