The general solution of the differential equation $x \left( \frac{dy}{dx} \right) = y \ln \left( \frac{y}{x} \right)$ 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 slope of the tangent at $(x, y)$ to a curve passing through $\left(1, \frac{\pi}{4}\right)$ is given by $\frac{y}{x}-\cos ^{2}\left(\frac{y}{x}\right)$,then the equation of the curve is

Let $f(x) = \sqrt{\lim_{r \rightarrow x} \left\{ \frac{2r^2 \left[(f(r))^2 - f(x)f(r)\right]}{r^2 - x^2} - r^3 e^{\frac{f(r)}{r}} \right\}}$ be differentiable in $(-\infty, 0) \cup (0, \infty)$ and $f(1) = 1$. Then the value of $ea$,such that $f(a) = 0$,is equal to:

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

Find the particular solution satisfying the given condition: $(x+y) dy + (x-y) dx = 0$; $y=1$ when $x=1$.