Reduce a trigonometric equation and apply the initial condition

$du/dx=2+dy/dx$, so $dy/dx=du/dx-2$. Therefore $du/dx-2=\sin^2u-3$, hence $du/dx=\sin^2u-1=-\cos^2u$.

Where $\cos u\ne0$, $\int\sec^2u\,du=\int-1\,dx$, giving $\tan u=-x+C$. The initial point $x=y=0$ gives u=0 and C=0. The continuous branch through zero is $u=\arctan(-x)$, so $y=\arctan(-x)-2x$. The excluded constant u-values $\pi/2+n\pi$ solve the transformed equation but do not pass through the origin.