H''(n) + (1/2)n*H'(n) = 0.
Boundary conditions: H(0) = 1, H(inf) = 0.
I really don't know where to start ... any help will be greatly appreciated.
First separate and integrate
$\displaystyle \frac{H''}{H'} = - \frac{1}{2} n\, dn$
so
$\displaystyle \ln H' = \ln c_1 - \frac{1}{4} n^2\;\; \Rightarrow \;\;H' = c_1 e^{-n^2/4}$
then integrate again (on the second integration we'll keep it in integral form)
$\displaystyle
H = c_1 \int_0^n e^{-t^2/4}dt + c_2
$
$\displaystyle
H(0) = 1\; \text{ gives}\; c_2 = 1
$ while $\displaystyle
H(\infty) = 0\; \text{ gives}\; c_1 \pi + 1 = 0
$
giving the solution as
$\displaystyle
H(n) = 1 - \frac{1}{\pi} \int_0^n e^{t^2/4}dt
$
or in terms of error functions
$\displaystyle
H(n) = 1 - \text{erf} \left(\frac{n}{2}\right)$.