91Ó°ÊÓ

The given expressions are$f\left(t\right)=\sin \mathrm{Ó\neg }t$.

A transformation of a function f(x) into the function g(t) that is useful especially in reducing the solution of an ordinary linear differential equation with constant coefficients to the solution of a polynomial equation.

The inverse Laplace transform of a function F(s) is the piecewise-continuous and exponentially-restricted real function f(t)

Use the function.

$f\left(t\right)=\sin \mathrm{Ó\neg }t$

And then

$$\begin{gathered} y\left(t\right)={∫}_0^l\sin \mathrm{Ó\neg }\left(t-t\text{'}\right)f\left(t\text{'}\right)dt \\ y\left(t\right)={∫}_0^l\sin \mathrm{Ó\neg }\left(t-t\text{'}\right)\left(\sin \mathrm{Ó\neg }t\right)dt \\ y\left(t\right)=\frac{1}{2\mathrm{Ó\neg }}{∫}_0^{l}2\sin \left(\mathrm{Ó\neg }t-\mathrm{Ó\neg }t\right)·\sin \mathrm{Ó\neg }tdt \\ y\left(t\right)=\frac{1}{2\mathrm{Ó\neg }}\left[{∫}_0^{l}2\sin \left(\mathrm{Ó\neg }t-\mathrm{Ó\neg }t\text{'}-\mathrm{Ó\neg }t\text{'}\right)-\cos \left(\mathrm{Ó\neg }t-\mathrm{Ó\neg }t-\mathrm{Ó\neg }t\right)\right]dt\text{'} \end{gathered}$$

Solve further

$$\begin{gathered} y\left(t\right)=\frac{1}{2\mathrm{Ó\neg }}\left[{∫}_0^t{\left\{\frac{1}{-2\mathrm{Ó\neg }}\sin \left(\mathrm{Ó\neg }t-2\mathrm{Ó\neg }t\text{'}\right)dt\right\}}_{t-0}^t-\cos \text{'}{∫}_0^{t}dt\text{'}\right] \\ y\left(t\right)=\frac{1}{2\mathrm{Ó\neg }}\left[{∫}_0^t\frac{1}{-2\mathrm{Ó\neg }}\left(\sin \left(\mathrm{Ó\neg }t-2\mathrm{Ó\neg }t\text{'}\right)-\sin \mathrm{Ó\neg }t-0\right)-\cos {\left\{t\right\}}_{t-0}^t\right] \\ y\left(t\right)=\frac{1}{2\mathrm{Ó\neg }{t}^2}\left[\sin \left(\mathrm{Ó\neg }t\right)-\mathrm{Ó\neg }t\cos \mathrm{Ó\neg }t\right] \end{gathered}$$

Thus, the value of function $y\left(t\right)={∫}_0^t\sin \left(t-t\text{'}\right)f\left(t\text{'}\right)dt$is equal to$y\left(t\right)=\frac{1}{2{\mathrm{Ó\neg }}^2}\left[\sin \left(\mathrm{Ó\neg }t\right)-\mathrm{Ó\neg }t\cos \mathrm{Ó\neg }t\right]$.