Chapter 11 - Series Solutions to Linear Differential Equations - 11.2 Series Solutions about an Ordinary Point - Problems - Page 740: 14

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Given: $y''+xy'+(2+x)y=0$ Let $y=\sum a_nx^n$, obtain: $y''+xy'+(2+x)y$$=\sum k(k-1)a_{k}x^{k-2}+\sum ka_{k}x^{k}+(a+x)\sum a_kx^k\\ =2a_2+2a_0+\sum [(k+2)(k+1)a_{k+2}+(k+2)a_{k}+a_{k-1}]x^{k}\\ =0$ Comparing $x^k$ on both sides: $a_2=-a_0\\ a_{k+2}=-\frac{1}{k+1}a_{k}-\frac{a_{k-1}}{(k+2)(k+1)}$ It gives: $a_2=-a_0\\ a_3=\frac{-a_0}{6}-\frac{a_1}{2}\\ a_4=\frac{1}{12}(4a_0+a_1)\\ a_5=\frac{1}{120}(11a_0+18a_1)\\ a_6=\frac{1}{120}(11a_0+18a_1)$ Therefore: $f(x)=a_0 (1-x^2-\frac{x^3}{6}+\frac{x^4}{3}+\frac{11x^5}{120}+...)+a_1(1-\frac{x^3}{2}+\frac{x^4}{12}-\frac{3x^5}{20}...)$ The independent solutions are: $y_1(x)=1-x^2-\frac{x^3}{6}+\frac{x^4}{3}+\frac{11x^5}{120}+...\\ y_2(x)=1-\frac{x^3}{2}+\frac{x^4}{12}-\frac{3x^5}{20}...$ Consequently, $\lim \frac{a_{k+1}}{a_{k}}=\frac{1}{R}$ Therefore, the radius of convergence of the series solution is $\frac{1}{R}=0 \rightarrow R=\infty$