Separable differential equation $y'e^x=\frac 1 {\ln(y)}$

Question

Hello everyone I am stuck with this equation with an initial value, i have no idea what to do anymore with this i tried to integrate both terms and got blocked

The equation: $$y'e^x=\frac{1}{\ln(y)}$$

PS: forgot to add the inital value it is $y(0)=e$

Answer 1

$$y'e^x=\frac{1}{\ln(y)}$$ It's separable $$\int \ln(y)dy=\int e^{-x}dx$$ $$y\ln(y)-y=-e^{-x}+K$$ For $y(0)=e \implies K=1 $

Therefore $$\boxed{y\ln(y)-y=1-e^{-x}}$$

Answer 2

We have $$y' e^{x} = \frac{1}{\ln(y)}$$ Rewritting, we can represent this as $$e^{-x} = y'\ln(y)$$ Integrating, we have \begin{align*} \int e^{-x}dx &= \int y'\ln(y) dy\\ \Rightarrow -e^{-x} &= y\ln(y) - y + C \end{align*} Can you finish it from here?