Question

Write the order and degree of the differential equation $y=x\frac{dy}{dx}+a\sqrt{1+{\left(\frac{dy}{dx}\right)}^2}$.

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ y=x\frac{dy}{dx}+a\sqrt{1+{\left(\frac{dy}{dx}\right)}^2} \\ y-x\frac{dy}{dx}=a\sqrt{1+{\left(\frac{dy}{dx}\right)}^2} \\ \mathrm{Squaring}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ {\left(y-x\frac{dy}{dx}\right)}^2={\left\{a\sqrt{1+{\left(\frac{dy}{dx}\right)}^2}\right\}}^2 \\ {y}^2+x^2{\left(\frac{dy}{dx}\right)}^2-2xy\frac{dy}{dx}=a^2\left\{1+{\left(\frac{dy}{dx}\right)}^2\right\} \\ {y}^2+{\left(\frac{dy}{dx}\right)}^2\left\{x^2-a^2\right\}-2xy\frac{dy}{dx}=a^2 \\ \mathrm{From}\mathrm{the}\mathrm{above}\mathrm{equation},\mathrm{we}\mathrm{see}\mathrm{that}\mathrm{the}\mathrm{highest}\mathrm{order}\mathrm{is}1. \\ \mathrm{So},\mathrm{its}\mathrm{order}\mathrm{is}1\mathrm{and}\mathrm{the}\mathrm{power}\mathrm{of}\mathrm{the}\mathrm{highest}\mathrm{order}\mathrm{derivative}\mathrm{is}2. \\ \mathrm{Thus},\mathrm{it}\mathrm{is}\mathrm{a}\mathrm{differential}\mathrm{equation}\mathrm{of}\mathrm{order}1\mathrm{and}\mathrm{degree}2. \end{gathered}$$