Ordinary differential equations of the form $M(x,y)dx+N(x,y)dy=0$ question An ODE of the form $M(x,y)dx+N(x,y)dy=0$ is called "good" if $\frac{\partial (M(x,y))}{\partial y}=\frac{\partial (N(x,y))}{\partial x}$
We are given the differential equation $(3x^2y+2xy+y^3)dx+(x^2+y^2)dy=0$. This ODE is not "good". We are asked to find $\mu (x,y)$ such that:
$$\mu (x,y)(3x^2y+2xy+y^3)dx+\mu (x,y)(x^2+y^2)dy=0,  (*)$$ 
is "good".
What I did:
if the equation $(*)$ is good then $\mu_y (x,y) M(x,y)+\mu (x,y)M_y (x,y)=\mu_x (x,y)N(x,y)+\mu (x,y)N_x (x,y)$
so we get
$\mu_y(x,y)(3x^2y+2xy+y^3)+\mu(x,y)(3x^2+2x+3y^2)=\mu_x(x,y) (x^2+y^2)+2x
\mu(x,y)$
And now I'm stuck.
Even if we were to guess $\mu_x(x,y)=0$ or $\mu_y(x,y)=0$ we will never get something like $\frac{\mu_y}{\mu}=\phi(y)$ or $\frac{\mu_x}{\mu}=\psi(x)$. $\mu$ seems to depend on both variables and unless the above restrictions apply (which they don't here) I don't know how to find $\mu$. Please help.
 A: Notice that:
$$
\frac{M_y - N_x}{N} = \frac{(3x^2 + 2x + 3y^2) - (2x)}{x^2 + y^2} = 3
$$
which is independent of $y$ (and also $x$, by conincidence). This suggests that we guess that $\mu_y = 0$ so that $\mu$ is a function of $x$ only. Thus, we obtain:
$$
\mu M_y = \mu_x N + \mu N_x \iff \frac{d\mu}{dx} = \frac{M_y - N_x}{N} \cdot \mu = 3\mu
$$
This ODE is separable/linear and can be easily solved to obtain $\mu(x) = e^{3x}$.
A: Solution 1
Hint
Make the ansatz $\mu(x)$(the reason is explained in Adriano's answer)
$$(\mu (x)M(x,y))_y=(\mu (x)N(x,y))_x\iff (3x^2+2x+3y^2)\mu (x)=\mu (x)_x (x^2+y^2)+2x\mu (x)\iff \frac{\mu (x)_x}{\mu (x)}=3 \iff \mu (x)=e^{3x}$$
Solution 2
Hint
$$3x^2y+(x^2+y^2)y'+y^3+2xy=0\stackrel{y\colon =xv}{\iff} (x^2+x^2v^2)(xv'+v)+x^3v^3+3x^3v++2x^2v\;\stackrel{\text{simplify}}{\iff}\; v'=\frac{-v^3-xv^3-3v-3xv}{x(v^2+1)}\iff v'=-\frac{v(x+1)(v^2+3)}{x(v^2+1)}\iff \frac{v'(v^2+1)}{(v^2+3)v}=-\frac{x+1}{x}$$
Notation in case 1;
$X(a,b)_a\colon =\frac{\partial X(a,b)}{\partial a}$
Notation in case 2;
$(•)'\colon=\frac{\mathrm{d}•}{\mathrm{d}x}$
