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(B) To find the degree, we first express the differential equation as a polynomial in its derivatives by eliminating fractional exponents. Given: $x\l

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$5$

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(B) To find the degree, we first express the differential equation as a polynomial in its derivatives by eliminating fractional exponents. Given: $x\left(\frac{d^2 y}{d x^2}\right)^{\frac{1}{3}}+2 x^2\left(\frac{d^2 y}{d x^2}\right)^{\frac{5}{3}}+7 \frac{d y}{d x}+y=0$. Multiply the entire equation by $\left(\frac{d^2 y}{d x^2}\right)^{-\frac{1}{3}}$ or simply isolate the terms with fractional powers to clear them. Let $D = \frac{d^2 y}{d x^2}$. The equation is $x D^{1/3} + 2x^2 D^{5/3} + 7 \frac{dy}{dx} + y = 0$. Multiply by $D^{-1/3}$: $x + 2x^2 D^{4/3} + (7 \frac{dy}{dx} + y) D^{-1/3} = 0$. To clear the fractional powers, we raise the terms to appropriate powers. The highest power of the highest order derivative $\frac{d^2 y}{d x^2}$ after rationalizing the exponents is $5$. Specifically, multiplying by $D^{1/3}$ gives $x D^{2/3} + 2x^2 D^2 + (7 \frac{dy}{dx} + y) D^{1/3} = 0$. Raising to the power of $3$ to eliminate the remaining fractional powers results in a polynomial where the highest power of the highest order derivative $\frac{d^2 y}{d x^2}$ is $5$. Thus, the degree is $5$.

The degree of the differential equation $x\left(\frac{d^2 y}{d x^2}\right)^{\frac{1}{3}}+2 x^2\left(\frac{d^2 y}{d x^2}\right)^{\frac{5}{3}}+7 \frac{d y}{d x}+y=0$ is:

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