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Differential EquationsAnswered question
hoperetauyk hoperetauyk 2022-05-28

I would need some help in solving the following differential equation:
u ( p ) + r + N λ p N λ p ( 1 p ) u ( p ) = r + N λ p N λ ( 1 p ) g .
I already know that the solution is of the form
g p + C ( 1 p ) ( 1 p p ) r / λ N ,,
where C is a constant.
Now this is a standard linear first-order differential equation. Letting
f ( p ) = r + N λ p N λ p ( 1 p ) ,     q ( p ) = r + N λ p N λ ( 1 p ) g ,     μ ( p ) = e f ( p ) d p ,
the solution should have the form
u ( p ) = C μ ( p ) + 1 μ ( p ) μ ( p ) q ( p ) d p ,
where C is a constant.
The calculation of μ ( p ) was not much of a problem. Indeed, we can rewrite f(p) as
f ( p ) = r N λ 1 p + ( 1 + r N λ ) 1 1 p
Using 1 p d p = ln ( p ) and 1 1 p d p = ln ( 1 p ), I found that
μ ( p ) = p r N λ ( 1 p ) 1 r N λ .
Where I have difficulties is calculating the term ∫μ(p)q(p)dp, where
μ ( p ) q ( p ) = r g N λ p r N λ ( 1 p ) 2 r N λ + p r N λ + 1 ( 1 p ) 2 r N λ .
How do I integrate this last term? Or did I miss something obvious, because in the paper I'm looking at they don't provide any details about how they get to the solution of the differential equation
Edit: small typo corrected in the last displayed equation.
Edit: there is a big typo in the differential equation: the rhs should have r + N λ rather than r + N λ p in the numerator. This probably makes it much simpler to solve. My bad.

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