The question being I must find the greens function for this one point boundary problem:
$y'' + 4y' + 9y = f(t)$, where $y(0) = y'(0) = 0$
I am slightly confused here. I know that I must put the equation in this form. I am unaware of how to achieve the dirac delta function, I know this is equal to $G(x,z)$,where this is greens function.
$Ly=δ(t−τ)$, $t<0$
Let's assume $f(t) = 0$ except for some impulse $I$ delivered "instantaneously" at time $\tau$; note that for this case $y(t) = y'(t) = 0$ for $t<\tau$.
The motion for $t>\tau$ is given by the solution to the homogeneous equation (the equation taking $f(t) = 0$ which is $$ y(t) = C_1 e^{-2t}\cos \left( \sqrt{5} t \right) + C_2 e^{-2t}\sin \left( \sqrt{5} t \right)=Ke^{-2t}\sin \left( \sqrt{5} (t - \delta) \right) $$
where $C_1$ and $C_2$ (or equivalently $K$ and $\delta$) are arbitrary constants.
As a result of the impulse at $t=\tau$, $y(t)$ is still zero immediately after $t=\tau$ but the velocity $y'$ is given by $$\lim_{\epsilon \to 0^+ }y'(\tau + \epsilon) = I$$ These two conditions (which are like boundary conditions at $t=\tau$) determine $C_1$ and $C_2$ and lead (for $t>\tau$) to $$y(t) = \frac{I}{\sqrt{5}} e^{-2(t-\tau)} \sin \left( \sqrt{5}(t-\tau) \right) $$ (It is easy to check that: $y(\tau) = 0$ because the argument of the sine is zero, and $y'(\tau+ |\epsilon|) = I$ because the $\sqrt{5}$ obtained by differentiating the sine function cancels the $\sqrt{5}$ in the denominator.)
This is the Green's function (use a capital G, it was named after the mathematician George Green, who presumedly was not himself the color green) for that differential equation. To find the solution in the general case, replace $I$ by $f(\tau) \, d\tau$ and integrate over $\tau$: $$ y(t) = \int_0^t \frac{f(\tau)}{\sqrt{5}} e^{-2(t-\tau)} \sin \left( \sqrt{5}(t-\tau) \right) d\tau $$
This Green's function $G(t,\tau)$ can be said to satisfy the differential equation $$ \frac{d^2G(t,\tau)}{dt^2} + 4 \frac{dG(t,\tau)}{dt} + 9G(t,\tau) = \delta(t-\tau) $$