The solution of the differential equation $e^{-x}(y+1) dy + (\cos^2 x - \sin 2x) y dx = 0$ at $x=0, y=1$ is

The differential equation $\frac{dx}{dy} = \frac{3y}{2x}$ represents a family of hyperbolas (except when it represents a pair of lines) with eccentricity:

Every curve represented by the general solution of $\frac{dy}{dx} = \frac{x \log x}{y^3 e^{y^2-5}}$ cuts every curve represented by the general solution of $\frac{dy}{dx} + \frac{y^3 e^{y^2-5}}{x \log x} = 0$ at an angle $\theta$. Then,$4\theta - \frac{\pi}{2} =$

Let $\frac{dy}{dx} = \frac{ax - by + a}{bx + cy + a}$,where $a, b, c$ are constants,represent a circle passing through the point $(2, 5)$. Then the shortest distance of the point $(11, 6)$ from this circle is

The general solution of the differential equation $\frac{dy}{dx} = \frac{2x-3y+4}{3x+2y-7}$ is

The curve amongst the family of curves represented by the differential equation,$(x^2 - y^2)dx + 2xy\, dy = 0$ which passes through $(1, 1)$,is