Solve this differential equation $$(xy^3 + y)dx + 2(x^2y^2 + x + y^4)dy = 0$$
I tried converting it to the form: $\frac{dy}{dx} + yp(x) = q(x)$ but couldn't. The equation is also not homogeneous. Keeping $\frac{dy}{dx}$ on one side will not render the numerator as the derivative of the denominator (with some manipulation) on the other side of the equation.
By multiplying both sides by $y$ (see Moo's comment) we have that $$0=y(xy^3 + y)dx + 2y(x^2y^2 + x + y^4)dy=d\left(\frac{y^2(2y^4+3x^2y^2+6x)}{6}\right).$$ Hence $$y^2(2y^4+3x^2y^2+6x)=C$$ where $C$ is an arbitrary constant. Have you any initial condition?