The order and degree of the differential equa | vedclass

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Question

(A) Given the differential equation: $y = px + \sqrt{a^{2}p^{2} + b^{2}}$,where $p = \frac{dy}{dx}$.Rearranging the terms,we get: $y - px = \sqrt{a^{2

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$1, 2$

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(A) Given the differential equation: $y = px + \sqrt{a^{2}p^{2} + b^{2}}$,where $p = \frac{dy}{dx}$.Rearranging the terms,we get: $y - px = \sqrt{a^{2}p^{2} + b^{2}}$.Squaring both sides:$(y - px)^{2} = a^{2}p^{2} + b^{2}$$y^{2} - 2pxy + p^{2}x^{2} = a^{2}p^{2} + b^{2}$Substituting $p = \frac{dy}{dx}$:$y^{2} - 2xy\left(\frac{dy}{dx}\right) + x^{2}\left(\frac{dy}{dx}\right)^{2} = a^{2}\left(\frac{dy}{dx}\right)^{2} + b^{2}$The highest derivative present is $\frac{dy}{dx}$,so the order is $1$.The highest power of the highest derivative is $2$,so the degree is $2$.Therefore,the order and degree are $1$ and $2$ respectively.

The order and degree of the differential equation $y=px+\sqrt{a^{2}p^{2}+b^{2}}$,where $p=\frac{dy}{dx}$,are respectively:

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