The general solution of the differential equation $\left(y^2+x+1\right) d y=(y+1) d x$ is

The difference of the order and degree of the differential equation $\left(\frac{d^2 y}{d x^2}\right)^{-\frac{7}{2}}\left(\frac{d^3 y}{d x^3}\right)^2-\left(\frac{d^2 y}{d x^2}\right)^{-\frac{5}{2}}\left(\frac{d^4 y}{d x^4}\right)=0$ is

If $x d y+\left(y+y^2 x\right) d x=0$ and $y=1$ at $x=1$, then

The solution of $x d y-y d x=\sqrt{x^2+y^2} d x$ when $y(\sqrt{3})=1$ is