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

The substitution required to reduce the differential equation $t^2 dx + (x^2 - tx + t^2) dt = 0$ to a differential equation which can be solved by the variables separable method is

If $\sin \left(\frac{y}{x}\right)=\log |x|+\frac{\alpha}{2}$ is the solution of the differential equation $x \cos \left(\frac{y}{x}\right) \frac{d y}{d x}=y \cos \left(\frac{y}{x}\right)+x$ and $y(1)=\frac{\pi}{3}$,then $\alpha^2$ is equal to

The solution of $\frac{d y}{d x}=\frac{y^2}{x y-x^2}$ is

The solution of the differential equation $x y^{\prime} = 2 x e^{-y / x} + y$ 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: