# Find the solutions to the $w''-z^2w=3z^2-z^4$ as Taylor series where $w(0)=0$ and $w'(0)=1$

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

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

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