Express $\frac{dt}{dx} = \frac{t}{x + t e^{-2x/t}}$ in the form of $\frac{dx}{dt} = \phi\left(\frac{x}{t}\right)$.

$A$ 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:

If $\frac{dy}{dx} = f(x, y)$ is a homogeneous differential equation,then the general form of $f(x, y)$ is

If $x^\alpha \frac{dy}{dx} = y^\beta(\gamma \log x + \delta \log y + 1)$ is a homogeneous differential equation,then

The particular solution of the differential equation,$x y \frac{dy}{dx} = x^2 + 2y^2$ when $y(1) = 0$ is

The general solution of ${y^2}\,dx + ({x^2} - xy + {y^2})\,dy = 0$ is