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.