$\displaystyle 3x^2y+2x^2-y^3-2y^2=\alpha{} \\[2ex] \textit{Differentiating this w.r.t x}\\[2ex] \displaystyle 3\left(2xy+x^2y^{'}\right)+4x-3y^2y^{'}-4y\ y^{'}=0 \\[2ex] \textit{To find the orthogonal trajectory, replacing }\\[2ex] \displaystyle y^{'}by-\frac{1}{y'} \\[2ex] \displaystyle 3\left(2xy-\frac{x^2}{y'}\right)+4x+\frac{3y^2}{y'}+\frac{4y}{y'}=0 \\[4ex] $

$\textit{Multiplying throughout by y'}\\[4ex] \displaystyle 3\left(2xyy^{'}-x^2\right)+4xy^{'}+3y^2+4y=0\ \\[2ex] \displaystyle 6xyy^{'}+4xy^{'}-3x^2+3y^2+4y=0 \\[2ex] \displaystyle \left(6xy+4x\right)\ \frac{dy}{dx}\ -3x^2+3y^2+4y=0 \\[2ex] \displaystyle \left(6xy+4x\right)\ dy+\left(-3x^2+3y^2+4y\right)dx=0 \\[2ex] $

$\textit{Hence the Differential equation is …