If $\mathrm{m}$ is order and $\mathrm{n}$ is degree of the differential equation…
If $\mathrm{m}$ is order and $\mathrm{n}$ is degree of the differential equation $\mathrm{y}=\frac{\mathrm{dp}}{\mathrm{dx}}+\sqrt{\mathrm{a}^2 \mathrm{p}^2-\mathrm{b}^2}$, where $\mathrm{p}=\frac{\mathrm{dp}}{\mathrm{dx}}$, then the value of $\mathrm{m}+\mathrm{n}$ is
2
3
4
5
Solution
$\begin{aligned}
& y=\frac{d p}{d x}+\sqrt{a^2 p^2-b^2}=\frac{d}{d x}\left(\frac{d y}{d x}\right)+\sqrt{a^2\left(\frac{d y}{d x}\right)^2-b^2} \\
& \therefore y-\frac{d^2 y}{d x^2}=\sqrt{a^2\left(\frac{d y}{d x}\right)^2-b^2}
\end{aligned}$
Squaring both sides, we write
$\begin{aligned}
& y^2+\left(\frac{d^2 y}{d x}\right)^2-2 y \frac{d^2 y}{d x^2}=a^2\left(\frac{d y}{d x}\right)^2-b^2 \\
& \therefore \text { order }=2 \text { and degree }=2
\end{aligned}$