Solve the ODE $yy''=y'$ Solve the ODE $yy''=y'$
Can anyone check my solution? And what is the answer? Thanks.
Attempt: 
\begin{align*}
y''=\frac{y'}{y} &\implies \frac{dy}{dx}=\ln|y|+c_1 \qquad\text{by integrating both side with respect to $x$} 
\\ &\implies \frac{dy}{\ln|y|+c_1}=dx  
\\ &\implies \int \frac{dy}{\ln|y|+c_1}=\int dx \qquad\text{(*)} 
\\ &\implies y(\ln|y|+c_2)=x+c_3 \qquad \text{by using integration by parts} 
\end{align*}
Here is the steps for the integral $I:=\int \frac{dy}{\ln|y|+c_1}$ :
Let $u=\ln|y|+c_1$ and $dv=dy$. Then $du=dy/y$ and $v=y$. So
$$I=uv-\int v du=y(\ln|y|+c_1)-(y+c_4)=y(\ln|y|+\underbrace{c_1-1}_\text{$c_2$})-c_4$$
So the equation (*) becomes
$$y(\ln|y|+c_2)=x+c \qquad \text{where $c=c_3+c_4$ }$$ I couldn't see what the wrong is. 

Hint:
Your integration of
$$
F=\int \frac{dy}{\ln y+c}
$$
is wrong. Using the substitution $\ln y +c=u$ we have:
$$
\ln y +c=u \quad \Rightarrow \quad y=\frac{e^u}{e^c}
$$
and
$$
\frac {1}{y}dy=du \quad \Rightarrow \quad dy=\frac{e^u}{e^c}du
$$
so the integral becomes:
$$
\int \frac{dy}{\ln y+c}=\frac{1}{e^c}\int \frac{e^u}{u}du
$$
this cannot be integrated with elementary functions, but only using the exponential-integral function $\mbox{Ei}(z)$ and gives:
$$
F=\frac{1}{e^c}\mbox{Ei}(y)+c_2
$$
