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. 
 A: 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:
$$
F=\frac{1}{e^c}\mbox{Ei}(y)+c_2
$$
A: $$y(t)y''(t)=y'(t)\Longleftrightarrow$$

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:
$$yv'(y)-1=0\Longleftrightarrow v'(y)=\frac{1}{y}\Longleftrightarrow\int v'(y)\space\text{d}y=\int\frac{1}{y}\space\text{d}y\Longleftrightarrow$$
$$v(y)=\ln|y|+\text{C}_2\Longleftrightarrow y'(t)=\ln|y(t)|+\text{C}_2\Longleftrightarrow\int\frac{y'(t)}{\ln|y(t)|+\text{C}_2}\space\text{d}t=\int1\space\text{d}t$$
A: To obtain the correct answer, you must first account for the possible solution $y=0.$ To obtain other solutions, you divide both sides of the equation by $y,$ resulting in $$y''=\frac{y'}{y}.$$ Now, you remember that if I have a function $f:\mathbb{R}\setminus\{0\}\rightarrow\mathbb{R}$ satisfying $$f'(x)=\frac1{x},$$ then the complete family of functions that satisfy it is given by the functions of the form $$f(x)=\begin{cases}\ln(-x)+A&x\lt0\\\ln(x)+B&x\gt0\end{cases}.$$ Therefore, we must consider two cases: $y\lt0$ and $y\gt0.$ If $y\lt0,$ then $$y'=\ln(-y)+A_-$$ while if $y\gt0,$ then $$y'=\ln(y)+A_-.$$ To proceed further, one must be able to evaluate $$\int\frac1{\ln(-y)+A_-}\,\mathrm{d}y$$ and $$\int\frac1{\ln(y)+A_+}\,\mathrm{d}y.$$ To evaluate $$\int\frac1{\ln(-y)+A_-}\,\mathrm{d}y,$$ let $y=-e^{z-A_-},$ hence $\mathrm{d}y=-e^{z-A}\,\mathrm{d}z,$ hence $$\int\frac1{\ln(-y)+A_-}\,\mathrm{d}y=-\int\frac{e^{z-A_-}}{z}\,\mathrm{d}z=\begin{cases}-e^{-A_-}[\ln(-z)-\operatorname{Ein}(-z)]+B_-&z\lt0\\-e^{-A_-}[\ln(z)-\operatorname{Ein}(-z)]+B_+&z\gt0\end{cases}=\begin{cases}-e^{-A_-}[\ln(-[\ln(-y)+A_-])-\operatorname{Ein}(-[\ln(-y)+A_-])]+B_-&y\gt-e^{-A_-}\\-e^{-A_-}[\ln([\ln(-y)+A_-])-\operatorname{Ein}(-[\ln(-y)+A_-])]+B_+&y\lt-e^{-A_-}\end{cases}.$$ An analogous mess can be done in the case when $y\gt0.$ I will not do that computation, but hopefully, you understand just how complicated this exercise is.
