How to solve the ordinary differential equation?

Question

$\displaystyle\frac{d²y}{dx^2}+ \frac{4}{y}\left(\frac{dy}{dx}\right)^2+2=0$

with $y(0) = 1$ and $\displaystyle\frac{dy}{dx} = 0$ for $x = 0$.

Answer 1

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 $c=\frac{_2F_1(1/2,5/9;14/9;1)}{5}$

Answer 2

HINT :

Use substitution : $v=y'$ , to get following equation :

$v'+\frac{4}{y} \cdot v=-2\cdot v^{-1}$

which is Bernoulli differential equation .