Question

$\sqrt{1+{\left(\frac{dy}{dx}\right)}^2}={\left(c\frac{d^{2}y}{d{x}^2}\right)}^{1/3}$

Answer 1

$$\begin{gathered} \sqrt{1+{\left(\frac{dy}{dx}\right)}^2}={\left(c\frac{d^{2}y}{d{x}^2}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.} \\ \mathrm{Squaring}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \Rightarrow 1+{\left(\frac{dy}{dx}\right)}^2={\left(c\frac{d^{2}y}{d{x}^2}\right)}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.} \\ \mathrm{Taking}\mathrm{cubes}\mathrm{of}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \Rightarrow {\left(c\frac{d^{2}y}{d{x}^2}\right)}^2={\left[1+{\left(\frac{dy}{dx}\right)}^2\right]}^3 \\ \Rightarrow c^2{\left(\frac{d^{2}y}{d{x}^2}\right)}^2=1+3{\left(\frac{dy}{dx}\right)}^2+3{\left(\frac{dy}{dx}\right)}^4+{\left(\frac{dy}{dx}\right)}^6 \end{gathered}$$

In this differential equation, the order of the highest order derivative is 2 and its power is 2. So, it is a differential equation of order 2 and degree 2.

It is a non-linear differential equation, as its degree is more than 1.