Given the ODE $(x+y+1)\,dx + (2x+2y+1)\,dy =0$, determine whether $e^{x+y}$ is an integrating factor or not.
I know we can multiply it to the ODE and check the exactness ($\partial M / \partial y = \partial N / \partial x$ for $M\,dx + N\,dy = 0 $ ODE) but the solution given to me was as follows :
For ODE $M\,dx + N\,dy = 0$, if for some function $v(x,y)$, if $\frac{(\partial M/\partial y) - (\partial N/ \partial x)}{N \cdot (\partial v/ \partial x) -M \cdot (\partial v/\partial x)}$ is again a function of $v$, say $f(v)$ then the integrating factor is $e^{\int f(v) dv}$
$e^{x+y} $ being a function of $x+y$, we take $v=x+y$ and $\frac{(\partial M/\partial y) - (\partial N/ \partial x)}{N \cdot (\partial v/ \partial x) -M \cdot (\partial v/\partial x)}$ turns out to be $\frac{-2}{x-y}$ which is a function of $v$ so the integrating factor is $\frac{1}{(x+y)^2} \neq e^{x+y}$ and hence $e^{x+y}$ is not an integrating factor.
End of solution.
Now the last line has clearly been established assuming the fact that there exists only one integrating factor of function $v$. So I want to know whether we have any such theorems available or not.
Your evaluation is wrong cause if $v=x+y$ the total derivative is $dv=dx+dy$ and you forgot to substitute either $dx$ or $dy$ in the original ODE
I chose $dy=dv-dx$ thus: $$(x+y+1)dx+(2x+2y+1)dy=0 \implies (v+1)dx+(2v+1)(dv-dx)=0\\ -vdx+(2v+1)dv=0$$
Then $$\frac {\partial M}{\partial v}=-1\qquad \& \qquad \frac{\partial N}{\partial x}=0$$
$$P(x)=\frac{M_v-N_x}{N}=\frac{-1}{2v+1}\qquad \text{No depends of x} $$ $$P(v)=\frac{N_x-M_v}{M}=-\frac1v\qquad \text{Depens only on V}$$
Our Integrating factor is $$\mu(v)=e^{\int P(v)}=-\frac1v$$ in fact we prove that the integrating factor depends of $x$ and $y$, $$\mu(x,y)=-\frac{1}{x+y}$$
If you need to, you can solve the ODE from here.