How comes that integrating factor = $1/(x\,y)$ I'm just curious how the integrating factor m(x,y) = $\frac{1}{x\,y}$ of $\frac{\mathrm{d}}{\mathrm{d}x}f(x,y) = x+y-\frac{x^2}{y}\,y'(x)$ was determined. It was just given. When I was introduced to this topic this week, my fist impression was that integrating factor just can depend on x or y. Because determinations like $m(x) = \exp(\int \alpha(x)\mathrm{d}x)$ whereas $\alpha(x) = \frac{\partial_y\partial_x(f(x,y))-\partial_x\partial_y(f(x,y))}{\partial_xf(x,y)}$ or such just seem to work this way. So loosely speaking: Are there other methods?
 A: You may have to guess, but in this case you can make it an educated guess.
If you multiply by $dx$ and thus rewrite the differential as
$M dx + N dy=(x+y)dx - (x^2/y)dy$
you see that if $x$ and $y$ were multiplied by a constant $\lambda$, you would get just the same differential multiplied by a constant:
$(\lambda x+\lambda y)d(\lambda x) - [(\lambda x)^2/\lambda y]d(\lambda y) = (\lambda^2)[(x+y)dx - (x^2/y)dy]$
A differential with this property is called homogeneous.
Homogeneous differentials tend to lend themselves to integrating factors having the form $x^ay^b$, which is itself homogeneous.  With that in mind let us try this type of integrating factor.  We then have the differential
$P dx +Q dy=x^ay^b(x+y)dx - x^ay^b(x^2/y)dy=(x^{a+1}y^b+x^ay^{b+1})dx - x^{a+2}y^{b-1}dy$
Applying the exactness criterion we have
$\partial P/\partial y = \partial Q/\partial x$
$bx^{a+1}y^{b-1}+(b+1)x^ay^b=-(a+2)x^{a+1}y^{b-1}$
Combining like terms reduces the exactness criterion to
$(a+b+2)x^{a+1}y^{b-1}+(b+1)x^ay^b=0$
So we must have $a+b+2=0,b+1=0$.  From this we get $b=-1$ and then $a=-1$, so we find the integrating factor $x^ay^b=1/(xy)$.
A: To add to what Oscar said, you don't necessarily have to guess the form of your integrating factor, but if you guess it right it will expedite your process. I'll lay out what I do when I'm trying to find an integrating factor.
To find the integrating factor for the expression
\begin{align}
f(x,y)+g(x,y)\frac{\mathrm{d}y}{\mathrm{d}x},
\end{align}
multiply by $m(x,y)$,
\begin{align}
m(x,y)f+m(x,y)g\frac{\mathrm{d}y}{\mathrm{d}x}
\end{align}
and assume that our equation is now a perfect differential of some function $\lambda$,
\begin{align}
mf+mg\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\mathrm{d}}{\mathrm{d}x}\lambda(x,y).
\end{align}
Writing out the chain rule for $\lambda$ gives
\begin{align}
mf+mg\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\partial\lambda}{\partial x}+\frac{\partial\lambda}{\partial y}\frac{\mathrm{d}y}{\mathrm{d}x}.
\end{align}
Giving two equations,
\begin{align}
mf=\frac{\partial\lambda}{\partial x},\\
mg=\frac{\partial\lambda}{\partial y}.
\end{align}
Assuming symmetry of second derivatives,
\begin{align}
\frac{\partial}{\partial y}mf&=\frac{\partial}{\partial x}mg\\
\frac{\partial m}{\partial y}f+m\frac{\partial f}{\partial y}&=\frac{\partial m}{\partial x}g+m\frac{\partial g}{\partial x}.
\end{align}
Rearranging gives the linear PDE
\begin{align}
f(x,y)\frac{\partial m}{\partial y}-g(x,y)\frac{\partial m}{\partial x}+\left(\frac{\partial f}{\partial y}-\frac{\partial g}{\partial x}\right)m(x,y)=0.
\end{align}
This PDE can be solved for the desired integrating factor, which in this specific case can be done via separation of variables.
