Answer
\[2x^2+3y^2=K\]
Work Step by Step
$y^2=cx^3$ _____(1)
Differentiate (1) with respect to $x$
\[2y\frac{dy}{dx}=3cx^2\]
From (1)
\[2y\frac{dy}{dx}=3\left(\frac{y^2}{x^3}\right)x^2=\frac{3y^2}{x}\]
$2\Large\frac{dy}{dx}=\Large\frac{3y}{x}$ ___(2)
Replace $\Large\frac{dy}{dx}$ by $-\Large\frac{dx}{dy}$ in (2) [For orthogonal trajectories]
\[-2\frac{dx}{dy}=\frac{3y}{x}\]
Separating variables,
\[-2x dx=3y dy\]
Integrating,
\[C_{1}-2\int x dx=3\int y dy\]
\[C_{1}-\frac{2x^2}{2}=\frac{3y^2}{2}\]
\[2x^2+3y^2=K\]
Where $K=2C_{1}$
Hence orthogonal trajectories of (1) is $2x^2+3y^2=K$
