Let \(\displaystyle \int_{0}^{x}\sqrt{1-\left(y'(t)\right)^{2}}\,dt=\int_{0}^{x}y(t)\,dt,\ 0\le x\le 3,\ y\ge 0,\ y(0)=0 \). Then at \( x=2 \), \( y''+y+1 \) is equal to
\( 2 \)
\( \sqrt{2} \)
\( 1 \)
\( {1}/{2} \)