Solve the following differential equation:
$\displaystyle 3x^2y^2 + \cos (xy) - xy \sin (xy) + \frac{dy}{dx}\{2x^3y - x^2\sin (xy)\} = 0$
$\displaystyle \frac{dy}{dx}\{2x^3y - x^2\sin (xy)\} = xy \sin (xy) -3x^2y^2 - \cos (xy) +xy \sin (xy) $
$\displaystyle \frac{dy}{dx} = \frac{xy \sin (xy) -3x^2y^2 - \cos (xy) }{2x^3y - x^2\sin (xy)} $
$\displaystyle \int \frac{dy}{dx} dx = \int\frac{xy \sin (xy) -3x^2y^2 - \cos (xy) }{2x^3y - x^2\sin (xy)} dx$
$\displaystyle \int dy = \int\frac{xy \sin (xy) -3x^2y^2 - \cos (xy) }{2x^3y - x^2\sin (xy)} dx$
$\displaystyle y+c = \int\frac{xy \sin (xy) -3x^2y^2 - \cos (xy) }{2x^3y - x^2\sin (xy)} dx$
There's my lazy contribution!
Note that the differential equaiton is exact( you can check the partials) i.e there is a function $\displaystyle z(x,y)$ such that
$\displaystyle dz=\frac{\partial z}{\partial x}dx+\frac{\partial z}{\partial y}dy$
so
$\displaystyle \frac{\partial z}{\partial x} = 3x^2y^2 + \cos (xy) - xy \sin (xy)$
Now we take the partial integral with respect to x and get
$\displaystyle z=x^3y^2+\frac{\sin(xy)}{y}+x\cos(xy)-\frac{\sin(xy)}{y}+g(y)$
Now we take the partial with respect to y to get
$\displaystyle \frac{\partial z}{\partial y}=2x^3y-xy\sin(xy)+g'(y)$
Now we see that $\displaystyle g'(y)=0$
So $\displaystyle g(y)=c$
So we get an implict solution
$\displaystyle x^3y^2+x\cos(xy)=c$