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. 

$$y(t)y''(t)=y'(t)\Longleftrightarrow$$

Let $r(y)=y'(t)$, so we get $y''(t)=\frac{\text{d}}{\text{d}t}\cdot\frac{\text{d}y(t)}{\text{d}t}=\frac{\text{d}v(y)}{\text{d}t}=v(y)v'(y)$:

$$yv'(y)v(y)=v(y)\Longleftrightarrow$$
$$v(y)\left(yv'(y)-1\right)=0$$
Now, you know that we've two solutions:


*

*$$v(y)=0$$

*$$yv'(y)-1=0$$


For the first one:
$$v(y)=0\Longleftrightarrow y'(t)=0\Longleftrightarrow\int\space y'(t)\text{d}t=\int0\space\text{d}t\Longleftrightarrow y(t)=\text{C}_1$$
For the second one:
$$yv'(y)-1=0\Longleftrightarrow v'(y)=\frac{1}{y}\Longleftrightarrow\int v'(y)\space\text{d}y=\int\frac{1}{y}\space\text{d}y\Longleftrightarrow$$
$$v(y)=\ln|y|+\text{C}_2\Longleftrightarrow y'(t)=\ln|y(t)|+\text{C}_2\Longleftrightarrow\int\frac{y'(t)}{\ln|y(t)|+\text{C}_2}\space\text{d}t=\int1\space\text{d}t$$
