How to solve exact equations by integrating factors? I know how to solve an exact equation like 
$$M(x,y) + N(x,y)y=0 $$
We just check $$\frac{\partial M}{\partial y} =\frac{\partial N}{\partial x} $$
If so, then it's just a little bit of algebra, taking anti-derivative and solve, such that $f = f(x,y) = c$, where we can obtain $f$ by just taking integal w.r.t. $x$ etc.
But now I'm stuck with integrating factors. I can do it if the integrating factor $\mu=\mu(x)$ depends only on one variable. But now I want to do another exercise, which uses a function of the form $\mu(x+y)$.
How can I solve $$(7x^3+3x^2y+4y) +(4x^3+x+5y)y'=0 $$ I'm given a hint that I should use a function of the form $\mu(x+y)$.
I guess I should just multiply by $(x+y)^m$ for an arbitrary $m$, but I don't know how to do the algebra after I multiplied...
I guess something like this  :
$$\frac{\partial (M \mu)}{\partial y}= m(x+y)^{m-1}(7x^3+3x^2y+4y) + (x+y)^m(3x^2+4) =$$ 
$$\frac{\partial (N\mu)}{\partial x} = m(x+y)^{m-1}(4x^3+x+5y) + (x+y)^m(12x^2+1) $$
How solve for $\mu$ ?
 A: Your integrating factor is
$$
\mu(x+y)=(x+y)^3.
$$
To find it you use the usual strategy expressing
$$
\mu_yM-\mu_xN=\mu(N_x-M_y).\tag{1}
$$
After some simplification you can find that
$$
\frac{N_x-M_y}{M-N}=\frac{3}{x+y}.
$$
Hence, if $\mu(x,y)=\mu(x+y)$ $(1)$ simplifies to
$$
\mu'_w=\frac{3\mu}{w},\quad w=x+y,
$$
from which you find your integrating factor.
A: Hint: Let you have an OE like $$M(x,y)dx+N(x,y)dy=0$$ and let $\mu=\mu(z(x,y))$ be an integrating factor for it. If $$\frac{M_y-N_x}{Nz_x-Mz_y}$$ be a function respect to $z$ then $$\mu=\exp\left(\int\frac{M_y-N_x}{Nz_x-Mz_y}dz\right)$$. Here $z=x+y$, so $z_x=1,z_y=1$ and therefore your integrating factor is $\mu=\exp\left(\int\frac{M_y-N_x}{N-M}dz\right)$. You can easily find $\mu$ by doing $$(\mu M)_y=(\mu N)_x$$ as well. For thsi problem $$\mu=\exp\left(\int\frac{3x^2+4-12x^2-1}{4x^3+x+5y-7x^3-3x^2y-4y}dz\right)=\exp\left(\int\frac{3(1-3x^2)}{(x+y)(1-3x^2)}dz\right)$$ which if $3x^2-1\neq0$ then $$\mu=\exp\left(\int\frac{3}{x+y}dz\right)=\exp\left(\int\frac{3}{z}dz\right)$$ I think you can do the rest.
