Differential equation $f''(x)+2 x f(x)f'(x) = 0$

Question

I am trying to solve,

$ f''(x)+2 x f(x)f'(x) = 0$

with boundary conditions $f(-\infty)=1$ and $f(\infty)=0$. I have found that for instance $f(x) = 3/2 x^{-2}$ but obviously it does not satisfy the proper boundary conditions. Any ideas for a solution?

Answer 1

(this is not an answer, but, this doesn't do well as a comment)

To solve $y''+2xyy'=0$ we might notice $(xy^2)' = y^2+ 2xyy'$ thus $$ y''+(xy^2)'-y^2 = 0 $$ or $$ (y'+xy^2)' - y^2 = 0 $$ I also tried multiplying by $y'$ to look for some energy equation, but, the presence of that $x$ is cramping my style. I bet someone can solve this.

Answer 2

Hint:

This belongs to a generalized Emden–Fowler equation.

$f''(x)+2xf(x)f'(x)=0$

$\dfrac{d^2f}{dx^2}=-2xf\dfrac{df}{dx}$

$\therefore\dfrac{d^2x}{df^2}=2fx\left(\dfrac{dx}{df}\right)^2$

Follow the method in http://science.fire.ustc.edu.cn/download/download1/book%5Cmathematics%5CHandbook%20of%20Exact%20Solutions%20for%20Ordinary%20Differential%20EquationsSecond%20Edition%5Cc2972_fm.pdf#page=377:

Let $\begin{cases}y=\dfrac{df}{dx}\\t=f^2\end{cases}$ ,

Then $\dfrac{d^2t}{dy^2}=2y^{-1}t^{-\frac{1}{2}}\left(\dfrac{dt}{dy}\right)^3$

$\therefore\dfrac{d^2y}{dt^2}=-2t^{-\frac{1}{2}}y^{-1}$

Which reduces to an Emden–Fowler equation.