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(C) Given the differential equation $y = \frac{dp}{dx} + \sqrt{a^2 p^2 - b^2}$,where $p = \frac{dy}{dx}$.Substituting $p = \frac{dy}{dx}$,we get $\fra

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$4$

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(C) Given the differential equation $y = \frac{dp}{dx} + \sqrt{a^2 p^2 - b^2}$,where $p = \frac{dy}{dx}$.Substituting $p = \frac{dy}{dx}$,we get $\frac{dp}{dx} = \frac{d}{dx}(\frac{dy}{dx}) = \frac{d^2y}{dx^2}$.Thus,the equation becomes $y = \frac{d^2y}{dx^2} + \sqrt{a^2(\frac{dy}{dx})^2 - b^2}$.Rearranging the terms,we have $y - \frac{d^2y}{dx^2} = \sqrt{a^2(\frac{dy}{dx})^2 - b^2}$.Squaring both sides to eliminate the square root,we get $(y - \frac{d^2y}{dx^2})^2 = a^2(\frac{dy}{dx})^2 - b^2$.Expanding the left side,$y^2 + (\frac{d^2y}{dx^2})^2 - 2y(\frac{d^2y}{dx^2}) = a^2(\frac{dy}{dx})^2 - b^2$.The highest order derivative present is $\frac{d^2y}{dx^2}$,so the order $m = 2$.The highest power of the highest order derivative is $2$,so the degree $n = 2$.Therefore,$m + n = 2 + 2 = 4$.

If $m$ is the order and $n$ is the degree of the differential equation $y = \frac{dp}{dx} + \sqrt{a^2 p^2 - b^2}$,where $p = \frac{dy}{dx}$,then the value of $m+n$ is

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