General solution of differential equation $\frac{dy}{dx} + y = 1$ $(y \neq 1)$ is

Let $y=y(x)$ be a solution curve of the differential equation,$(1-x^2 y^2) dx = y dx + x dy$. If the line $x = 1$ intersects the curve $y = y(x)$ at $y = 2$ and the line $x = 2$ intersects the curve $y = y(x)$ at $y = \alpha$,then a value of $\alpha$ is

Let $y = y(x)$ be the solution of the differential equation $\frac{dy}{dx} = (1+x^2)(1-y+y^2)$, with the initial condition $y(0) = \frac{1}{2}$. Then $(2y(1) - 1)$ is equal to:

General solution of the differential equation $(y^3+y)(x^2+1) dy = (xy^4+2y^2x) dx$ is (where $C$ is a constant of integration.)

The particular solution of $\frac{y}{x} \frac{dy}{dx} = \frac{1+y^2}{1+x^2}$ when $x=2, y=1$ is

The solution of $(\text{cosec } x \log y) dy + (x^2 y) dx = 0$ is