find series solution of the differential equation y"+3x^2y'-xy=0

Answer 1

well, if y = sum(a_n x^n)
y' = sum(n*a_n x^(n-1))
y" = sum(n(n-1)*a_n x^(n-2))

so, from the DE, we have

sum(n(n-1)*a_n x^(n-2)) + 3x^2*sum(n*a_n x^(n-1)) - x*sum(a_n x^n)

sum(n(n-1)*a_n x^(n-2)) + 3sum(n*a_n x^(n+1)) - sum(a_n x^(n+1))

sum(n(n-1)*a_n x^(n-2)) + sum((3n-1)*a_n x^(n+1))

If these are all summed from 0, they don't match up. So, adjust the index variable so all the x's go from x^n

sum[2,∞],(n(n-1)*a_n x^(n-2))
= sum[0,∞],((n+1)(n+2)*a_n+2 x^n)

sum[0,∞]((3n-1)*a_n x^(n+1))
= sum[1,∞]sum((3(n-1)-1)*a_n-1 x^n)
= sum[1,∞]sum((3n-4)*a_n-1 x^n)

So we have

2a_2 + sum[1,∞]((n+1)(n+2)*a_n+2 + (3n-4)*a_n-1) x^n

Now just solve the recurrence relations, as discussed at

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