The solution of frac{dy}{dx} = (frac{x}{y})^{ | vedclass

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(B) Given differential equation is $\frac{dy}{dx} = (\frac{x}{y})^{-1/3}$.By simplifying the right side,we get $\frac{dy}{dx} = (\frac{y}{x})^{1/3} =

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

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(B) Given differential equation is $\frac{dy}{dx} = (\frac{x}{y})^{-1/3}$.By simplifying the right side,we get $\frac{dy}{dx} = (\frac{y}{x})^{1/3} = \frac{y^{1/3}}{x^{1/3}}$.Separating the variables,we have $y^{-1/3} dy = x^{-1/3} dx$.Integrating both sides,we get $\int y^{-1/3} dy = \int x^{-1/3} dx$.Applying the power rule for integration $\int x^n dx = \frac{x^{n+1}}{n+1} + C$,we get $\frac{y^{2/3}}{2/3} = \frac{x^{2/3}}{2/3} + C'$.Multiplying by $\frac{2}{3}$,we get $y^{2/3} = x^{2/3} + \frac{2}{3}C'$.Let $c = \frac{2}{3}C'$,then $y^{2/3} - x^{2/3} = c$.

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

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