# Asymptotic behavior: (d^2 u)/(dp^2) = [1 - (p_0)/p + {L(L+1)}/(p^2)]u

#### Viona

##### New member
Hello every one,

I was reading in a physics book where we want to solve this differential equation:

I guess it is difficult to solve directly so they examined asymptotic behavior:

The thing what I did not get is equation (4.60). It will be nice if some one helped me to understand how they did this step.
Thanks.

#### JeffM

##### Elite Member
Hello every one,

I was reading in a physics book where we want to solve this differential equation:
View attachment 10618
I guess it is difficult to solve directly so they examined asymptotic behavior:
View attachment 10619
View attachment 10620
View attachment 10621

View attachment 10622
The thing what I did not get is equation (4.60). It will be nice if some one helped me to understand how they did this step.
Thanks.
I think they are implicitly defining something. Very obscure if that is what is going on. That is, I guess they are doing this:

$$\displaystyle \text {Define } \nu( \rho ) = \dfrac{\mu ( \rho )}{ \rho^{(L + 1)} e^{- \rho}} \implies \mu ( \rho ) = \rho^{(L+1)} e^{- \rho}\nu ( \rho ).$$

This is why I could not stand physics.

#### Viona

##### New member
I think they are implicitly defining something. Very obscure if that is what is going on. That is, I guess they are doing this:

$$\displaystyle \text {Define } \nu( \rho ) = \dfrac{\mu ( \rho )}{ \rho^{(L + 1)} e^{- \rho}} \implies \mu ( \rho ) = \rho^{(L+1)} e^{- \rho}\nu ( \rho ).$$

This is why I could not stand physics.
Interesting guess. It works for me. Thanks