review the conditions for the series to exist. In this case, it will exist for x=0, so we want
y = ∑anx^n
y' = ∑nanx^(n-1)
y" = ∑n(n-1)anx^(n-2)
The DE then says
∑n(n-1)anx^(n-2) + x*∑nanx^(n-1) + ∑anx^n = 0
∑n(n-1)anx^(n-2) + ∑nanx^n + ∑anx^n = 0
∑n(n-1)anx^(n-2) + ∑(n+1)anx^n = 0
Now shift y" so the powers of x add up
∑(n+2)(n+1)an+2x^n + ∑(n+1)anx^n = 0
∑[(n+2)(n+1)an+2 + (n+1)an]x^n = 0
For the series to be zero for all x, we have the recurrence relation
(n+2)(n+1)an+2 + (n+1)an = 0
an+2 = -an/(n+2)
At this point, just start listing the coefficients and see where you go. There is a similar DE used as an example at
http://tutorial.math.lamar.edu/Classes/DE/SeriesSolutions.aspx
determine a series if solution of the equation y"(x)+x.y'(x)+y(x)=0
show me step plz
1 answer