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

Let a smooth curve $y=f(x)$ be such that the slope of the tangent at any point $(x, y)$ on it is directly proportional to $\left(\frac{-y}{x}\right)$. If the curve passes through the points $(1, 2)$ and $(8, 1)$,then $\left| y \left(\frac{1}{8}\right) \right|$ is equal to

Let $y=y(x)$ be the solution of the differential equation $e^{x} \sqrt{1-y^{2}} dx + \frac{y}{x} dy = 0$,with $y(1) = -1$. Then the value of $(y(3))^{2}$ is equal to:

The solution of the differential equation $\sin \left( \frac{dy}{dx} \right) = a$ with the initial condition $y(0) = 1$ is:

Let $\alpha |x| = |y| e^{xy-\beta}$,where $\alpha, \beta \in \mathbb{N}$,be the solution of the differential equation $x dy - y dx + xy(x dy + y dx) = 0$ with the initial condition $y(1) = 2$. Then $\alpha + \beta$ is equal to:

$a, b, c, d$ are real numbers. The general solution of $\frac{dy}{dx} = \frac{ax+b}{cy+d}$ represents a family of straight lines,when