I have eqn: $\frac{dx}{dt} = -y(t)$ and $\frac{dy}{dt} = x(t)$
I know that $(x(0),y(0))= (1,0)$.
I want to solve eqn and show that it admits an invariant $I = x(t)^2 + y(t)^2$.
I know $x' = -y$,
$y' = x$,
$x^{\prime\prime} = -y' = -x$
I know general solution of $x" = -x$ is $x = a\sin x = b\cos x$.
I know $x(0) = a\sin 0 + b\cos 0 = 1$ So $b = 1$
How can I show $a = 1$? (I think it should!)
I tried $x' = a\cos x - b\sin x$ since $y = -x$ but it just gives $ a = 0$.