Question 1165588
<pre>
{{{2xy^3  -  3y  -  (3x + alpha*x^2y^2  -  2*alpha*y)*"y'" = 0}}}

{{{2xy^3  -  3y  +  (-3x - alpha*x^2y^2  +  2*alpha*y)*"y'" = 0}}}

{{{2xy^3  -  3y  -  (-3x - alpha*x^2y^2  +  2*alpha*y)*expr(dy/dx) = 0}}}

{{{(2xy^3  -  3y)dx  +  (-3x - alpha*x^2y^2  +  2*alpha*y)dy = 0}}}

M(x,y)=2xy³-3y 
N(x,y)=-3x-αx²y²+2αy

∂M/∂y = 6xy²-3
∂N/∂x = -3-2αxy²

These will be equal and the differential equation will be exact iff:

-2α = 6
  α = -3

I guess you didn't want to solve the exact differential equation.

2α-α² = 2(-3)-(-3)² = -6-(9) = -15

Edwin</pre>