Let the solution curve of the differential equation $x \frac{dy}{dx} - y = \sqrt{y^2 + 16x^2}$ with the initial condition $y(1) = 3$ be $y = y(x)$. Then the value of $y(2)$ is equal to:

Consider the differential equation $\frac{dy}{dx} = \frac{y^3}{2(xy^2 - x^2)}$.
Statement $-1:$ The substitution $z = y^2$ transforms the above equation into a first-order homogeneous differential equation.
Statement $-2:$ The solution of this differential equation is $y^2 e^{-y^2/x} = C$.

$A$ curve passes through the point $\left(1, \frac{\pi}{6}\right)$. Let the slope of the curve at each point $(x, y)$ be given by $\frac{dy}{dx} = \frac{y}{x} + \sec \left(\frac{y}{x}\right)$,where $x > 0$. Then the equation of the curve is:

The solution of $x \frac{d y}{d x} = y(\log y - \log x + 1)$ is

Let the solution curve of the differential equation $x dy = (\sqrt{x^{2}+y^{2}}+y) dx$,$x > 0$,intersect the line $x = 1$ at $y = 0$ and the line $x = 2$ at $y = \alpha$. Then the value of $\alpha$ is.

The solution of the differential equation $y \sin \left(\frac{x}{y}\right) dx = \left\{x \sin \left(\frac{x}{y}\right) - y\right\} dy$ satisfying $y\left(\frac{\pi}{4}\right) = 1$ is