In my book I have the following problem:
$y'' + 4y = sin(t) + u_\pi(t)sin(t-\pi)$ where $y(0)=0, y'(0)=0$ and $u$ is the heaviside function.
The answer in the back of the book is:
$y=\frac 16((2sin(t)-sin(2t)) - u_\pi(t)(2sin(t)+sin(2t))$
I don't understand why the $(2sin(t)+sin(2t)$ terms that are being multiplied by the heaviside function aren't shifted by $\pi$?
Shouldn't the answer be $y=\frac 16((2sin(t)-sin(2t)) - u_\pi(t)(2sin(t-\pi)+sin(2(t-\pi))$?
I worry that I'm not correctly understanding what it actually means to shift a function.
Thanks for any and all help.
Recall,
$$\sin(t -\pi) = - \sin t$$
We then have:
$$\mathscr{L}( u_\pi(t) \sin(t-\pi) ) = \mathscr{L}( -u_\pi(t) \sin(t) ) = -e^{-\pi s} \mathscr{L}(- \sin(t)) = e^{-\pi s} \dfrac{1}{s^2+1}$$
So, for the LHS, we have:
$$\mathscr{L} ( y'' + 4y) = s^2 y(s) - sy(s) - y'(0) + 4y(s) = (s^2 + 4)y(s)$$
For the RHS, we have:
$$\mathscr{L} (\sin t + u_\pi(t)\sin(t-\pi)) = \mathscr{L} (\sin t -u_\pi(t) \sin(t)) = \dfrac{e^{-\pi s}+1}{s^2+1}$$
Putting this together yields:
$$y(s) = \dfrac{e^{-\pi s}+1}{(s^2+1)(s^2+4)}$$
The partial fraction expansion is then:
$$y(s) = \dfrac{e^{-\pi s}+1}{(s^2+1)(s^2+4)} = \dfrac{1}{3(s^2+1)} - \dfrac{1}{3(s^2+4)} + \dfrac{e^{-\pi s}}{3(s^2+1)} - \dfrac{e^{-\pi s}}{3(s^2+4)}$$