I have problems following the calculations in Griffiths' Introduction to Quantum Mechanics (Chapter 4.2.1):
If you apply the Schrödinger equation to the Coulomb potential you get the following differential equation:
$$\tag{4.56}
\frac{d^2u}{d\rho^2} ~=~ [1-\frac{\rho_0}{\rho} + \frac{\ell(\ell+1)}{\rho^2}]u.
$$
A few steps later my books substitutes $u(\rho)$ without explaining why this is allowed:
The next step is to peel of the asymptotic behavior, introducing a new function $v(\rho)$:
$$\tag{4.60}
u(\rho) ~=~ \rho^{\ell+1}e^{-\rho}v(\rho).
$$
How is this function still related to the Hydrogen atom? To me it looks like something total random.
Best Answer
As Lubos points out, you are allowed to make any substitutions you want as long as the new function is equivalent to the old one. In this case, substituting $u(\rho)=\rho^l e^{-\rho}v(\rho)$ is allowed because you can recover $u$ if you know $v$.
The question is, though, why would you choose exactly such a form? If you're just given it outright, it does look pretty random. However, there's method behind it.
Essentially, you've got an equation that's got very particular behaviour at both of its endpoints. Take, for example the $\rho\ll1$ limit: then the equation reads $$ \frac{d^2u}{d\rho^2}\approx\frac{\ell(\ell+1)}{\rho^2}u, $$ whose solutions are $u=\rho^l$ and $u=\rho^{-(l+1)}$ (which we reject as unphysical). Of course, this says nothing exact about the full solution but it does inform us about its asymptotic behaviour as $\rho\rightarrow0$.
Similarly, in the $\rho\gg l$ limit you have $\frac{d^2u}{d\rho^2}\approx u$, so $u=e^{-\rho}$ (as $u=e^\rho$ is unphysical). Again, this only gives information about the asymptotics about $\rho\rightarrow\infty$.
So what have our limit games gained us? Formally, we know nothing about the exact solution, but we do know that it has to behave something like $u=\rho^le^{-\rho}\times\mathrm{some\, function}$. Plus, this new function will be better behaved than $u$ at the endpoints - constant at the origin and up to polynomial at infinity.
Thus, we make an informed guess and substitute $u=\rho^le^{-\rho}v(\rho)$, transforming the equation to the new variables, and hoping as we do so that the result will be a simpler equation. As it happens, it is both simpler and better behaved, with better-behaved singularities, and its solutions are now simple enough (the complexity being absorbed into the prefactors just discussed) that they are in fact polynomials.
In general, the exact form of the substitution will depend on the exact problem. One usually tries to do this kind of asymptotic analysis to try and distil away, slowly, the problem's complexity.