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.

DE: $y'' + 2y'/x + y^5 = 0$, $y(\infty) = 0$

Proof: Notice the DE is scale-invariant under $ x \leftrightarrow ax$ and $y \leftrightarrow a^{-1/3}y$. Therefore, if We make the substitution $y = x^{-1/3}u(x)$, then the DE reduces to:

$x^{-1/3}u'' + \frac{4}{3}x^{-4/3}u' - \frac{2}{9}x^{-7/3}u + x^{-5/3}u^5 = 0$ .... $(I)$

Is this approach correct? Notice $(I)$ can be solved by letting $u = x^m$. Is there a better way to approach this problem?

Thanks.

share|improve this question
5  
In differential equations, whatever works is the best way. –  Euler....IS_ALIVE Oct 13 '12 at 3:53
    
The question in your question isn't the question in the title. Would you like to edit to bring them into line? –  Gerry Myerson Oct 13 '12 at 4:05
    
sure............. –  Chasky Oct 13 '12 at 4:09
    
The accepted answer does not appear to be correct. –  Bob Terrell Oct 21 '12 at 12:50
    
Thanks, I made an attempt to correct it –  Valentin Oct 22 '12 at 18:26

1 Answer 1

up vote 1 down vote accepted

Notice that if you assign dimension $m$ to $y$, $dy$, $d^2y$ and dimension 1 to $x$, $dx$, then the first two terms have dimension $m-2$ and the last one $5m$. Setting $m=-\frac{1}{2}$ makes the expression homogeneous in a generalised sense. If we now let $$x=e^{\xi}\qquad y=ue^{m\xi}$$ we arrive at an equation which is invariant with respect to shift in $\xi$, hence does not depend on $\xi$ explicitly. $$\frac{dy}{dx}=e^{(m-1)\xi}\left(\frac{du}{d\xi}+mu\right)=e^{-\frac{3}{2}\xi}\left(u'_{\xi}-\frac{1}{2}u\right)$$ $$\frac{d^2y}{dx^2}=e^{(m-2)\xi}\left(u''_{\xi}+(2m-1)u'_{\xi}+m(m-1)u\right)\\=e^{-\frac{5}{2}\xi}\left(u''_{\xi}-2u'_{\xi}+\frac{3}{4}u\right)$$

EDIT: I was not careful enough in the previous version when inserting the above results in the original equation $$e^{-\frac{5}{2}\xi}\left(u''_{\xi}-2u'_{\xi}+\frac{3}{4}u\right)+e^{-\frac{5}{2}\xi}\left(2u'_{\xi}-u\right)+e^{-\frac{5}{2}\xi}u^5=0$$ Hence $$u''_{\xi}-\frac{1}{4}u+u^5=0$$

share|improve this answer

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.