Solve $x''+4x = \delta(t) + \delta(t-\pi)$, $x(0) = 0, x'(0) = 0$
My Work So Far
Begin with conversion:
$$s^2X(s) -sx(0) - x'(0) + 4X(s) = e^{-0s} + e^{-\pi s}$$
Now clean up:
$$s^2X(s) + 4X(s) = 1 + e^{-\pi s}$$
$$X(s) = \frac{1}{s^2+4}+e^{-\pi s}\frac{1}{s^2+4}$$
Massage into recognizable Laplace shapes:
$$X(s) = \frac{1}{2}\frac{2}{s^2+2^2}+e^{-\pi s}\frac{1}{2}\frac{2}{s^2+2^2}$$
Now ready to find inverse Laplace to solve:
$$x(t) = \frac{1}{2}\cdot \sin(2t) + \frac{1}{2}\cdot u(t-\pi)\sin(2(t-\pi))$$
$$x(t) = \frac{1}{2}(\sin(2t) + u(t-\pi)\sin(2(t-\pi)))$$
And that's my answer. It's very similar to the correct answer, which is:
$$x(t) = \frac{1}{2}(\sin(2t) + u(t-\pi)\color{red}{\sin(2t)})$$
The $(t-\pi)$ has converted to just $t$, which doesn't make sense to me, as the formula I used is:
$$\mathcal{L}^{-1}\{e^{-\pi s}F(s)\}= u(t-\pi)f(t-\pi)$$
Why would $t-\pi$ suddenly become $t$?
I have checked your work several times. You have not made any mistake. You can simplify your answer further by noting that $f(x)=\sin(2x)$ has a period of $\pi$ radians. Thus, it follows that: $$\sin(2(t-\pi))=\sin(2t-2\pi)=\sin(2t)$$ Hence, the two answers are equivalent: $$x(t) = \frac{1}{2}(\sin(2t) + u(t-\pi)\color{blue}{\sin(2(t-\pi))})=\frac{1}{2}(\sin(2t) + u(t-\pi)\color{blue}{\sin(2t)})$$ Where $u(x)$ is the Heaviside step function.