91Ó°ÊÓ

The given expressions are$f\left(t\right)=e^{-t}$.

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.

$$\begin{gathered} f\left(t\right)=e^{-t}y\left(t\right) \\ ={∫}_0^t\frac{1}{\mathrm{Ó\neg }}·\sin \mathrm{Ó\neg }\left(t-t\text{'}\right)f\left(t\text{'}\right)dt\text{'}y\left(t\right) \\ ={∫}_0^t\frac{1}{\mathrm{Ó\neg }}·\sin \mathrm{Ó\neg }\left(t-t\text{'}\right)·e^{-t}dty\left(t\right) \\ ={∫}_0^t{e}^{-t}\frac{1}{\mathrm{Ó\neg }}·\sin \mathrm{Ó\neg }\left(t-t\text{'}\right)dty\left(t\right) \end{gathered}$$

Solve further

$$\begin{gathered} =\frac{1}{\mathrm{Ó\neg }}{\left[\frac{e^{-t}}{{\left(-1\right)}^2-\left({\mathrm{Ó\neg }}^2\right)}\left\{\left(-1\right)\sin \left(\mathrm{Ó\neg }t-\mathrm{Ó\neg }t\right)-\left(-\mathrm{Ó\neg }\right)\cos \left(\mathrm{Ó\neg }t\right)\right\}\right]}_{t-0}^t \\ y\left(t\right)=\frac{1}{\mathrm{Ó\neg }}{\left[\frac{e^{-t}}{{\left(-1\right)}^2-\left({\mathrm{Ó\neg }}^2\right)}\left\{-\sin \left(\mathrm{Ó\neg }t-\mathrm{Ó\neg }{t}^t\right)+\mathrm{Ó\neg }\cos \left(\mathrm{Ó\neg }t-\mathrm{Ó\neg }t\right)\right\}\right]}_{t-0}^t \\ =\frac{1}{\mathrm{Ó\neg }}{\left[\frac{e^{-t}}{{\left(-1\right)}^2-\left({\mathrm{Ó\neg }}^2\right)}\left\{-\sin \mathrm{Ó\neg }\left(\mathrm{Ó\neg }t-\mathrm{Ó\neg }t\right)+\mathrm{Ó\neg }\cos \left(\mathrm{Ó\neg }t-\mathrm{Ó\neg }t\right)\right\}-\frac{e^{-0}}{{\left\{-1\right\}}^2-{\left(\mathrm{Ó\neg }\right)}^2}\left\{-\sin \mathrm{Ó\neg }\left(\mathrm{Ó\neg }t-0\right)+\mathrm{Ó\neg }\cos \left(\mathrm{Ó\neg }t-0\right)\right\}\right]}_{t-0}^t \\ y\left(t\right)=\frac{1}{\mathrm{Ó\neg }\left(1+{\mathrm{Ó\neg }}^2\right)}\left[\mathrm{Ó\neg }{e}^{-1}+\sin \mathrm{Ó\neg }t-\mathrm{Ó\neg }\cos \mathrm{Ó\neg }t\right] \end{gathered}$$

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