Use fourier transform to solve second-order differential equation -- an "easy" integral? I have scoured the internet for a fully-explained solution to this problem but have found none:
The problem asks to solve this differential equation for $y(t)$ using Fourier Transforms, and then consider cases where $b > w_0$, $b < w_0$, and $b = w_0$
$$
\frac{d^2y(t)}{dt^2} + 2b\frac{dy(t)}{dt} + w_0^2y(t) = \delta(t)
$$
So far I have, by differential properties of Fourier Transforms, converting the function of $t$ to a function of $w$,
$$
-w^2\hat{y}(w) + 2biw\hat{y}(w) + w_0^2\hat{y}(w) = \frac{1}{\sqrt{2\pi}}
$$
So algebraically,
$$
\hat{y}(w) = \frac{\frac{1}{\sqrt{2\pi}}}{-w^2 + 2biw + w_0^2}
$$
And the Inverse Fourier Transform of $\hat{y}(w)$ can yield $y(t)$ as
$$
y(t) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}{\frac{e^{-iwt} \frac{1}{\sqrt{2\pi}}}{-w^2 + 2biw + w_0^2}dw}
$$
My professor says that this is an "easy integral; just do it". I have found no way of doing it thusfar, even for the simplest(?) case of $b = w_0$. I don't even know if this is of the right form (i.e. using the "easiest" definition of FT to solve the problem). He looked at it for like 2 seconds.
All help is appreciated!

Such kind of integral usually evaluate via residue.
$$
y(t) = -\frac{1}{2\pi}\int_{-\infty}^{\infty}{\frac{e^{-iwt} }{(w-ib)^2 - w_0^2-b^2}dw}
$$
The function $\frac{e^{-iwt} }{(w-ib)^2 - w_0^2-b^2}$ is a holomorphic function. Tis function has two residue $w_{\pm}=i b \pm\sqrt{w_0^2+b^2}$
You should consider two different case $t>0$ and $t<0$
In the first case $t>0$ you should close path in the lower half plane and the integral equal zero becouse of all residue is located in the upper half-plane.
In the second case $t<0$ one can close path in the upper half plane and using  Cauchy's theorem we get
$$
y(t) = -\frac{i}{2}{\frac{e^{-iw_+t}-e^{-iw_-t} }{w_+-w_-}}
$$
Thus$$
y(t) = -\frac{i}{2}{\frac{e^{-iw_+t}-e^{-iw_-t} }{w_+-w_-}}\theta(-t)$$
where $\theta(x)$ is a Heaviside step function.
