The differential equation left( frac{d^2y}{dx | vedclass

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(D) The given differential equation is $\left( \frac{d^2y}{dx^2} \right)^2 - \left( \frac{dy}{dx} \right)^{1/2} = y^3$. To find the degree,we must fir

Vedclass · Accepted Answer

$4$

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(D) The given differential equation is $\left( \frac{d^2y}{dx^2} ight)^2 - \left( \frac{dy}{dx} ight)^{1/2} = y^3$. To find the degree,we must first express the equation as a polynomial in terms of its derivatives by removing fractional exponents. Rearranging the equation,we get $\left( \frac{d^2y}{dx^2} ight)^2 - y^3 = \left( \frac{dy}{dx} ight)^{1/2}$. Squaring both sides to eliminate the fractional exponent,we obtain $\left( \left( \frac{d^2y}{dx^2} ight)^2 - y^3 ight)^2 = \frac{dy}{dx}$. Expanding the left side,the highest order derivative present is $\frac{d^2y}{dx^2}$,which is of order $2$. The power to which the highest order derivative is raised in this polynomial form is $2 	imes 2 = 4$. Therefore,the degree of the differential equation is $4$.

The differential equation $\left( \frac{d^2y}{dx^2} \right)^2 - \left( \frac{dy}{dx} \right)^{1/2} = y^3$ has the degree

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