- #1

chwala

Gold Member

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## Homework Statement:

- ##y"+y'\frac {1}{z}+y[\frac {z^2-n^2}{z^2}]=0##

## Relevant Equations:

- power series

let ##y= \sum_{k=-∞}^\infty a_kz^{k+c}##

##y'=\sum_{k=-∞}^\infty (k+c)a_kz^{k+c-1}##

##y"=\sum_{k=-∞}^\infty (k+c)(k+c-1)a_kz^{k+c-2}##

therefore,

##y"+y'\frac {1}{z}+y[\frac {z^2-n^2}{z^2}]=0## =##[\sum_{k=-∞}^\infty [(k+c)^2-n^2)]a_k + a_k-2]z^{k+c} ##

it follows that,

##(k+c)^2-n^2)]a_k + a_k-2=0##

if ##k=0, →c^2-n^2=0, a_0≠0##

##c^2=n^2, →c=±n ##(roots)

now considering

i am ok with the rest of the steps...

##y'=\sum_{k=-∞}^\infty (k+c)a_kz^{k+c-1}##

##y"=\sum_{k=-∞}^\infty (k+c)(k+c-1)a_kz^{k+c-2}##

therefore,

##y"+y'\frac {1}{z}+y[\frac {z^2-n^2}{z^2}]=0## =##[\sum_{k=-∞}^\infty [(k+c)^2-n^2)]a_k + a_k-2]z^{k+c} ##

it follows that,

##(k+c)^2-n^2)]a_k + a_k-2=0##

if ##k=0, →c^2-n^2=0, a_0≠0##

##c^2=n^2, →c=±n ##(roots)

now considering

**case 1**, if ##c=n##, and given k=1, then ##a_1=0##...this is the part that i am not getting.i am ok with the rest of the steps...