Question

$\sqrt[3]{\frac{d^{2}y}{d{x}^2}}=\sqrt{\frac{dy}{dx}}$

Answer 1

$$\begin{gathered} \sqrt[3]{\frac{d^{2}y}{d{x}^2}}=\sqrt{\frac{dy}{dx}} \\ \Rightarrow {\left(\frac{d^{2}y}{d{x}^2}\right)}^{\frac{1}{3}}={\left(\frac{dy}{dx}\right)}^{\frac{1}{2}} \\ \mathrm{Taking}\mathrm{cubes}\mathrm{of}\mathrm{both}\mathrm{the}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \Rightarrow \frac{d^{2}y}{d{x}^2}={\left(\frac{dy}{dx}\right)}^{\frac{3}{2}} \\ \mathrm{Squaring}\mathrm{both}\mathrm{the}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \Rightarrow {\left(\frac{d^{2}y}{d{x}^2}\right)}^2={\left(\frac{dy}{dx}\right)}^3 \\ \Rightarrow {\left(\frac{d^{2}y}{d{x}^2}\right)}^2-{\left(\frac{dy}{dx}\right)}^3=0 \end{gathered}$$

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.

Thus, it is a non-linear differential equation, as its degree is 2, which is greater than 1.