Question

(x^2+4) y"+ xy = 0 . find the power series solution.. @amistre64

Answer 1

and given that .. \[x _{0} = 0\]

Answer 2

let y=sum0 an x^n
y'=sum1 an nx^(n-1)
y''=sum2 an n(n-1) x^(n-2)

plug and play

Answer 3

(x^2+4) y"+ xy = 0

(x^2+4) sum2 an n(n-1) x^(n-2) + x sum0 an x^n = 0

x^2 sum2 an n(n-1) x^(n-2) + sum2 4an n(n-1) x^(n-2) + sum0 an x^(n+1) = 0

sum2 an n(n-1) x^n + sum2 4an n(n-1) x^(n-2) + sum0 an x^(n+1) = 0

adjust to the smallest; in this case, n-2

sum4 a{n-2} (n-2)(n-3) x^(n-2) + sum2 4an n(n-1) x^(n-2) + sum3 a{n-3} x^(n-2) = 0

and pull out the summations to line up the indexes

Answer 4

\[\sum_4 a_{n-2} (n-2)(n-3) x^{n-2} \\+ \sum_2 4a_n n(n-1) x^{n-2} \\+ \sum_3 a_{n-3} x^{n-2} = 0\]

\[\sum_4 a_{n-2} (n-2)(n-3) x^{n-2} \\
+ 4a_2 2(2-1) x^{2-2}+4a_3 3(3-1) x^{3-2}+\sum_4 4a_n n(n-1) x^{n-2} \\
+ a_{3-3} x^{3-2} + \sum_4 a_{n-3} x^{n-2} = 0\]

\[8a_2 +24a_3~x+ a_{0}~x\\+\sum_4a_{n-3}+a_{n-2} (n-2)(n-3)+ 4a_n n(n-1) x^{n-2} = 0\]

Answer 5

with any luck, i kept the errors in that to a minimum :)

Answer 6

ouh okay2, thankyou @amistre64 ;)