Question

The degree of the differential equation $\sqrt{1+\frac{d^{2}y}{d{x}^2}}=\frac{dy}{dx}+x$ is ______________.

Answer 1

Given: $\sqrt{1+\frac{d^{2}y}{d{x}^2}}=\frac{dy}{dx}+x$

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

Here, the power of the highest order derivative is 1.

Hence, the degree is 1.