Solve $\ln(y-x) y’ = y \ln (y)$

Question

This was a problem I worked on for a while before realising I misread the homework question (it was supposed to be $\ln(y-x) y’ = y \ln (y)$.)

Can anyone tell me if this can be solved (maybe series solutions)?

Answer 1

I suppose that the equation is, as in the original post,$$(y\log(y)-x)y’=y \log(y)$$ Switching variables, it becomes $$(y\log(y)-x)=x'y \log(y)$$ Let $x=\frac z {\log(y)}$ and the equation becomes $$z'=\log(y)\implies z=y(\log(y)-1)+C$$ that is to say $$x=y+\frac{C-y}{\log (y)}$$