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:

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

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

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

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

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

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

$$a(i\omega {)}^{2}Y(\omega )+bi\omega Y(\omega )+cY(\omega )=F(\omega )$$

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

Simplifying, we get:

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

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

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

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:

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

Therefore, the particular solution to the differential equation is:

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

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