Integrating Factor of $(x\ln(y) + xy)\mathrm{d}x + (y\ln(x) + xy)\mathrm{d}y$

Question

Last day I try to prove that the equation $$ (x\ln(y) + xy)\mathrm{d}x + (y\ln(x) + xy)\mathrm{d}y =0 $$ is not exact, really easy task, but I want to go further, I tried to find a integrating factor such that the equation become exact, but is so strange, I try every method that I know, but I get nothing, later I catch some hope with the factor $R = \ln\big(\ln(x+y)\big)$, but then nothing again. There's some method to find this integrating factor? Do you know any? Thank you so much!

Answer 1

Both coefficient terms are separable, so that you can separate terms in $x$ and $y$. $$ \frac{x\,dx}{\ln x+x}+\frac{y\,dy}{\ln y+y}=0. $$

Answer 2

The ODE: $$(xln(y) + xy)dx + (yln(x) + xy)dy=0~~~~(1) $$ can be re written as $$\int \frac{xdx}{x+\ln x}+\int \frac{ydy}{y+\ln y}=c~~~~(2)$$ by dividing (1) by $(x+\ln x)(y+\ln y)$ (2) is separable yest the integrals are not doable. So $[(x+\ln x)(y+\ln y)]^{-1}$ can be taken as the integrating factor for (1).