Question

Problems 2 and 3, use (12.6) to solve (12.1) when $\mathit{f}\left(t\right)$is as given.$\mathit{f}\left(t\right)\mathbf{=}\mathit{s}\mathit{i}\mathit{n}\mathit{\omega }\mathit{t}$

Short answer

Answer

The value of function $y\left(t\right)-{\int }_0^t\sin \omega \left(t-t\text{'}\right)f\left(t\text{'}\right)dl$is equal to$y\left(t\right)=\frac{1}{2{\omega }^2}\left[\sin \left(\omega t\right)-\omega t\cos \omega t\right]$

Step-by-step solution

Given information

The given expressions are$f\left(t\right)=\sin \omega t$.

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.

$f\left(t\right)=\sin \omega t$

And then

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

Solve further

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

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