<s>[INST] Orthogonal trajectories for a given family of curves Find the orthogonal trajectories for the given family of curves:
$$y^2=Cx^3-2$$
Derivative with respect to x, and finding the value of C:
$$2yy' = C3x^2 $$
$$C=\frac{y^2+2}{x^3}$$
Replacing C, and solving for y':
$$2yy' = \frac{3y^2+6}{x}$$
$$y'=\frac{3y^2+6}{2xy}$$
Finding the orthogonal and separating terms:
$$y' = -\frac{2xy}{3y^2+6}$$
$$\frac{(3y^2+6)}{y}y'=-2x$$
$$\frac{(3y^2+6)}{y}dy=-2xdx$$
Integrating:
$$\int{\frac{(3y^2+6)}{y}dy}=\int{-2xdx}$$
$$\frac{3y^2}{2}+6log(y)=-x^2+C$$
$$C=\frac{3y^2}{2}+6log(y)+x^2$$
Was I close? We weren't given solutions, so I'd appreciate a check on my answer. :)
 [/INST] That's what I got
$$
y^2=Cx^3-2 \\
2yy'= 3Cx^2 \\
y'=\frac{3Cx^2}{2y}
$$
Orthogonal curve
$$
y'=-\frac {2y}{3Cx^2} \\
\frac {y'}{y} = -\frac 2{3Cx^2} \\
\ln |y| = \frac 2{3Cx} \\
|y| = \exp \left (\frac 2{3Cx} \right)
$$
and here some plots

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