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!