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

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
up vote 1 down vote accepted

your equation is $yy''+4(y')^{2}+2y=0$




$(4(y')^{2}+yy'')=\frac{z''}{5y^{3}}$ If we put it to your equation




$\int z'z'' dx=-10\int z^{4/5}z'dx$

$ (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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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

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