# Differential Equations - Elementary Application

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

### Rewriting the equation,

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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