How to solve the ordinary differential equation? $\displaystyle\frac{d²y}{dx^2}+ \frac{4}{y}\left(\frac{dy}{dx}\right)^2+2=0$
with $y(0) = 1$ and $\displaystyle\frac{dy}{dx} = 0$ for $x = 0$.

your equation is $yy''+4(y')^{2}+2y=0$
$z=y^5$
$z'=5y^{4}y'$
$z''=20y^{3}(y')^{2}+5y^{4}y''=5y^{3}(4(y')^{2}+yy'')$
$(4(y')^{2}+yy'')=\frac{z''}{5y^{3}}$
If we put it to your equation
$\frac{z''}{5y^{3}}+2y=0$
$z''=-10y^{4}=-10z^{4/5}$
$z'z''=-10z^{4/5}z'$
$\int z'z'' dx=-10\int z^{4/5}z'dx$
$ (z')^{2}/2 =-(50/9) z^{9/5}+m$
$ (z')^{2} =-(100/9) z^{9/5}+k$
$ z' =\sqrt{-(100/9) z^{9/5}+k}$
$z'=5y^{4}y'=\sqrt{-(100/9) y^{9}+k}$
if $x=0$ and
$y'(0)=0$  and $y(0)=1$  then $k=100/9$
$\frac{5y^{4}y'}{\sqrt{-(100/9) y^{9}+100/9}}=1$
$\int \frac{y^{4}y'}{\sqrt{1-y^{9}}} dx=\frac{2x}{3}+c$
I asked to wolfram that the solution is expressed by hypergeometric functions
the solution
$y^{5} \frac{_2F_1(1/2,5/9;14/9;y^{9})}{5}=\frac{2x}{3}+c$
if $x=0$ and
$y(0)=1$  then $c=\frac{_2F_1(1/2,5/9;14/9;1)}{5}$
