Chapter 11 - Series Solutions to Linear Differential Equations - 11.2 Series Solutions about an Ordinary Point - Problems - Page 740: 3

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Given: $y''-2xy'-2y=0$ Let $y=\sum a_nx^n$, obtain: $y''-2xy'-2y$$=\sum k(k-1)a_{k}x^{k-2}-2\sum ka_{k}x^k-2\sum a_kx^k\\ =2a_2-2a_0+\sum [k(k+1)(k+2)a_{k+2}-2ka_k-2a_k]x^k\\ =0$ Comparing $x^k$ on both sides: $a_2=a_0\\ a_{k+2}=\frac{3k}{(k+1)(k+2)}a_k$ It gives: $a_2=a_0\\ a_4=\frac{1}{2}a_0\\ a_6=\frac{1}{6}a_0\\ a_{2k}=\frac{1}{k!}$ and: $a_3=\frac{2}{3}a_1\\ a_5=\frac{4}{15}a_1\\ a_7=\frac{8}{105}a_1\\ a_{2k+1}=\frac{2^k}{1.3.5...(2k+1)}$ Therefore: $f(x)=a_0\sum \frac{x^{2k}}{k!}+a_1 \sum \frac{2^{2k+1}(k+1)!x^{2k+1}}{(2k+1)!}$