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