The order and degree of the differential equation $y=p x+\sqrt{a^{2} \mathbf{p}^{2}+b^{2}}$, where…
The order and degree of the differential equation $y=p x+\sqrt{a^{2} \mathbf{p}^{2}+b^{2}}$, where $\mathrm{p}=\frac{\mathrm{d} y}{\mathrm{~d} x}$ are respectively
$1, 2$
$3,1$
2,1
$1,3$
Solution
Given
$\begin{array}{r}
y=p x+\sqrt{a^{2} p^{2}+b^{2}} \\
\therefore y-p x=\sqrt{a^{2} p^{2}+b^{2}}
\end{array}$
On squaring both side we get
$y^{2}-2 p x y+p^{2} x^{2}=a^{2} p^{2}+b^{2}$
$\mathrm{y}^{2}-2 \mathrm{xy} \frac{\mathrm{dy}}{\mathrm{dx}}+\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^{2} \mathrm{x}^{2}=\mathrm{a}^{2}\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^{2}+\mathrm{b}^{2} \quad \ldots\left[\mathrm{p}=\frac{\mathrm{dy}}{\mathrm{dx}}, \text { given }\right]$
Thus order is 1 and degree is 2.