【微分積分】5-4-4 常微分方程式のべき級数解｜問題集

\(\displaystyle y=\sum_{n=0}^{\infty}a_nx^n\)とおくと、
\(\displaystyle y'=\sum_{n=1}^{\infty}na_nx^{n-1}\)
\(\displaystyle y''=\sum_{n=2}^{\infty}n(n-1)a_nx^{n-2}\)
すなわち
\(\displaystyle y''-2xy'-4y\)
\(\displaystyle =\sum_{n=2}^{\infty}n(n-1)a_nx^{n-2}\)
\(\displaystyle \ \ \ -2x\sum_{n=1}^{\infty}na_nx^{n-1}-4\sum_{n=0}^{\infty}a_nx^n\)
\(\displaystyle =\sum_{n=0}^{\infty}(n+2)(n+1)a_{n+2}x^n\)
\(\displaystyle \ \ \ -2x\sum_{n=0}^{\infty}na_nx^n-4\sum_{n=0}^{\infty}a_nx^n\)
\(\displaystyle =\sum_{n=0}^{\infty}\{(n+2)(n+1)a_{n+2}-2na_n-4a_n\}x^n\)
\(x^n\)の係数が\(0\)にならなければならないので、
\((n+2)(n+1)a_{n+2}-2na_n-4a_n=0\)
\(\displaystyle a_{n+2}=\frac{2}{n+1}a_n\)
初期条件より
\(y(0)=a_0=1\)
\(y'(0)=a_1=0\)
すなわち
\(\displaystyle a_{2m}=\frac{2}{2m-1}a_{2m-2}=\frac{2^m}{(2m-1)!!}\)
よって、
\(\displaystyle y=\sum_{n=0}^{\infty}a_nx^n\)
\(\displaystyle =\sum_{m=0}^{\infty}a_{2m}x^{2m}\)
\(\displaystyle =\sum_{m=0}^{\infty}\frac{2^m}{(2m-1)!!}x^{2m}\)
\(\displaystyle =\sum_{m=0}^{\infty}\frac{(2x^2)^m}{(2m-1)!!}\)