# 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 $x = 0$.

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Note that $y=\displaystyle-\frac{1}{9}x^2$ is a solution to the ODE $\displaystyle\frac{d²y}{dx^2}+ \frac{4}{y}\left(\frac{dy}{dx}\right)^2+2=0$. However, it does not satisfy the initial condition $y(0)=1$. –  Paul Feb 1 '12 at 12:17
I'm trying to find a way to solve it with $(y y')' = (y')^2 +y y''$ –  Pedro Tamaroff Feb 3 '12 at 2:13

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

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HINT :

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

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

which is Bernoulli differential equation .

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I think it should be mentioned that in this method of solution, $\nu$ is considered as a function of $y$, and $\nu'=\dfrac{d\nu}{dy}$. This is standard in equations where the $x$ (the independent variable) does not appear explicitly in the equation. –  Julián Aguirre Feb 1 '12 at 13:45
@JuliánAguirre,You are correct...I thought that it was obvious... –  pedja Feb 1 '12 at 13:58
It was obvious to me, but perhaps not to someone who is beginning to study differential equations. That's why I made the comment. –  Julián Aguirre Feb 1 '12 at 15:05
@JuliánAguirre,And that's why I upvoted your comment... –  pedja Feb 1 '12 at 15:12
the solution of the Bernoulli equation is $v y^4 = -2 \int{\frac{y^4}{v}}dy+C$. How to solve it further? Thanks to all of you. –  Mia Feb 1 '12 at 16:38