Solving ODE $g(\eta)''+\eta/2 \cdot g(\eta)'-n\cdot g(\eta)=0$; BC: $g(0)=1$ and $g(\infty)=0$ I'm lost! How would I solve $$g(\eta)''+\eta/2 \cdot g(\eta)'-n\cdot g(\eta)=0$$
with BC $$g(0)=1$$ and $$g(\infty)=0$$
 A: An attempt at solution.
Let's make a transformation for some $a$:
$$g(\eta)=f(\eta)e^{a\eta^2}$$
$$g'=(f'+ 2a \eta f)e^{a\eta^2}$$
$$g''=(f''+ 2a \eta f'+2a f+2a \eta f'+4a^2 \eta^2 f)e^{a\eta^2}=(f''+ 4a \eta f'+2a (1+2a\eta^2)  f)e^{a\eta^2}$$
After substitution we obtain:
$$f''+ 4a \eta f'+2a (1+2a\eta^2)  f+\frac{1}{2} \eta (f'+ 2a \eta f)-nf=0$$
Or:
$$f''+ (4a+1/2) \eta f'+(2a-n+a(1+4a)\eta^2)  f=0$$
We can now pick $a$ to get rid of the first derivative:
$$a=-\frac{1}{8}$$
$$f(\eta)=g(\eta) e^{\eta^2/8}$$
Which gives us:
$$f''-\left(n+\frac{1}{4}+\frac{1}{16}\eta^2 \right)  f=0$$

$$f''-\frac{1}{16}\left(4(4n+1)+\eta^2 \right)  f=0 \tag{1}$$

This equation has a general solution in the form of parabolic cylinder functions.
So in principle, we have solved the problem, and now only need to apply the boundary conditions.

However, the boundary conditions pose some challenge. The first one is simple enough.
$$f(0)=1$$
But the second one we can't really guess, because for $\eta \to \infty$ the exponential factor blows up, and we don't know enough about $g(x)$ to figure out what is the limit:
$$\lim_{\eta \to \infty} g(\eta) e^{\eta^2/8} = ?$$
Unless there's some way to apply this boundary condition correctly to (1), we are back to the drawing board.

Update:
I tried to use the Mathematica ODE solver with the second initial condition:
$$f(\eta_m)=0, \qquad \eta_m  \gg 1$$
For example $\eta_m =100$. The convergence is very good, i.e. the solutions for different but large  $\eta_m$ agree very well.
So I think this may be the way to solve the original equation.
