Solve the System of Differential Equations

Firstly notice $x'(0)=1$ and $y'(0)=0$ . First we solve the $y$ set. To do this we differentiate the second equation to substitute into the first. This produces $y'' + y = -1.$ Taking the Laplace transform of this differential equation produces $\mathcal{L}(y) (s^2 +1) = \frac{-1}{s}$ $\mathcal{L}(y) = \frac{s}{s^2+1}-\frac{1}{s}$ then taking the inverse Laplace transform gives $y = \mathcal{L}^{-1}\Big(\frac{s}{s^2+1} \Big)-\mathcal{L}^{-1}\Big( \frac{1}{s}\Big)$ $y = \cos(t) -1.$ The exact same process for the $x$ equation produces $\mathcal{L}(x) (s^2+1) = 1$ $\mathcal{L}(x) = \frac{1}{s^2+1}$ which obviously has the inverse transform $x = \sin(t)$ so $(x,y) = (\sin(t), \cos(t)-1)$ and at $t=\pi$ we get $(x,y) = (\sin(\pi), \cos(\pi)-1) = (0,-2)$ therefore $\boxed{a+b = -2}$ .