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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?


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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............. – ILoveMath 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
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 $$u''_{\xi}-\frac{1}{4}u+u^5=0$$

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