Question

solve the following equation if it is exact: (dy/dx)=(2xy-3x^2-2)/ (6y^2-x^3+3)

Answer 1

find the antiderivative?

Answer 2

I don't think this is exact, since if you arrange it in the appropriate form,\[(6y^2-x^3+3)dy=(2xy-3x^2-2)dx\]\[=(3x^2-2xy+2)dx+(6y^2-x^3+3)dy=0\]is exact if and only if\[\frac{\partial (3x^2-2xy+2)}{\partial y}=\frac{\partial (6y^2-x^3+3)}{\partial x}\]

Answer 3

it is a differential equation and I think it can be turned into a ratio-dependent differential equation and then into a separable differential equation however I don't know how to put it in a ratio dependent form

Answer 4

\[LHS = -2x \neq -3x^2 = RHS\]