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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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    I'm trying to find a way to solve it with $(y y')' = (y')^2 +y y''$2012-02-03

2 Answers 2

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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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    @Mia,your solution is incorrect...2012-02-01