Let where .
So then
We expand in a pertubation series
Inserting this into (1)
Inserting into initial condition
Now let
Then
That is
Subtracting 1 from both sides and diving the rest with we get
One more time let
Then
if we repeat the procedure we realise that also
We now determine ...
subsequently by solving the equations that we get by comparing terms containing the same power of .
I'm unsure what to do now, as for I need to integrate
however my equation for is:
i.e. really complicated, and coupled with other functions.
Does anybody have any idea what I can do from here?
Thanks in advance!