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

Find the solution of the following differential equation: $\{x \cos (y/x) + y \sin (y/x)\} y dx = \{y \sin (y/x) - x \cos (y/x)\} x dy$.

The curve satisfying the differential equation $(x^2 - y^2) \, dx + 2xy \, dy = 0$ and passing through the point $(1, 1)$ is

Let $y=y(x)$ be the solution of the differential equation $x \tan \left(\frac{y}{x}\right) d y=\left(y \tan \left(\frac{y}{x}\right)-x\right) d x$ for $-1 \leq x \leq 1$,with the initial condition $y\left(\frac{1}{2}\right)=\frac{\pi}{6}$. Then the area of the region bounded by the curves $x=0$,$x=\frac{1}{\sqrt{2}}$,and $y=y(x)$ in the upper half plane is:

For two events $A$ and $B$,$P(B) \neq 0$ and $P(A \mid B) = 1$,then which of the following is true?

The equation of the curve satisfying $x dy - y dx = \sqrt{x^2 - y^2} dx$ with the condition $y(1) = 0$ is: