Question

The degree of the differential equation $\sqrt{1+{\left(\frac{dy}{dx}\right)}^2}$ = x is ______________.

Answer 1

Given: $\sqrt{1+{\left(\frac{dy}{dx}\right)}^2}$ = x

To find the degree of the differential equation, the differential equation must be free from fractions and radicals.
Thus,
$$\begin{gathered} \sqrt{1+{\left(\frac{dy}{dx}\right)}^2}=x \\ \mathrm{Squaring}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \Rightarrow {\left(\sqrt{1+{\left(\frac{dy}{dx}\right)}^2}\right)}^2={\left(x\right)}^2 \\ \Rightarrow 1+{\left(\frac{dy}{dx}\right)}^2=x^2 \end{gathered}$$

Here, the power of the highest order derivative is 2.

Hence, the degree is 2.