Chapter 11 - Series Solutions to Linear Differential Equations - 11.2 Series Solutions about an Ordinary Point - Problems - Page 740: 6

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Given: $y''+xy'+3y=0$ Let $y=\sum a_nx^n$, obtain: $y''+xy'+3y$$=\sum k(k-1)a_{k}x^{k-2}+x\sum ka_{k}x^{k-1}+3\sum a_kx^k\\ =2a_2+3a_0+\sum [(k+2)(k+1)a_{k+2}+(k+3)a_k]x^{k+1}\\ =0$ Comparing $x^k$ on both sides: $2a_2+3a_0=0\\ a_{k+2}=-\frac{k+3}{(k+1)(k+2}a_k$ It gives: $a_2=-\frac{3}{2}a_0\\ a_4=\frac{5}{8}a_0\\ a_6=-\frac{7}{48}a_0\\ a_{2k}=(-1)^{k+1}\frac{2k-1}{2^{k-1}(k-1)!}a_0$ and: $a_3=-\frac{2}{3}a_1\\ a_7=\frac{1}{5}a_1\\ a_{10}=-\frac{4}{105}a_1$ Therefore: $f(x)=a_0 (1-\frac{3x^2}{2}+\frac{5x^4}{8}-\frac{7x^6}{4}+...)+a_1(x-\frac{2x^3}{3}+\frac{x^5}{5}-\frac{4x^{7}}{105}...)$