Lane Emden equation variation $y''(x) + \frac{4}{x}y'(x) +k^2 y(x) =0$ does anyone know how to solve this variation of the lane emden equations?
$$y''(x) + \frac{4}{x}y'(x) +k^2 y(x) =0$$
Im told the solution by wolfram is, using A, B constant
$$\frac{A}{\sqrt{k}} \frac{ (\sin(kx)/x - x \cos(kx))}{ x^3} + \frac{B}{\sqrt{k}} \frac{ (-x \sin(kx) - \cos(kx))} {x^3}$$
Ive tried method of frobenius with a sum of $ a_n x^{n+c}$ which gave me the indicial equation
$c^2 + 3c =0 $ so $ c= 0, -3$
And reccurence relation
$a_{n+2 }= -\frac{a_n}{(n+c+2)(n+c+5)}$
And now am stuck aiming to wolframs answer.
Thanks!
PS I have also tried substitution $$z=x^2 y$$ which made the equation
$z'' +( k^2 - 2/x^2 )z =0$
If that helps!
 A: You have the equation
$$xy''+4y'+k^{2}xy=0$$
Let
$$\xi=kx$$
and let
$$y(x)=x^{-3/2}\eta(\xi)$$
and you get the Bessel equation
$$\xi^{2}\eta''+\xi\eta'+(\xi^{2}-9/4)\eta=0$$
The solution is
$$\eta(\xi)=c_{1}J_{3/2}(\xi)+c_{2}Y_{3/2}(\xi)=\sqrt{\frac{2\xi}{\pi}}[c_{1}j_{1}(\xi)+c_{2}y_{1}(\xi)]$$
Where $J, Y$ are the ordinary Bessel Functions and $j, y$ are the spherical ones. Using
$$j_{1}(\xi)=\frac{\sin(\xi)}{\xi^{2}}-\frac{\cos(\xi)}{\xi}$$
and
$$y_{1}(\xi)=-\frac{\cos(\xi)}{\xi^{2}}-\frac{\sin(\xi)}{\xi}$$
You have get your result
A: Well, we have:
$$\text{y}''\left(x\right)+\frac{\text{n}}{x}\cdot\text{y}'\left(x\right)+\text{k}^2\cdot\text{y}\left(x\right)=0\tag1$$
Now, take the Laplace transform of both sides:
$$\mathscr{L}_x\left[\text{y}''\left(x\right)+\frac{\text{n}}{x}\cdot\text{y}'\left(x\right)+\text{k}^2\cdot\text{y}\left(x\right)\right]_{\left(\text{s}\right)}=$$
$$\mathscr{L}_x\left[\text{y}''\left(x\right)\right]_{\left(\text{s}\right)}+\text{n}\cdot\mathscr{L}_x\left[\frac{1}{x}\cdot\text{y}'\left(x\right)\right]_{\left(\text{s}\right)}+\text{k}^2\cdot\mathscr{L}_x\left[\text{y}\left(x\right)\right]_{\left(\text{s}\right)}=\mathscr{L}_x\left[0\right]_{\left(\text{s}\right)}\tag2$$
Use:


*

*$$\mathscr{L}_x\left[\text{y}''\left(x\right)\right]_{\left(\text{s}\right)}=\text{s}^2\cdot\text{Y}\left(\text{s}\right)-\text{s}\cdot\text{y}\left(0\right)-\text{y}'\left(0\right)\tag3$$

*$$\mathscr{L}_x\left[\frac{1}{x}\cdot\text{y}'\left(x\right)\right]_{\left(\text{s}\right)}=\int_\text{s}^\infty\mathscr{L}_x\left[\text{y}'\left(x\right)\right]_{\left(\sigma\right)}\space\text{d}\sigma=\int_\text{s}^\infty\left(\sigma\cdot\text{Y}\left(\sigma\right)-\text{y}\left(0\right)\right)\space\text{d}\sigma\tag4$$

*$$\mathscr{L}_x\left[\text{y}\left(x\right)\right]_{\left(\text{s}\right)}=\text{s}\cdot\text{Y}\left(\text{s}\right)-\text{y}\left(0\right)\tag5$$

*$$\mathscr{L}_x\left[0\right]_{\left(\text{s}\right)}=0\tag6$$


So, we get:
$$\text{s}^2\cdot\text{Y}\left(\text{s}\right)-\text{s}\cdot\text{y}\left(0\right)-\text{y}'\left(0\right)+\text{n}\cdot\int_\text{s}^\infty\left(\sigma\cdot\text{Y}\left(\sigma\right)-\text{y}\left(0\right)\right)\space\text{d}\sigma+\text{k}^2\cdot\text{Y}\left(\text{s}\right)=0\tag7$$
A: (Not a full answer, just working so far):
$xy''(x) + 4y'(x)+k^2 xy(x)=0$
Now $\mathcal{L}[xy'']+4\mathcal{L}[y']+k^2\mathcal{L}[xy]=0$
Given
$$\mathcal{L}[xy]=\int_0^\infty e^{-sx}xy(x)dx=-\frac{d}{ds}\mathcal{L}[y] \; \forall y $$
And using
$$\mathcal{L}[y']=s\mathcal{L}[y]-y(0)
$$
We get, defining $Y(s) = \mathcal{L}[y]$
$$-(k^2+s^2)Y'(s)+2sY(s) -5y(0)=0
$$
Which I'm about to solve!
