- #1
tiredryan
- 51
- 0
I am reading through a textbook and came across this part of the solution. I am wondering if anyone can give me a suggestion on how one goes from the top equation to the bottom equation. Is this something that is found in some book of derivatives or is it solved by hand? I am completely confused how this specific partial derivative is solved. Thanks
"[...]
[tex]
\left[\frac{\partial}{\partial R^2}+\frac{sin \phi}{R^2}\frac{\partial}{\partial\phi}\left(\frac{1}{sin\phi}\frac{\partial}{\partial\phi}\right)\right]^2 \psi = 0
[/tex]
This is satisfied by
[tex]
\psi = sin^2 \phi f(R)
[/tex]
if
[tex]
\left(\frac{d^2}{dR^2}-\frac{2}{R^2}\right)^2 f(R) = 0
[/tex]
The solution of the prior equation is
[tex]
f(R) = \frac{A}{R} + BR + CR^2 + DR^4
[/tex]
[...]"
"[...]
[tex]
\left[\frac{\partial}{\partial R^2}+\frac{sin \phi}{R^2}\frac{\partial}{\partial\phi}\left(\frac{1}{sin\phi}\frac{\partial}{\partial\phi}\right)\right]^2 \psi = 0
[/tex]
This is satisfied by
[tex]
\psi = sin^2 \phi f(R)
[/tex]
if
[tex]
\left(\frac{d^2}{dR^2}-\frac{2}{R^2}\right)^2 f(R) = 0
[/tex]
The solution of the prior equation is
[tex]
f(R) = \frac{A}{R} + BR + CR^2 + DR^4
[/tex]
[...]"