Finding $f$ from $f'$

Question

I have a function $f:(1,+\infty)\rightarrow(0,+\infty)$ that derives and for which is true that: $x\cdot\ln(x)\cdot f'(x)=f(x)+x\cdot\ln(x)\cdot f(x)$ and $f(e)=e^e.$ I want to find $f$ so from the given equality I reach at this point: $$x\cdot\ln(x)\cdot f'(x)=f(x)+x\cdot\ln(x)\cdot f(x)\iff\frac{f'(x)}{f(x)}=\frac{(\ln(x))'}{\ln(x)}\iff(\ln(f(x)))'=(\ln(\ln(x)))'$$ but how can I continue?

Answer 1

I have rechecked it and I think it should be $$\frac{f'(x)}{f(x)} = \frac{1}{x\cdot \ln x}+1$$ At this point, let's say $y = f(x)$ and therefore $\frac{dy}{dx} = f'(x)$. Then we have $$\frac{dy}{dx}\cdot \frac{1}{y} = \frac{1}{x\cdot \ln x}+1$$ Multiplying both sides with $dx$, we get $$\frac{dy}{y} = \bigg(\frac{1}{x\cdot \ln x}+1\bigg)dx$$ Now we can integrate both sides to get $$\int\frac{dy}{y} = \int\bigg(\frac{1}{x\cdot \ln x}+1\bigg)dx$$ which yields $$\ln|y| = x+\ln(\ln(x))+C\ ,C \in \mathbb{R}$$ Since codomain of $y$ is defined on positive real numbers, we can get rid of the absolute value and $$y = e^{x+\ln(\ln(x))+C} = e^x \cdot e^{\ln(\ln(x))} \cdot e^C = e^x \cdot \ln(x) \cdot e^C$$ Now in order to find the constant $C$, we can use the information $f(e) = e^e$. Putting $x = e$, we get $$e^e \cdot e^C = e^e \implies C = 0$$ So finally, we have $$f(x) = e^x \cdot \ln(x)$$