What is the differential equation of the orthogonal trajectories of the family of curves x2−3xy−2y2=C?

Given family of Curve is $x^2-3xy-2y^2=C$
Since $x^2-3xy-2y^2=c$
Diff. both side with respect to x, we get
$\begin{array}{}\frac{d}{dx}[x^2-3xy-2y^2]=\frac{d}{dx}[c]\\ \Rightarrow 2x-3(x\frac{dy}{dx}+y)-4y\frac{dy}{dx}=0\\ \Rightarrow 2x-3x\frac{dy}{dx}-3y-4y\frac{dy}{dx}=0\\ \Rightarrow \frac{dy}{dx}(-3x-4y)=3y-2x\\ put\frac{dy}{dx}=-\frac{dx}{dy}\\ -\frac{dx}{dy}(-3x-4y)=3y-2x\\ \Rightarrow (3x+4y)dx=(3y-2x)dy\\ \Rightarrow (3x+4y)dx+(2x-3y)dy=0\end{array}$
Hence $(3x+4y)dx+(2x-3y)dy=0$ is the differential equation of the orthogonal trajectories of family of given curve.