For the differential equation sqrt{frac{d^2 y | vedclass

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Question

(D) Given the differential equation: $\sqrt{\frac{d^2 y}{d x^2}}=\sqrt[3]{\left[y \frac{d y}{d x}+x \sin \left(\frac{d y}{d x}\right)\right]^2}$ Raisi

Vedclass · Accepted Answer

Order is $2$ and degree is not defined

Vedclass · Answer

(D) Given the differential equation: $\sqrt{\frac{d^2 y}{d x^2}}=\sqrt[3]{\left[y \frac{d y}{d x}+x \sin \left(\frac{d y}{d x}ight)ight]^2}$ Raising both sides to the power of $6$ to remove the radicals: $\left(\frac{d^2 y}{d x^2}ight)^{\frac{6}{2}} = \left[y \frac{d y}{d x}+x \sin \left(\frac{d y}{d x}ight)ight]^{\frac{2 	imes 6}{3}}$ $\Rightarrow \left(\frac{d^2 y}{d x^2}ight)^3 = \left[y \frac{d y}{d x}+x \sin \left(\frac{d y}{d x}ight)ight]^4$ The order of a differential equation is the highest derivative present, which is $\frac{d^2 y}{d x^2}$, so the order is $2$. The degree of a differential equation is the power of the highest derivative when the equation is expressed as a polynomial in derivatives. Since the term $\sin \left(\frac{d y}{d x}ight)$ is a transcendental function of the derivative, the equation cannot be expressed as a polynomial in derivatives. Therefore, the degree is not defined.

For the differential equation $\sqrt{\frac{d^2 y}{d x^2}}=\sqrt[3]{\left[y \frac{d y}{d x}+x \sin \left(\frac{d y}{d x}\right)\right]^2}$, find the order and degree.

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