Nonlinear 2nd order ODE I have been looking at numerical solutions to the following nonlinear Bessel-type ODE:
$$
xy'' + 2 y' = y^2 - k^2,
$$
where k is a constant. In general, $y = \pm k$ is an asymptotic solution, and as $x$ approaches 0, the solution diverges.  
Is there a way to get a better analytical handle of the solutions? I can't hope for a closed-form solution, but if there are any other insights that would be great.
EDIT:
I can transform the equation to a Bessel standard form, with a nonlinear inhomogeneous term. Let $xy+kx=w$, then:
$$xy'+y+k=w', \space 2y'+xy''=w''$$
$$y'=\frac{w'}{x}-\frac{w}{x^2}, \space xy''=w''-2\frac{w'}{x}-2\frac{w}{x^2}$$
Substituting,
$$w''=\frac{(w-kx)^2}{x^2}-k^2$$
$$x^2w''=w^2-2kxw$$
Let $v=kx$,
$$v^2w_{vv}''=w^2-2vw$$
Let $z=\sqrt{v}$,
$$w_{vv}=\frac{1}{4v}w_{zz}-(\frac{1}{4v^{\frac{3}{2}}})w_z=w^2-2z^2w$$
$$v^2w_{vv}=\frac{1}{4}vw_{zz}-\frac{1}{4}\sqrt{v}w_{z}=w^2-2z^2w$$
$$z^2w''-zw'+8z^2w=4w^2$$
Let $w=W/2$ and $z=Z/2$
$$Z^2W''_{ZZ}-ZW'_{Z}+Z^2W=W^2$$
 A: Hint:
Let $u=xy$ ,
Then $\dfrac{du}{dx}=x\dfrac{dy}{dx}+y$
$\dfrac{d^2u}{dx^2}=x\dfrac{d^2y}{dx^2}+\dfrac{dy}{dx}+\dfrac{dy}{dx}=x\dfrac{d^2y}{dx^2}+2\dfrac{dy}{dx}$
$\therefore\dfrac{d^2u}{dx^2}=\dfrac{u^2}{x^2}-k^2$
You can consider as two members Emden-Fowler type nonlinear ODE and follow the method in http://www.sciencepubco.com/index.php/ijamr/article/download/723/628
A: $$x\frac{d^2y}{dx^2}+2\frac{dy}{dx}=y^2-k^2$$
Change of function : $y=kY \quad\to\quad x\frac{d^2Y}{dx^2}+2\frac{dY}{dx}=kY^2-k$
Change of variable : $x=\frac{X}{k}$
$$X\frac{d^2Y}{dX^2}+2\frac{dY}{dX}=Y^2-1$$
There is no parameter in the new ODE which simplifies the study.
For example, searching a particular solution around $(X=0\:,\;Y=0)$ on the form of series expansion , since the function $Y(X)$ is odd, the identification of the coefficients of the terms $X^{2n+1}$ leads to :
$$Y(X)= -\frac{1}{2}X+\frac{1}{48}X^3+\frac{1}{1440}X^5+O(X^6)$$
Bringing it back into the ODE gives the order of deviation : $X\frac{d^2Y}{dX^2}+2\frac{dY}{dX}-(Y^2-1)=O(X^6)$
