Solve the second order differential equation using two different methods I'm trying to solve the differential equation $yy''=y'^2 (1+\log(y)$ using two different method and arriving at the same result. However in both methods I get stuck. It is important to use both methods and compare solutions so please don't write any completely different solutions since they won't be of much use to me.
First Method
Since there is no $x$ we can use the substitution $p(x)=y'$ in which case $y''= \frac{dp}{dx} = \frac{dp}{dy} \cdot \frac{dy}{dx} = \frac{dp}{dy} \cdot p$. So the equation becomes:
$y p\frac{dp}{dy} = p^2 (1 + \log(y))$ and can be easily separated and integrated:
$$\int{\frac{dp}{p}} = \int{\frac{1+\log(y)}{y} dy}$$ or equivalenty $\frac{(1+\log{y})^2}{2}+A=\log{p}$. But now it seems pretty impossible to get $p$ out of the logarithm and integrate again.
Second Method
In my book it is recommended to use the substitution $y=e^z$ , $z=z(x)$. I can see how this would be useful for the logarithms but it doesn't seem to do as much as I had hoped. I probably have made a mistake somewhere. Here is my approach:
$$y=e^z, \ y'=e^z z', \ y''=e^z z' z' + e^z z'' = e^z z'^2 + e^z z''$$
So the equation becomes:
$$ e^z (e^z z'^2 + e^z z'') = (e^z z') (1+z)$$
$$e^z (z'^2 + z'') = z' (1+z)$$
Which still is of second order and I wasn't able to solve.
 A: $$yy′′=y′^2(1+\log(y)) $$
$$(e^z)(e^z)''=(e^z)'^2(1+\log (e^z))$$
$$e^z(e^z(z')^2+e^zz'')=(e^zz')^2(1+z)$$
$$((z')^2+z'')=(z')^2(1+ z)$$
$$z''=(z')^2z$$
$$\frac {z''}{z'}=z'z$$
$$(\ln z')'=\frac 12 (z^2)'$$
Integrate to reduce the order.
You forgot a square on the right side...Look here
So the equation becomes:
$$e^z(e^zz′^2+e^zz′′)=\color{red}{(e^zz′)^2}(1+z)$$
A: $$y\frac{d^2y}{dx^2}=\left(\frac{dy}{dx}\right)^2(1+\ln(y))$$
FIRST METHOD :
The substitution $p(x)=y'$ is not correct. It should be 
$$p(y)=y'(x)=\frac{dy}{dx}$$
$$y''(x)=\frac{d^2y}{d^2x}=\frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{d}{dx}\left(p(y)\right)=\frac{d}{dy}\left(p(y)\right)\frac{dy}{dx}=p'(y)\:p(y)$$
$$yp'p=p^2(1+\ln(y))$$
$$\frac{p'(y)}{p(y)}=\frac{1+\ln(y)}{y}$$
$$\ln(p)=\int \frac{1+\ln(y)}{y}dy=\frac12(1+\ln(y))^2+\text{constant}$$
$$p(y)=\frac{dy}{dx}=c\:e^{\frac12(1+\ln(y))^2}$$
$$\boxed{x=C\int e^{-\frac12(1+\ln(y))^2}dy}$$
The integral cannot be expressed with a finite number of elementary functions. A special function is required. To put it simple we have to be satisfied with the implicit solution which is the inverse function of the function defined by the above integral. 
SECOND METHOD :
The begining of your calculus is correct, but there is a mistake in the equation $ e^z (e^z z'^2 + e^z z'') = (e^z z') (1+z)$ which should be 
$$ e^z (e^z z'^2 + e^z z'') = (e^z z')^2 (1+z)$$
After simplification :
$$z''=(z')^2 z$$
$$\frac{z''}{z'}=z z'$$
$$\ln(z')=\frac12 z^2+\text{constant}$$
$$z'=c_1\:e^{\frac12 z^2}$$
With $z=\ln(y)$
$$\frac{y'}{y}=e^{\frac12 \ln(y)^2+c_2}$$
$$y'=e^{\ln(y)+\frac12 \ln(y)^2+c_2}$$
$$y'=c\:e^{\frac12(1+\ln(y))^2}$$
This is the same equation than above, which of course leads to the same above result.
