The order and degree of the differential equa | vedclass

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Question

(B) Given equation: $\sqrt{1 + \frac{1}{(\frac{dy}{dx})^2}} = (\frac{d^2y}{dx^2})^{\frac{3}{2}}$Simplify the left side: $\sqrt{\frac{(\frac{dy}{dx})^2

Vedclass · Accepted Answer

$2, 3$

Vedclass · Answer

(B) Given equation: $\sqrt{1 + \frac{1}{(\frac{dy}{dx})^2}} = (\frac{d^2y}{dx^2})^{\frac{3}{2}}$Simplify the left side: $\sqrt{\frac{(\frac{dy}{dx})^2 + 1}{(\frac{dy}{dx})^2}} = (\frac{d^2y}{dx^2})^{\frac{3}{2}}$$\frac{\sqrt{(\frac{dy}{dx})^2 + 1}}{|\frac{dy}{dx}|} = (\frac{d^2y}{dx^2})^{\frac{3}{2}}$Square both sides to remove the fractional exponent: $\frac{(\frac{dy}{dx})^2 + 1}{(\frac{dy}{dx})^2} = (\frac{d^2y}{dx^2})^3$Multiply by $(\frac{dy}{dx})^2$: $(\frac{dy}{dx})^2 + 1 = (\frac{d^2y}{dx^2})^3 (\frac{dy}{dx})^2$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 a polynomial in derivatives 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 $\sqrt{1 + \frac{1}{(\frac{dy}{dx})^2}} = (\frac{d^2y}{dx^2})^{\frac{3}{2}}$ are, respectively:

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