Solving of the second-order nonlinear differential equation I'm solving differential equation $2yy''=y^2+y'^2$. I guess it necessary to reduce an order. I try to write equation in terms of $y'=u$. I get the first-order equation, and after i let $u=zy$. But then i have trouble with solving. Is that way right or i make mistake?
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Write the equation as $\ds{{y'' \over y} = \half + {y'^{2} \over 2y^{2}}}$. With $\ds{z \equiv \ln\pars{y}}$:

\begin{align}
{y'' \over y}&={1 \over y}\,\totald{}{x}\pars{y\,\totald{\ln\pars{y}}{x}}
=\bracks{\totald{\ln\pars{y}}{x}}^{2} + \totald[2]{\ln\pars{y}}{x}
= z'^{2} + z''
\\[3mm]
{y' \over y} &= \totald{\ln\pars{y}}{x} = z'
\end{align}

Then,
$$
z'^{2} + z'' = \half + \half\,z'^{2}\quad\imp\quad z''=\half\pars{1 - z'^{2}}
\quad\imp\quad{\dd z' \over 1 - z'^{2}} = \half\,\dd x
$$

$$
{\rm arctanh}\pars{z'} = \half\,x + A\,,\qquad A\ \mbox{is a constant}
$$

$$
z' = \tanh\pars{\half\,x + A}\quad\imp\quad
z=2\ln\pars{\root{B}\cosh\pars{{x \over 2} + A}}\,,
\qquad B\ \mbox{is a constant}
$$

$$
\color{#66f}{\large y}= \expo{z}
=\color{#66f}{\large B\cosh^{2}\pars{\half\,x + A}}
$$

