My question is where did I go wrong? I cannot seem to duplicate the answer given in text.
The Problem: Solve the following differential equation using Laplace Transforms given that $y(0)=0$, $y'(0)=0$, $r(t)=sin(2t)$ on the interval $(0<t<\pi)$. The value of the function $r(t)$ is zero everywhere else.
$$y''+y=r(t)$$
My solution:
Express our problem using the unit step function...
$$y''+y=sin(2t)u(t)-sin(2t)u(t-\pi)$$
Taking Laplace transforms renders the following subsidiary equation...
$$Y(s^2+1)=\frac{2}{s^2+4}-\frac{e^{-\pi s}}{s}\frac{2}{s^2+4}$$
Solving for Y...
$$Y=\frac{2}{(s^2+4)(s^2+1)}-\frac{e^{-\pi s}}{s}\frac{2}{(s^2+4)(s^2+1)}$$
We now need to decompose each of the two terms into partial fractions...
An online calculator was used for this. It can be found here.
$$Y= \left[\frac{2}{3(s^2+1)}-\frac{2}{3(s^2+4)}\right]_1-\left[-\frac{2s}{3(s^2+1)}+\frac{s}{6(s^2+4)}+\frac{1}{2s}\right]_2\cdot e^{-\pi s}$$
We can now take inverse Laplace Transforms but being careful to keep all terms within their original brackets...
$$y(t)= \left[\frac{2}{3}sin(t)-\frac{1}{3}sin(2t)\right]_1-\left[-\frac{2}{3}cos(t-\pi)+\frac{1}{12}cos(2(t-\pi))+\frac{1}{2}u(t-\pi)\right]_2$$
We can now simplify some of the terms in bracket $2$...
$$y(t)= \left[\frac{2}{3}sin(t)-\frac{1}{3}sin(2t)\right]_1-\left[\frac{2}{3}cos(t)-\frac{1}{12}cos(2t)+\frac{1}{2}\right]_2$$
The answers:
When $0<t<\pi$ we have the terms in bracket $1$...
$$y(t)=\frac{2}{3}sin(t)-\frac{1}{3}sin(2t)$$
When $t>\pi$ we have the terms in bracket $1$ minus bracket 2...
$$y(t)=\frac{2}{3}sin(t)-\frac{1}{3}sin(2t)-\frac{2}{3}cos(t)+\frac{1}{12}cos(2t)-\frac{1}{2}$$
The answer in text:
When $0<t<\pi$ the text has the following which agrees with my solution for this part of the problem...
$$y(t)=\frac{2}{3}sin(t)-\frac{1}{3}sin(2t)$$
When $t>\pi$ the text has the following which DOES NOT agree with my solution for this part of the problem...
$$y(t)=\frac{4}{3}sin(t)$$
My question:
What did I do wrong and how can I fix it?
The calculation of the Laplace transform of $r$ is wrong: the Laplace transform of a product is not the product of the Laplace transforms. $$ (\mathcal{L}r)(x)=\int_0^\pi e^{-sx}\sin(2\,x)\,dx=\frac{2-2\,e^{-\pi s}}{4+s^2}. $$