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TopNewestFor the differential equation \[ \frac{dy}{dx} \; = \; \frac{y(x+y^3)}{x(y^3-x)} \] try the substitution \(y \,=\, x^{\frac13}u\), so that \[ \begin{array}{rcl}\displaystyle x^{\frac13}\frac{du}{dx} + \tfrac13x^{-\frac23}u & = & \displaystyle\frac{x^{\frac13}u(x + xu^3)}{x(xu^3-x)} \; = \; x^{-\frac23}\frac{u(u^3+1)}{u^3-1} \\ xu' + \tfrac13u & = & \displaystyle\frac{u(u^3+1)}{u^3-1} \\ xu' & = & \displaystyle\frac{2u(u^3+2)}{3(u^3-1)} \\ \displaystyle\int \Big(\frac{3u^2}{u^3+2} - \frac{1}{u}\Big)\,du \; = \; \int \frac{2(u^3-1)}{u(u^3+2)}\,du & = & \displaystyle\int \frac{4}{3x}\,dx \; = \; \tfrac43\ln x + c \\ \displaystyle \ln \Big(\frac{u^3+2}{u}\Big) & = & \tfrac43\ln x + c \\ [2ex] \displaystyle \frac{u^3+2}{u} & = & Ax^{\frac43} \\ u^3 + 2 & = & Aux^{\frac43} \\ xu^3 + 2x & = & Aux^{\frac73} \\ y^3 + 2x & = & Ax^2y \end{array} \]

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thanku sir

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I think the answer is x=(-y^3)

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i think tis is nt the answer ,,,the answer is given in the book..bt no steps..@shiva raj

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