Series solution of xy^{' '}+2y^{'}-xy=0 I get r(r+1)=0, (r+1)(r+2)c_1=0 and c

estroishave4 2022-05-03 Answered
Series solution of xy+2yxy=0
I get r(r+1)=0,(r+1)(r+2)c1=0 and
cn+1=cn1(n+1+r)(n+r+2)
The first equation gives the indicial roots r=1 and r=0. The case for r=0 is fine.
For r=1, I don't see how c1=0 is implied by the second equation. The way I see it, since 1+r=0 when r=1,c1 is not necessarily zero here, but this leads to a different solution to what is in the text book. What am I missing? Why is c1 forced to be zero?
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Answers (1)

4dorts3x
Answered 2022-05-04 Author has 9 answers
As you already found a basis solution for r=0, you need just one other. Setting c1 in the case r=1 to some non-zero value will just add a multiple of the first, r=0, basis solution, so nothing new gets found.
Note that your DE can be written as (xy)(xy)=0,
so that the solution you find are y1(x)=sinh(x)x and y2(x)=cosh(x)x
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