the solution of dydx=yx1/3 is

dydx=y1/3x1/3⇒1y1/3dy=1x1/3dx integrating on both sides ∫1y1/3dy=∫1x1/3dx ⇒∫y -1/3dy=∫x-1/3 dx ⇒y23-x23=c

the solution of dydx=yx1/3 is

dydx=y1/3x1/3⇒1y1/3dy=1x1/3dx integrating on both sides ∫1y1/3dy=∫1x1/3dx ⇒∫y -1/3dy=∫x-1/3 dx ⇒y23-x23=c

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the solution of $\frac{dy}{dx} = {(\frac{y}{x})}^{1 / 3}$ is

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$x^{2 / 3} + y^{2 / 3} = c$

$y^{2 / 3} - x^{2 / 3} = c$

$x^{1 / 3} + y^{1 / 3} = c$

$y^{1 / 3} - x^{1 / 3} = c$

answer is B.

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Detailed Solution

$\frac{dy}{dx} = \frac{y^{1 / 3}}{x^{1 / 3}} \Rightarrow \frac{1}{y^{1 / 3}} dy = \frac{1}{x^{1 / 3}} dx i n t e g r a t i n g o n b o t h s i d e s \int \frac{1}{y^{1 / 3}} dy = \int \frac{1}{x^{1 / 3}} dx \Rightarrow \int y^{^{- 1 / 3}} d y = \int x^{- 1 / 3} dx \Rightarrow y^{\frac{2}{3}} - x^{\frac{2}{3}} = c$