The slope of the tangent at $(x, y)$ to the curve passing through $(2, 1)$ is $\frac{x^2+y^2}{2xy}$. Find the equation of the curve.

The general solution of the differential equation $x \cos \frac{y}{x}(y d x+x d y)=y \sin \frac{y}{x}(x d y-y d x)$ is

The solution curve of the differential equation $y \frac{dx}{dy} = x(\log_e x - \log_e y + 1)$,$x > 0, y > 0$ passing through the point $(e, 1)$ is

Let the solution curve $y = y(x)$ of the differential equation $\left[\frac{x}{\sqrt{x^{2}-y^{2}}}+e^{\frac{y}{x}}\right] x \frac{dy}{dx} = x + \left[\frac{x}{\sqrt{x^{2}-y^{2}}}+e^{\frac{y}{x}}\right] y$ pass through the points $(1, 0)$ and $(2\alpha, \alpha)$,where $\alpha > 0$. Then $\alpha$ is equal to:

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

Show that the differential equation $\left(1+e^{\frac{x}{y}}\right) dx+e^{\frac{x}{y}}\left(1-\frac{x}{y}\right) dy=0$ is a homogeneous equation and find its solution.