Chapter 11 - Series Solutions to Linear Differential Equations - 11.2 Series Solutions about an Ordinary Point - Problems - Page 740: 7

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Given: $y''-x^2y'-3xy=0$ Let $y=\sum a_nx^n$, obtain: $y''-x^2y'-3xy$$=\sum k(k-1)a_{k}x^{k-2}-x^2\sum ka_{k}x^{k-1}-3x\sum a_kx^k\\ =2a_2+\sum [(k+2)(k+3)a_{k+3}-(k+3)a_k]x^{k+1}\\ =0$ Comparing $x^k$ on both sides: $a_2=0\\ a_{k+3}=\frac{1}{k+2}a_k$ It gives: $a_3=\frac{1}{2}a_0\\ a_6=\frac{1}{10}a_0\\ a_9=\frac{1}{80}a_0$ and: $a_4=\frac{1}{3}a_1\\ a_7=\frac{1}{18}a_1\\ a_{10}=\frac{1}{162}a_1$ Therefore: $f(x)=a_0 (1+\frac{x^3}{2}+\frac{x^6}{10}+\frac{x^9}{80}+...)+a_1(x+\frac{x^4}{3}+\frac{x^7}{18}+\frac{x^{10}}{162}...)$