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Let $\lambda$ be the first positive value for which $y=0$ where $y(x)$ satisfy the following differential equation $$ y''+\frac{2}{x}y'+y^n=0,\qquad\text{where }n\in\mathbb{R},\ y(0)=1,\text{ and }\ y'(0)=0. $$ This equation is known as the Lane-Emden equation and has analytic solutions for $n=0,1,5$.

For $n=0$, $$ y(x)=1-\frac{x^2}{6}\Rightarrow\lambda=\sqrt{6} $$

For $n=1$, $$ y(x)=\frac{\sin{x}}{x}\Rightarrow\lambda=\pi $$

And for $n = 5$, $$ y(x)=\frac{1}{\sqrt{1+\frac{x^2}{3}}}\Rightarrow\lambda=\infty $$

We want to show that $\lambda$ increases steadily as $n$ goes from $0$ to $5$. We can verify numerically that it is indeed true but it has not yet been proved analytically.

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After the change of variable $x=\frac 1z$ your equation will become a 'Emden-Fowler equation' i.e. of type $y''(z)=A\;z\;y(z)^m$ (Polyanin 'Handbook of Exact Solutions for Ordinary Differential Equations' 2.3) more exactly : $$y''(z)=-\frac {y(z)^n}{z^4}$$ Of course this is not a proof but it could make it easier... –  Raymond Manzoni Aug 5 '12 at 10:24

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