Using Fourier transform to find the particular solution of a differential equation

Sure, using the properties of the Fourier Transform, we can indeed simplify the process of solving the differential equation. Let's apply this to the same differential equation:

$$a \frac{d^2 y(t)}{dt^2} + b \frac{dy(t)}{dt} + c y(t) = f(t)$$

The Fourier Transform of a function $g(t)$ is given by:

$$G(\omega) = \int_{-\infty}^{\infty} g(t) e^{-i\omega t} dt$$

And the Fourier Transform of the $n$-th derivative of $g(t)$ is given by:

$$\left(\frac{d^n g(t)}{dt^n}\right)^{\mathcal{F}} = (i\omega)^n G(\omega)$$

Applying the Fourier Transform to both sides of the differential equation, we get:

$$a (i\omega)^2 Y(\omega) + b i\omega Y(\omega) + c Y(\omega) = F(\omega)$$

where $Y(\omega)$ and $F(\omega)$ are the Fourier Transforms of $y(t)$ and $f(t)$ respectively. 

Simplifying, we get:

$$(-a\omega^2 + b i\omega + c) Y(\omega) = F(\omega)$$

Isolating $Y(\omega)$, we obtain:

$$Y(\omega) = \frac{F(\omega)}{-a\omega^2 + b i\omega + c}$$

To find the particular solution $y(t)$ in the time domain, we take the inverse Fourier Transform of $Y(\omega)$. The inverse Fourier Transform is given by:

$$g(t) = \frac{1}{2\pi} \int_{-\infty}^{\infty} G(\omega) e^{i\omega t} d\omega$$

Therefore, the particular solution to the differential equation is:

$$y(t) = \frac{1}{2\pi} \int_{-\infty}^{\infty} \frac{F(\omega)}{-a\omega^2 + b i\omega + c} e^{i\omega t} d\omega$$

The calculation of this integral will depend on the specific form of $F(\omega)$ and the coefficients $a$, $b$, and $c$.