How to solve this ODE (integration factor?) Im trying to solve the following ODE:
$(x+y+1) dx + (2x +2y -1) dy = 0$
In the theory of my book these presented with the form
$P(x,y) dx + Q(x,y) dy = 0$
So for my example we have
$P(x,y) = x +y +1 , \, \, \, Q(x,y) = 2x + 2y -1$
Thus we notice that
$\dfrac{\partial}{\partial y}P(x,y) = 1 \neq \dfrac{\partial}{\partial x}Q(x,y) = 2$
So the ODE is not exact. Then I would try to use an integrate factor if the following expresion depends only on $x$
$\dfrac{1}{P(t,x)} \left( \dfrac{\partial}{\partial y} P(x,y) - \dfrac{\partial}{\partial x} Q(x,y)\right) = - \dfrac{1}{x+y+1}$
but as you can see, that is not the case. Have I done something wrong? How can I solve this ODE?
 A: Hint: substitute $w(x)=y(x)+x$.
A: $\newcommand{\+}{^{\dagger}}
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$\ds{\large\tt Hint}$:

$$
\mbox{With}\quad x \equiv u + v\quad\mbox{and}\quad y \equiv u - v\quad \mbox{you'll get}\quad
{3u \over 1 - u}\,\dd u + \dd v = 0
$$

