The particular solution of the differential equation $(1+y^2) dx - xy dy = 0$ with the condition $y(1) = 0$ represents:

The solution of $\frac{dy}{dx} = \frac{e^x(\sin^2 x + \sin 2x)}{y(2\log y + 1)}$ is

Let $\alpha |x| = |y| e^{xy-\beta}$,where $\alpha, \beta \in \mathbb{N}$,be the solution of the differential equation $x dy - y dx + xy(x dy + y dx) = 0$ with the initial condition $y(1) = 2$. Then $\alpha + \beta$ is equal to:

The equation of the curve that passes through the point $(1, 2)$ and satisfies the differential equation $\frac{dy}{dx} = \frac{-2xy}{x^2 + 1}$ is

If the solution of the differential equation $(2x+3y-2)dx+(4x+6y-7)dy=0$ with $y(0)=3$ is $\alpha x+\beta y+3 \log_e|2x+3y-\gamma|=6$,then $\alpha+2\beta+3\gamma$ is equal to

If the solution curve $y=y(x)$ of the differential equation $(1+y^2)(1+\log_e x) dx + x dy = 0, x>0$ passes through the point $(1,1)$ and $y(e) = \frac{\alpha-\tan(3/2)}{\beta+\tan(3/2)}$,then $\alpha+2\beta$ is equal to: