The order and degree of the differential equation \(y=\frac{d p}{d x} x=\sqrt{a^{2} p^{2}+b^{2}}, \quad\) where \(\mathrm{p}=\frac{\mathrm{dy}}{\mathrm{dx}}\) (here \(\mathrm{a}\) and \(\mathrm{b}\) are arbitrary constants) respectively are

Given differential equation is
\(\quad \mathrm{y}=\left(\frac{\mathrm{dp}}{\mathrm{dx}}\right) \mathrm{x}+\sqrt{\mathrm{a}^{2} \mathrm{p}^{2}+\mathrm{b}^{2}},\left(\mathrm{p}=\frac{\mathrm{dy}}{\mathrm{dx}}\right)\) \(\Rightarrow \quad \mathrm{y}=\left[\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)\right] \cdot \mathrm{x}+\sqrt{\mathrm{a}^{2}\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^{2}+\mathrm{b}^{2}}\) \(\Rightarrow \quad\left(\mathrm{y}-\mathrm{x} \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}\right)=\sqrt{\mathrm{a}^{2}\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^{2}+\mathrm{b}^{2}}\) \(\Rightarrow \quad\left(\mathrm{y}-\mathrm{x} \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}}\right)^{2}=\mathrm{a}^{2}\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^{2}+\mathrm{b}^{2}\)
Hence, order \(=2\) and degree \(=2\)