Find $f(x)$ for $f'(x) = f(x) \ln(f(x))$ and $f(0) = 1$ $$f'(x) = f(x) \ln\big(f(x)\big)$$
$$f(0) = 1, \qquad  f(x) > 0$$
I am studying for finals on my own and this exercise is really bothering me because I can't seem to solve it. If I divide by $f(x)$ I get $\ln'f(x) = \ln f(x)$ but then I don't know what to do. Any help is appreciated! 
 A: Let $y=e^{u}$. Then
\begin{eqnarray}
dy&=&e^{u}du \\
u&=&\ln y
\end{eqnarray}
From which (with a little rearrangement)
\begin{equation}
\frac{e^{u}du}{u.e^{u}}=dx
\end{equation}
Thus
\begin{equation} \ln u = x+c
\end{equation}
Hence,
\begin{equation}
u=ae^{x}
\end{equation}
Or
\begin{equation}
y=e^{ae^{x}}
\end{equation}
A: writing your equations as $$\frac{dy}{dx} = y \ln y $$ you can see that it is separable. so $$\frac{dy}{y\ln y} = dx \to x =  \int_1^y \frac{dy}{y \ln y} = \int_1^y \frac{d \,( \ln y)}{\ln y} = \ln (\ln y)\big|_1^y = diverges$$
there is trouble with the initial value $f = 1$ at $x = 0.$
A: $$\frac{f'(x)}{f(x)}=\ln{f(x)}$$
$$ \frac{d\ln{f(x)}}{dx}=\ln{f(x)}$$
Call $g(x)=\ln(f(x))$
$$g'(x)=g(x)$$
$$g(x)=ce^{x}$$
A: With $f(x)>0$, let $g(x)=\ln(f(x))$, i.e. $f(x)=e^{g(x)}$.
The equation turns to 
$$e^{g(x)}g'(x)=e^{g(x)}g(x),$$
which can be simplified as
$$g'(x)-g(x)=0,$$ with $g(0)=0$.
Multiplying by $e^{-x}$,
$$g'(x)e^{-x}-g(x)e^{-x}=\left(g(x)e^{-x}\right)'=0,$$
and 
$$g(x)e^{-x}=C^{st}=g(0)e^{-0}=0,$$ hence
$$g(x)=0,f(x)=1.$$
A: I am the original poster but because I am curious about this and wanted to format my text properly, I logged in from my computer. (What is the formatting for the <=> symbol btw?)
So what I finally did was:
Let $g(x) = lnf(x)$ (1)
$g'(x) = g(x)$
$e^xg(x) - e^xg'(x) = 0$
$e^xg(x) + e^x(-g'(x)) = 0$
$(e^xg(x))' = 0$
$e^xg(x) = c$
From (1)=> $g(0) = lnf(0) = 0$
So $c = e^0g(0) = 1 * 0 = 0$
<=> $e^xg(x) = 0$
$e^xlnf(x) = 0$
$lnf(x) = 0$
$f(x) = 1$ for every $xεR$
Your answers really helped me and led me to the answer. Is it correct though?
A: Cheating to death, we notice that as $f(x)$ and $e^{x}$ are nonzero and
$$\frac{{f'(x)}-f(x)\ln(f(x))}{f(x)e^{x}}=\frac{f'(x)}{f(x)}e^{-x}-\ln(f(x))e^{-x}=\left(\ln(f(x))e^{-x}\right)'=0.$$
Then,
$$\ln(f(x))e^{-x}=C^{st}=\ln(f(0))e^{-0}=0,$$
and $$f(x)=1.$$
