The slope of a curve at any point is the reciprocal of twice the ordinate at the point and it passes through the point $(4, 3)$. The equation of the curve is

The particular solution of the differential equation $(2x - 2y + 3)dx - (x - y + 1)dy = 0$ when $x = 0, y = 1$ is

The solution of $\frac{dy}{dx} = \frac{x \log x^2 + x}{\sin y + y \cos y}$ is

The general solution of the differential equation $y \log y \, dx - x \, dy = 0$ is . . . . . . .

Find the particular solution of the differential equation $\log \left(\frac{d y}{d x}\right)=3 x+4 y$ given that $y=0$ when $x=0$.

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