Question

(Xy^2 -x^2 )dx + ( 3x^2 y^2 + x^2 y-2x^3 + y^2) dy =0

Answer 1

$(xy^2-x^2)dx+(3x^2y^2+x^2y-2x^3+y^2)dy=0\\ P(x,y)=xy^2-x^2\\ Q(x,y)=3x^2y^2+x^2y-2x^3+y^2\\ \frac{\partial P}{\partial y}=2xy\\ \frac{\partial Q}{\partial x}=6xy^2+2xy-6x^2\\ \frac{\partial P}{\partial y}\neq\frac{\partial Q}{\partial x}\\ \Psi(y)=\frac{\frac{\partial P}{\partial y}-\frac{\partial Q}{\partial x}}{-P}=\\ =\frac{2xy-(6xy^2+2xy-6x^2)}{-(xy^2-x^2)}=\frac{-6xy^2+6x^2}{x^2-xy^2}=\frac{6(xy^2-x^2)}{xy^2-x^2}=6\\ \mu(y)=e^{\int\Psi(y)dy}=e^{\int6dy}=e^{6y}$

we multiply the equation by $e^{6y}$

$(xy^2-x^2)e^{6y}dx+\\ +(3x^2y^2+x^2y-2x^3+y^2)e^{6y}dy=0\\ P_1(x,y)=(xy^2-x^2)e^{6y}\\ Q_1(x,y)=(3x^2y^2+x^2y-2x^3+y^2)e^{6y}\\ \frac{\partial P_1}{\partial y}=6e^{6y}(xy^2-x^2)+e^{6y}(2xy)=\\ =e^{6y}(6xy^2+2xy-6x^2)\\ \frac{\partial Q_1}{\partial x}=e^{6y}(6xy^2+2xy-6x^2)\\ \frac{\partial P_1}{\partial y}=\frac{\partial Q_1}{\partial x}$

Then there is a function $u(x,y)$ such that

$\frac{\partial u}{\partial x}=(xy^2-x^2)e^{6y}\\ \frac{\partial u}{\partial y}=(3x^2y^2+x^2y-2x^3+y^2)e^{6y}\\ u(x,y)=\int(xy^2-x^2)e^{6y}dx=\\ =(\frac{x^2}{2}y^2-\frac{x^3}{3})e^{6y}+\phi(y)\\ \frac{\partial{u(x,y)}}{\partial{y}}=6e^{6y}(\frac{x^2y^2}{2}-\frac{x^3}{3})+yx^2e^{6y}+\phi'(y)=\\ =e^{6y}(3x^2y^2-2x^3+yx^2)+\phi'(y)\\ e^{6y}(3x^2y^2-2x^3+yx^2)+\phi'(y)=\\ =e^{6y}(3x^2y^2+x^2y-2x^3+y^2)\\ \phi'(y)=y^2e^{6y}\\ \phi(y)=y^2\frac{e^{6y}}{6}-y\frac{e^{6y}}{18}+\frac{e^{6y}}{108}$

 solutions of the equation

$(\frac{x^2}{2}y^2-\frac{x^3}{3})e^{6y}+y^2\frac{e^{6y}}{6}-y\frac{e^{6y}}{18}+\frac{e^{6y}}{108}=C$