Need help for finding an integrating factor that makes a differential exact and solving it

Question

When finding the integrating factor for $$e^x(x+1)dx+(ye^y-xe^x)dy=0$$ I used $$\frac{N_x-M_y}{M}=\frac{xe^x-e^x}{e^x(x+1)}$$ and got $\frac{x-1}{x+1}$. Then my integrating factor when solving for $dy$ is $e^{(x-1)y/(x+1)}$. When multiplying the original equation I got $$e^{(x-1)y/(x+1)}e^x(x+1)dx+e^{(x-1)y/(x+1)}(ye^y-xe^x)dy=0$$ but when finding $M_y$ for $e^{(x-1)y/(x+1)}e^x(x+1)dx$, I got $$(x-1)e^{((x-1)*y)/(x+1)+x)}.$$ For $N_x$ when using $e^{(x-1)y/(x+1)}(ye^y-xe^x)dy$, I got $$(((2*x*y+x+1)*e^y+(x-x^2)*e^x)*e^{(((x-1)*y)/(x+1)))/(x+1)}$$ which they are not the same. I tried integrating $\frac{x-1}{x+1}$ with $dx$ but I still don't get the same answer. So what did I do wrong?

Answer 1

It seems to me that you have made a mistake $N_x=(ye^y-xe^x)_x=-xe^x-e^x=-e^x(x+1)$, then you have $\frac{N_x-M_y}{M}=-1$. And therefore its integrating factor will be $\mu(y)=e^{-y}$. Multiplying the ODE by the integrating factor we have: $$ e^{x-y}(x+1)dx+(y-xe^{x-y})dy=0 $$ Thus having an exact ODE: Hence there is an $F$ such that $F_x=M$ and $F_y=N$. I think that from there you can continue on your own.