The degree of the differential equation (1 + | vedclass

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(D) To find the degree,we must first express the differential equation as a polynomial in terms of its derivatives.Given equation: $(1 + \frac{dy}{dx}

Vedclass · Accepted Answer

$1$

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(D) To find the degree,we must first express the differential equation as a polynomial in terms of its derivatives.Given equation: $(1 + \frac{dy}{dx})^2 = (\frac{d^3y}{dx^3})^{1/3}$.Raise both sides to the power of $3$ to eliminate the fractional exponent: $((1 + \frac{dy}{dx})^2)^3 = ((\frac{d^3y}{dx^3})^{1/3})^3$.This simplifies to: $(1 + \frac{dy}{dx})^6 = \frac{d^3y}{dx^3}$.The highest order derivative present is $\frac{d^3y}{dx^3}$,which is of order $3$.The degree of a differential equation is the power of the highest order derivative when the equation is written as a polynomial in its derivatives.Here,the power of $\frac{d^3y}{dx^3}$ is $1$. Therefore,the degree is $1$.

The degree of the differential equation $(1 + \frac{dy}{dx})^2 = (\frac{d^3y}{dx^3})^{1/3}$ is . . . . . . .

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