The order and the degree of the differential | vedclass

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(B) Given the differential equation: $y = px + \sqrt{a^2p^2 + b^2}$,where $p = \frac{dy}{dx}$. To find the order and degree,we first eliminate the squ

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$1, 2$

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(B) Given the differential equation: $y = px + \sqrt{a^2p^2 + b^2}$,where $p = \frac{dy}{dx}$. To find the order and degree,we first eliminate the square root by squaring both sides: $\sqrt{a^2p^2 + b^2} = y - px$ Squaring both sides: $a^2p^2 + b^2 = (y - px)^2$ $a^2p^2 + b^2 = y^2 + p^2x^2 - 2xyp$ Rearranging the terms: $(x^2 - a^2)p^2 - 2xyp + (y^2 - b^2) = 0$ Substituting $p = \frac{dy}{dx}$: $(x^2 - a^2)\left(\frac{dy}{dx}\right)^2 - 2xy\left(\frac{dy}{dx}\right) + (y^2 - b^2) = 0$ The highest order derivative present is $\frac{dy}{dx}$,so the order is $1$. The highest power of the highest order derivative is $2$,so the degree is $2$. Thus,the order and degree are $1$ and $2$ respectively.

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

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