Chapter 11 - Series Solutions to Linear Differential Equations - 11.2 Series Solutions about an Ordinary Point - Problems - Page 740: 8

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Given: $y''+2x^2y'+2xy=0$ Let $y=\sum a_nx^n$, obtain: $y''+2x^2y'+2xy$$=\sum k(k-1)a_{k}x^{k-2}+2x^2\sum ka_{k}x^{k-1}+2x\sum a_kx^k\\ =2a_2+\sum [(k+2)(k+3)a_{k+3}+2(k+1)a_k]x^{k+1}\\ =0$ Comparing $x^k$ on both sides: $2a_2=0\\ a_{k+3}=-\frac{2(k+1)}{(k+2)(k+3)}a_k$ It gives: $a_3=-\frac{1}{3}a_0\\ a_6=\frac{4}{45}a_0\\ a_9=-\frac{7}{405}a_0$ and: $a_3=-\frac{1}{3}a_1\\ a_7=\frac{5}{63}a_1\\ a_{10}=-\frac{8}{567}a_1$ Therefore: $f(x)=a_0 (1+\frac{x^3}{3}+\frac{4x^6}{45}-\frac{7x^9}{405}+...)+a_1(x-\frac{x^4}{3}+\frac{5x^7}{63}-\frac{8x^{10}}{567}...)$