The general solution of $x \frac{dy}{dx} = y - x \tan \left( \frac{y}{x} \right)$ 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:

The solution of the differential equation $3 x y' - 3 y + (x^2 - y^2)^{1/2} = 0$,satisfying the condition $y(1) = 1$ is

Express $\frac{dt}{dx} = \frac{t}{x + t e^{-2x/t}}$ in the form of $\frac{dx}{dt} = \phi\left(\frac{x}{t}\right)$.

The solution of the differential equation $(x^2 - xy)dy = (xy + y^2)dx$ is

The solution of the differential equation $y \frac{dy}{dx} = x \left[ \frac{y^2}{x^2} + \frac{\phi(y^2/x^2)}{\phi'(y^2/x^2)} \right]$ is (where $c$ is a constant):