The degree of the given differential equation | vedclass

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(D) To find the degree of a differential equation,it must first be expressed as a polynomial in terms of its derivatives. Given equation: $\left( \fra

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$6$

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(D) To find the degree of a differential equation,it must first be expressed as a polynomial in terms of its derivatives. Given equation: $\left( \frac{d^2y}{dx^2} \right)^3 = \left( 1 + \frac{dy}{dx} \right)^{1/2}$. To eliminate the fractional exponent,square both sides of the equation: $\left( \left( \frac{d^2y}{dx^2} \right)^3 \right)^2 = \left( \left( 1 + \frac{dy}{dx} \right)^{1/2} \right)^2$. This simplifies to: $\left( \frac{d^2y}{dx^2} \right)^6 = 1 + \frac{dy}{dx}$. The degree of a differential equation is the highest power of the highest order derivative present in the equation after it has been made free from radicals and fractions. The highest order derivative here is $\frac{d^2y}{dx^2}$,and its power is $6$. Therefore,the degree of the differential equation is $6$.

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

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