solve (x-1) (d^2 y)/〖dx〗^2 +2dy/dx=0 using series method

Answer 1

(x-1)y" + 2y' = 0

If we expand about x=0, we assume

y = ∑a_nx^n
y' = ∑na_nx^(n-1)
y" = ∑n(n-1)_nx^(n-2)

Plug these into our DE and we have

(x-1)∑n(n-1)_nx^(n-2) + 2∑na_nx^(n-1) = 0

∑n(n-1)_nx^(n-1) - ∑n(n-1)_nx^(n-2) + 2∑na_nx^(n-1) = 0

∑n(n-1)_nx^(n-2) + ∑(2n+n-1)a_nx^(n-1) = 0

∑(n+1)(n+2)_n+2x^n
+ ∑(3n+2)a_n+1x^n

∑((n+1)(n+2)_n+2+(3n+2)a_n+1)x^n = 0

So, now we have the recurrence relation

(n+1)(n+2)_n+2+(3n+2)a_n+1 = 0

I expect you can take it from here; if you get stuck, read the excellent article at

http://tutorial.math.lamar.edu/Classes/DE/SeriesSolutions.aspx