So i am struggling with this problem.
The following DGL is given:
$$ T'' - \ \omega ^{2} _{0} T = \delta _{0} $$
So first i have to Fourier transform the equation. Then i get the following:
$$ -\omega^{2} \widehat{T} - \ \omega ^{2} _{0} \widehat{T} = \frac{1}{ \sqrt{2 \pi } }$$
Then i have this:
$$ \Longrightarrow \widehat{T} = - \frac{1}{ \sqrt{2 \pi } } \frac{1}{ ( \ \omega ^{2} _{0}+\omega^{2} ) } = - \frac{1}{2\omega _{0}} \widehat{g}(\omega)$$
with
$$ g(t)= e^{-\omega_{0}|t|} $$
$$ \widehat{g}(\omega) = \frac{2 \omega_{0}}{ \sqrt{2 \pi } } \frac{1}{\omega ^{2} _{0} + \omega ^{2}} $$
Now i would transform it back:
$$ \Longrightarrow T = - \frac{1}{2\omega _{0}}g(t)$$
to solve the equation, but i don't know of this is right.
With $T(t) = -\frac{1}{2\omega_0} e^{-\omega_0|t|}$ we have $$T'(t) = \frac12 e^{-\omega_0|t|} \operatorname{sign}(t),$$ $$T''(t) = -\frac12 \omega_0 e^{-\omega_0|t|} + (T'(0+)-T'(0-)) \, \delta(t) = -\frac12 \omega_0 e^{-\omega_0|t|} + \delta(t).$$
We therefore have $$T''(t) - \omega_0^2 T(t) = \left( -\frac12 \omega_0 e^{-\omega_0|t|} + \delta(t) \right) - \omega_0^2 \left( -\frac{1}{2\omega_0} e^{-\omega_0|t|} \right) = \delta(t).$$
Thus $T$ satisfies your differential equation.
There are actually more solutions, but those are not Fourier transformable: $$T(t) = A e^{\omega_0 t} + B e^{-\omega_0 t} -\frac{1}{2\omega_0} e^{-\omega_0|t|},$$ where $A$ and $B$ are some constants.