\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 $x = 0$.
\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 $c=\frac{_2F_1(1/2,5/9;14/9;1)}{5}$
HINT :
Use substitution : v=y' , to get following equation :
v'+\frac{4}{y} \cdot v=-2\cdot v^{-1}
which is Bernoulli differential equation .