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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.

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    $\begingroup$ "using integration by parts" Could you expand on this step? Note that $\frac{d}{dy}\left(y(\ln|y|+c_2)\right)\ne\frac{1}{\ln|y|+c_1}$. $\endgroup$ – Did Jul 3 '16 at 10:30
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    $\begingroup$ Have you tried going from your solution back to the original version? In general that is the safest way to find out if you arrived at the correct solution. $\endgroup$ – Chinny84 Jul 3 '16 at 10:31
  • $\begingroup$ The integral isn't expressible using standard functions. It requires the "logarithmic integral". $\endgroup$ – Yves Daoust Jul 3 '16 at 10:52
  • $\begingroup$ Let $u=ye^{-y'}$, then $u'=y'e^{-y'}-yy''e^{-y'}=0$. Hence, $u=ye^{-y'}=constant=C_1$. It follows that $e^{y'}={C_2}y$⇒$y'=C_3+lny$. However, I can't proceed integrating it to get the solution. I typed this comment on my phone and I'm not very familiar with the syntax of Latex. Please bear me if there are some typos. $\endgroup$ – Huang Jul 3 '16 at 11:08
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    $\begingroup$ Your mistake is that you are applying the formula for $$\int uv'$$ to $$\int\frac{v'}{u}.$$ $\endgroup$ – Did Jul 3 '16 at 11:35
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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:

$$ F=\frac{1}{e^c}\mbox{Ei}(y)+c_2 $$

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  • $\begingroup$ Actually I don't think the first step shown by OP is right. Integrating $frac{y'}{y}$ with respect to x doesn't lead to $ln y+C$. Right? $\endgroup$ – Huang Jul 3 '16 at 11:15
  • $\begingroup$ Why not? $\frac{d}{dx} \ln (y(x))=\frac{y'(x)}{y}$. $\endgroup$ – Emilio Novati Jul 3 '16 at 12:12
  • $\begingroup$ Why didn't you use the absolute value |y| in the integration F? Doesn't it effect the answer? $\endgroup$ – Ergin Suer Jul 3 '16 at 17:00
  • $\begingroup$ Yes, of course, but since the integrand is symmetric with respect to the origin, we simply have a factor $\mbox{sign} (x)$ in the result. $\endgroup$ – Emilio Novati Jul 3 '16 at 17:05
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$$y(t)y''(t)=y'(t)\Longleftrightarrow$$


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:

  1. $$v(y)=0$$
  2. $$yv'(y)-1=0$$

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:

$$yv'(y)-1=0\Longleftrightarrow v'(y)=\frac{1}{y}\Longleftrightarrow\int v'(y)\space\text{d}y=\int\frac{1}{y}\space\text{d}y\Longleftrightarrow$$ $$v(y)=\ln|y|+\text{C}_2\Longleftrightarrow y'(t)=\ln|y(t)|+\text{C}_2\Longleftrightarrow\int\frac{y'(t)}{\ln|y(t)|+\text{C}_2}\space\text{d}t=\int1\space\text{d}t$$

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