Consider the Emden equation y'' + (2)/(x)y' + y^α = 0 , (E)…

Consider the Emden equation $y'' + \dfrac{2}{x}y' + y^\alpha = 0$, (E), with $\alpha\in\mathbb{R}$. 1° Show that if $f(x)$ solves (E), then so does $g(x)=A^{2/(\alpha-1)}f(Ax)$, for any $A\gt 0$. 2° For $\alpha=5$, find a particular solution of $y''+\dfrac{2}{x}y'+y^5=0$.

Step 1: 1° — Setting up the scaling ansatz

Suppose $f$ solves (E). Seek a rescaled solution of the form $g(x)=\lambda f(Ax)$, with $\lambda$ to be determined, and write $\xi=Ax$. Then

\[ g'(x) = \lambda A f'(\xi), \qquad g''(x) = \lambda A^2 f''(\xi). \]

Step 2: Substituting into (E)

Substituting into $g''+\dfrac{2}{x}g'+g^\alpha$ and using $\dfrac{2}{x}=\dfrac{2A}{\xi}$: \[g''+\frac{2}{x}g'+g^\alpha = \lambda A^2 f''(\xi) + \lambda A\cdot\frac{2A}{\xi}f'(\xi) + \lambda^\alpha f(\xi)^\alpha = \lambda A^2\left[f''(\xi)+\frac{2}{\xi}f'(\xi)\right] + \lambda^\alpha f(\xi)^\alpha.\]

Since $f$ solves (E) at $\xi$, the bracket equals $-f(\xi)^\alpha$, so this reduces to

\[ g''+\frac{2}{x}g'+g^\alpha = \bigl(\lambda^\alpha - \lambda A^2\bigr)f(\xi)^\alpha. \]

Step 3: Determining $\lambda$

For $g$ to solve (E) for every $x$, the coefficient must vanish: $\lambda^\alpha = \lambda A^2$, i.e. $\lambda^{\alpha-1}=A^2$, i.e.

\[ \lambda = A^{2/(\alpha-1)}. \]

Hence $g(x) = A^{2/(\alpha-1)}f(Ax)$ is again a solution of (E), for any $A\gt 0$.

\[f \text{ solves (E)} \ \Longrightarrow\ g(x) = A^{2/(\alpha-1)}f(Ax) \text{ solves (E), for every } A\gt 0\]

Step 4: 2° — A particular solution for $\alpha=5$

For $\alpha=5$ the scaling law gives $\lambda=A^{2/4}=A^{1/2}$: a solution rescales as $x\mapsto Ax$, $y\mapsto A^{1/2}y$. Guided by this symmetry, try the two-parameter ansatz

\[ y = (ax^2+c)^{-1/2}, \qquad D := ax^2+c. \]

Then \[y' = -\frac{ax}{D^{3/2}}, \qquad y'' = \frac{2a^2x^2-ac}{D^{5/2}}, \qquad \frac{2}{x}y' = -\frac{2a}{D^{3/2}} = -\frac{2aD}{D^{5/2}}.\]

Step 5: Substituting into the equation

Combining the first two terms over the common denominator $D^{5/2}$: \[y''+\frac{2}{x}y' = \frac{2a^2x^2-ac-2aD}{D^{5/2}} = \frac{2a^2x^2-ac-2a^2x^2-2ac}{D^{5/2}} = \frac{-3ac}{D^{5/2}}.\]

Also $y^5=D^{-5/2}$, so the full equation $y''+\frac{2}{x}y'+y^5=0$ becomes

\[ \frac{-3ac+1}{D^{5/2}} = 0 \quad\Longrightarrow\quad ac = \frac{1}{3}. \]

So $y=(ax^2+c)^{-1/2}$ solves the $\alpha=5$ Emden equation for any $a,c$ with $ac=\tfrac13$. The simplest choice, $a=1$, $c=\tfrac13$, gives

\[ y = \left(x^2+\frac13\right)^{-1/2}. \]

Step 6: Verification

Substituting $y=(x^2+\tfrac13)^{-1/2}$ directly into $y''+\dfrac{2}{x}y'+y^5$ confirms the residual is exactly $0$. The general symbolic argument was checked independently in two ways: (i) for arbitrary $\alpha$, expanding $g''+2g'/x+g^\alpha$ and confirming that $\lambda=A^{2/(\alpha-1)}$ makes the exponents $2\alpha/(\alpha-1)$ and $2+2/(\alpha-1)$ exactly equal, forcing the bracket $\lambda^\alpha-\lambda A^2$ to vanish identically; (ii) at $\alpha=5$ specifically, using the explicit family $f(x)=(ax^2+c)^{-1/2}$, confirming that $g(x)=A^{1/2}f(Ax)$ also solves the $\alpha=5$ equation exactly, for symbolic $A$.

\[y = \left(x^2+\frac13\right)^{-1/2}\]