Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

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

share|improve this question
    
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

2 Answers 2

up vote 1 down vote accepted

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

share|improve this answer

HINT :

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

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

which is Bernoulli differential equation .

share|improve this answer
2  
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.