The integral curve satisfying $y' = \frac{x^2 + y^2}{x^2 - y^2}$ with $y(1) = 2$ has a slope at the point $(1, 0)$ equal to:

The general solution of the differential equation $(x + y)dx + xdy = 0$ is

If a curve $y=f(x),$ passing through the point $(1,2),$ is the solution of the differential equation $2 x^{2} dy=\left(2 xy+y^{2}\right) dx,$ then $f\left(\frac{1}{2}\right)$ is equal to

The general solution of the differential equation $(x-y-1) dy = (x+y+1) dx$ is

If $\frac{dy}{dx} = \frac{y + x \tan(\frac{y}{x})}{x}$, then $\sin(\frac{y}{x})$ is equal to

If $(x^2+y^2) dy = xy dx$,with $y(x_0) = e$ and $y(1) = 1$,then $x_0$ has the value: