I have to find $x(t)$ and $y(t)$ knowing that: $$\begin{cases} x'= y-\int_0^tx(u)\cos(t-u)\,du \\ y'=x+\cos(t) - 1 \end{cases}$$ Where $x=x(t)$, $y=y(t)$ and $x(0)=1$, $y(0)=0$.
I've noticed that the integral in the first equation is the convolution between $x(t)$ and $cos(t)$, so I can apply the Laplace Transform to determine the originals $x$ and $y$. $$\begin{cases} sF(s)-1=G(s)-F(s)\frac{s}{s^2+1} \\ sG(s)=F(s)+\frac{s}{s^2+1}-\frac{1}{s} \end{cases}$$ Is there another way to find the original functions $x$ and $y$?
Thanks!
Considering the Laplace Transform is fine. From your work, we have $$\begin{cases} sF(s)-1=G(s)-F(s)\frac{s}{s^2+1} \\ sG(s)=F(s)+\frac{s}{s^2+1}-\frac{1}{s} \end{cases}$$ and by evaluating $s\cdot I+II$ we get
$$ s^2F(s)-s=F(s)+\frac{s}{s^2+1}-\frac{1}{s}-F(s)\frac{s^2}{s^2+1}$$ that is $$s\left(s+\frac{s}{s^2+1}-\frac{1}{s}\right)F(s)=s+\frac{s}{s^2+1}-\frac{1}{s}$$ and we find that $F(s)=1/s$. Hence $$G(s)=\frac{F(s)}{s}+\frac{1}{s^2+1}-\frac{1}{s^2}=\frac{1}{s^2+1}.$$ Therefore $x(t)=1$ and $y(t)=\sin(t)$.
P.S. It is also easy to check that they verify $$\begin{cases} x'= y-\int_0^tx(u)\cos(t-u)\,du \\ y'=x+\cos(t) - 1 \end{cases}$$