Orthogonal Trajectories y^3 - 3 = ln |Cx|

Please check if the answer I solved is correct. My answer: y(3x^2 + c)=2

Rewriting the equation,

\begin{eqnarray*} y^3 - 3 &=& \ln C + \ln |x|\\ y^3 - 3 &=& C + \ln |x|\\ y^3 - \ln |x| &=& C\\ 3y^2 y' - \dfrac{1}{x} &=& 0\\ y' &=& \dfrac{1}{3xy^2}\\ y'_o &=& -3xy^2\\ \dfrac{dy}{y^2} &=& -3x dx\\ \dfrac{-1}{y} &=& \dfrac{-3x^2}{2} + C\\ \dfrac{1}{y} &=& \dfrac{3x^2}{2} + C\\ 2 &=& 3xy^2 + Cy \end{eqnarray*}

You are correct with $\boxed{y(3x^2+C)=2}$

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Rewriting the equation,

\begin{eqnarray*}

y^3 - 3 &=& \ln C + \ln |x|\\

y^3 - 3 &=& C + \ln |x|\\

y^3 - \ln |x| &=& C\\

3y^2 y' - \dfrac{1}{x} &=& 0\\

y' &=& \dfrac{1}{3xy^2}\\

y'_o &=& -3xy^2\\

\dfrac{dy}{y^2} &=& -3x dx\\

\dfrac{-1}{y} &=& \dfrac{-3x^2}{2} + C\\

\dfrac{1}{y} &=& \dfrac{3x^2}{2} + C\\

2 &=& 3xy^2 + Cy

\end{eqnarray*}

You are correct with $\boxed{y(3x^2+C)=2}$

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