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 sum of the degree and order of the differential equation $\sqrt{\frac{d^2 y}{d x^2}}=\sqrt[5]{\frac{dy}{d x}-5}$ is

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

The order and the degree of the differential equation $\left( \frac{d^2s}{dt^2} \right)^2 + 3\left( \frac{ds}{dt} \right)^3 + 4 = 0$ are

The degree of the differential equation $y = a(1 - e^{-x/a})$,where $a$ is a parameter,is: