The degree of the differential equation $\frac{d^2y}{dx^2} + \left( \frac{dy}{dx} \right)^3 + 6y = 0$ is

Assertion $(A)$: The order of the differential equation of a family of circles with a constant radius is $2$.
Reason $(R)$: An algebraic equation having two arbitrary constants is the general solution of a second-order differential equation.

The order and degree of the differential equation $\left(1+\frac{dy}{dx}\right)^{\frac{1}{3}}=\sqrt{\frac{d^2y}{dx^2}}$ are respectively.

If the order and degree of the differential equation $x \frac{d^2 y}{d x^2} = \left(1 + \left(\frac{d^2 y}{d x^2}\right)^2\right)^{-1/2}$ are $k$ and $l$ respectively,then $k, l$ are the roots of

Consider the following differential equations.
$D_1: y=4 \frac{dy}{dx}+3x \frac{dx}{dy}$
$D_2: \frac{d^2y}{dx^2}=\left(3+\left(\frac{dy}{dx}\right)^2\right)^{\frac{4}{3}}$
$D_3: \left[1+\left(\frac{dy}{dx}\right)\right]^2=\left(\frac{dy}{dx}\right)^2$
The ratio of the sum of the orders of $D_1, D_2$ and $D_3$ to the sum of their degrees is

The order of the differential equation $y = c_{1} e^{c_{2}+x} + c_{3} e^{c_{4}+x}$ is