Answer
\[5y^2+x^2=K\]
Work Step by Step
Given that, $y=cx^5$ _____(1)
Differentiating both side with respect to $x$
\[\frac{dy}{dx}=5cx^4\]
From (1)
$\frac{dy}{dx}=5(\frac{y}{x^5})x^4=\frac{5y}{x}$ ____(2)
Replace $\frac{dy}{dx}$ by $\frac{-dx}{dy}$ in (2) [ For orthogonal trajectories]
$\frac{-dx}{dy}=\frac{5y}{x}$
$-xdx = 5ydy$
Integrating, $-\int xdx= \int 5ydy$
$k-\frac{x^2}{2}=\frac{5y^2}{2}$
$5y^2+x^2=K$, where $K = 2k$
Hence family of orthogonal trajectries to (1) is $5y^2+x^2=K$.
