How to compute this Laplace Transform!

Question

How to solve this using the Laplace transform? $$ y''+4y = u_{2\pi}(t)\sin(t-2\pi), \qquad y(0)=0,\,y'(0)=0 $$

And how to compute $y\left(\frac{\pi}{2}\right)$ and $y\left(\frac{5\pi}{2}\right)$ ?

Answer 1

Assuming that $u_{2\pi}(t)$ is the function that equals $1$ for $t\in[0,2\pi]$ and zero otherwise, by assuming $g(s)=\int_{0}^{+\infty}y(t)e^{-st}\,dt$ and applying the properties of the Laplace transform we get:

$$ (4+s^2) g(s) = \frac{1-e^{-2\pi s}}{1+s^2} $$ hence $y$ is given by the inverse Laplace transform of $\frac{1-e^{-2\pi s}}{(1+s^2)(4+s^2)}$, that equals

$$ y(t)=\frac{2}{3}\sin(t)\sin^2\left(\frac{t}{2}\right) $$ over $[0,2\pi]$ and zero anywhere else. It follows that $y\left(\frac{\pi}{2}\right)=\frac{1}{3}$ and $y\left(\frac{5\pi}{2}\right)=0$.