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We need to find the solutions of the

$w''-z^2w=3z^2-z^4$

where

$w(0)=0;w'(0)=1$

I wrote down the series that we can use to find the answer ($w$ as Taylor series):

$w=\sum_{n=0}^\infty C_nz^n$

$w'=\sum_{n=0}^\infty nC_nz^{n-1}$

$w''=\sum_{n=0}^\infty n(n-1)C_nz^{n-2}$

It is easy to find $C_0$ and $C_1$:

$w(0)=C_0=0$

$w'(0)=C_1=1$

I found this problem in Isaac Aramanovich "Collection of problems on the theory of functions of a complex variable", problem #3.112

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1 Answer 1

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Now insert into the equation and compare the coefficients of equal power $$ z^n:~~~~(n+2)(n+1)c_{n+2}-c_{n-2}=3\delta_{n,2}-\delta_{n,4} $$ with $c_n=0$ for $n<0$. This then allows you to compute the coefficients step-by-step.

This gives equations $$ 2c_2=0\\ 6c_3=0\\ 12c_4-c_0=3\\ 20c_5-c_1=0\\ 30c_6-c_2=-1\\ 42c_7-c_3=0\\ ... $$

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  • $\begingroup$ Okay, but what is $3δ_{n,2}−δ_{n,4}$ here? I don't know this function $\endgroup$
    – Egor
    May 29, 2020 at 10:03
  • $\begingroup$ That is just the usual notation for the Kronecker delta, essentially the components of a unit matrix. $\endgroup$ May 29, 2020 at 10:44
  • $\begingroup$ It also gives the wrong answer $\endgroup$
    – Egor
    May 29, 2020 at 12:07
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    $\begingroup$ That you would have to explain. Up to now I get $w(z)=z+\frac14z^4+\frac1{20}z^5-\frac1{30}z^6+\frac1{224}z^8+...$ $\endgroup$ May 29, 2020 at 12:15
  • $\begingroup$ It should be $z+z^3+\frac{z^5}{4*5}+\frac{z^9}{4*5*8*9}+...$ $\endgroup$
    – Egor
    May 29, 2020 at 12:34

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