$$ \text { The general solution of the differential equation }(1+\tan y)(d x-d y)+2 x d y=0 \text { is } $$

The sum of the order and degree of the differential equation $$\left(\frac{d^2 y}{d x^2}\right)^5+\frac{4\left(\frac{d^2 y}{d x^2}\right)^3}{\left(\frac{d^3 y}{d x^3}\right)}+\frac{d^3 y}{d x^3}=x^2-1$$ is

$$ \text { If } \frac{d y}{d x}=y+3>0 \text { and } y(0)=2 \text { then } y(\log 2) \text { is equal to } $$

The particular solution of $$\frac{d y}{d x}+\sqrt{\frac{1-y^2}{1-x^2}}=0$$, when $$x=0, y=\frac{1}{2}$$ is