The solution of the differential equation $x y^2 d y - (x^3 + y^3) d x = 0$ is

$A$ 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:

Solve the differential equation $y e^{\frac{x}{y}} dx = \left( x e^{\frac{x}{y}} + y^2 \right) dy$ where $y \neq 0$.

The slope of the tangent at $(x, y)$ to a curve passing through a point $(2, 1)$ is $\frac{x^2 + y^2}{2xy}$,then the equation of the curve is

The differential equation $y^{\prime} = \frac{y}{x + \sqrt{xy}}$ has a general solution given by (where $C$ is a constant of integration):

The general solution of the differential equation $\frac{d y}{d x}=\frac{x+7 y+3}{3 x+5 y+9}$ is