The order and degree of the differential equa | vedclass

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(A) Given differential equation is $y = x \frac{dp}{dx} + \sqrt{a^{2} p^{2} + b^{2}}$,where $p = \frac{dy}{dx}$.Substituting $p = \frac{dy}{dx}$,we ge

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$2, 2$

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(A) Given differential equation is $y = x \frac{dp}{dx} + \sqrt{a^{2} p^{2} + b^{2}}$,where $p = \frac{dy}{dx}$.Substituting $p = \frac{dy}{dx}$,we get $y = x \frac{d}{dx} \left( \frac{dy}{dx} \right) + \sqrt{a^{2} \left( \frac{dy}{dx} \right)^{2} + b^{2}}$.This simplifies to $y = x \frac{d^{2}y}{dx^{2}} + \sqrt{a^{2} \left( \frac{dy}{dx} \right)^{2} + b^{2}}$.Rearranging the terms,we have $y - x \frac{d^{2}y}{dx^{2}} = \sqrt{a^{2} \left( \frac{dy}{dx} \right)^{2} + b^{2}}$.Squaring both sides,we get $\left( y - x \frac{d^{2}y}{dx^{2}} \right)^{2} = a^{2} \left( \frac{dy}{dx} \right)^{2} + b^{2}$.The highest order derivative present is $\frac{d^{2}y}{dx^{2}}$,so the order is $2$.The power of the highest order derivative after rationalizing is $2$,so the degree is $2$.Thus,the order and degree are $2$ and $2$ respectively.

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

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