Orthogonal Trajectories

y^3 - 3 = ln |Cx|

Please check if the answer I solved is correct.

My answer: y(3x^2 + c)=2

Engineering Mathematics

- I need guidance in designing a beam supporting specified ultimate moment of 1100 kN.m (doubly reinforced beam)
- I need guidance in solving the ultimate moment capacity (doubly reinforced beam)
- I need guidance in solving the balance steel area
- Please help me solve this problem using WSD (Working Stress Design) Method
- Physics: Uniform Motion
- Rescue at Sea
- Using two pumps
- Emptying a Tank
- Speed of a Plane
- Range of an Airplane

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 $\boxed{y(3x^2+C)=2}$