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 $\boxed{y(3x^2+C)=2}$
More information about text formats
Follow @iMATHalino
MATHalino
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 $\boxed{y(3x^2+C)=2}$
Add new comment