The general solution of the differential equation $\frac{dy}{dx} = \frac{x^2}{y^2}$ is

If a curve passes through the point $(1,1)$ and at any point $(x, y)$ on the curve,the product of the slope of its tangent and $x$ coordinate of the point is equal to the $y$ coordinate of the point,then the curve also passes through the point

Find the general solution of the differential equation $\frac{dy}{dx} + \sqrt{\frac{1-y^2}{1-x^2}} = 0$.

The particular solution of the differential equation $\frac{dy}{dx} = y \tan x$ with the initial condition $y(0) = 1$ is:

The general solution of $\left(\left(1+x^2\right) y \sin x-2 x y\right) d x-\log y^{1+x^2} d y=0$ is

The solution of $\frac{dy}{dx} = \frac{1}{x}$ is