The order and degree of the differential equa | vedclass

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(A) Given differential equation is $\sqrt{1+\frac{1}{(\frac{dy}{dx})^2}} = (\frac{d^2y}{dx^2})^{3/2}$.Squaring both sides,we get $1+\frac{1}{(\frac{dy

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$2, 3$

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(A) Given differential equation is $\sqrt{1+\frac{1}{(\frac{dy}{dx})^2}} = (\frac{d^2y}{dx^2})^{3/2}$.Squaring both sides,we get $1+\frac{1}{(\frac{dy}{dx})^2} = (\frac{d^2y}{dx^2})^3$.Multiplying both sides by $(\frac{dy}{dx})^2$,we get $(\frac{dy}{dx})^2 + 1 = (\frac{d^2y}{dx^2})^3 (\frac{dy}{dx})^2$.The highest order derivative present is $\frac{d^2y}{dx^2}$,so the order is $2$.The power of the highest order derivative after making the equation free from radicals and fractions is $3$,so the degree is $3$.Thus,the order and degree are $2$ and $3$ respectively.

The order and degree of the differential equation $\sqrt{1+\frac{1}{(\frac{dy}{dx})^2}} = (\frac{d^2y}{dx^2})^{3/2}$ are respectively:

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