Question

In Problems 2 and 3, use (12.6) to solve (12.1) when $\mathit{f}\left(t\right)$is as give$\mathit{f}\left(t\right)\mathbf{=}{\mathit{e}}^{\mathbf{-}\mathbf{t}}$

Short answer

Answer

The value of value of $y\left(t\right)={\int }_0^t\frac{1}{\omega }·\sin \omega \left(t-t\text{'}\right)f\left(t\text{'}\right)dt\text{'}$is$y\left(t\right)=\frac{1}{\omega \left(1+{\omega }^2\right)}\left[\omega e^{-t}+\sin \omega t\right]$

Step-by-step solution

Given information

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

Definition of Laplace Transformation

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)

Solve the given function

Use the function.

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

Solve further

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

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