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!