Chapter 11 - Series Solutions to Linear Differential Equations - 11.2 Series Solutions about an Ordinary Point - Problems - Page 740: 13

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Given: $y''+2y'+4xy=0$ Let $y=\sum a_nx^n$, obtain: $y''+2y'+4xy$$=\sum k(k-1)a_{k}x^{k-2}+2\sum ka_{k}x^{k-1}+4\sum a_kx^k\\ =2a_2+2a_1+\sum [(k+2)(k+3)a_{k+3}+2(k+2)a_{k+2+4a_k}]x^{k+1}\\ =0$ Comparing $x^k$ on both sides: $a_2=-a_1\\ a_{k+3}=-\frac{2}{(k+3)}a_{k+2}-\frac{4a_k}{(k+2)(k+3)}$ It gives: $a_2=-a_1\\ a_3=\frac{2a_1}{3}-\frac{2a_0}{3}\\ a_4=\frac{1}{3}(a_0-2a_1)\\ a_5=\frac{1}{15}(-2a_0+7a_1)$ and: $a_3=-\frac{1}{3}a_1\\ a_7=\frac{5}{63}a_1\\ a_{10}=-\frac{8}{567}a_1$ Therefore: $f(x)=a_0 (1-\frac{2x^3}{3}+\frac{x^4}{3}-\frac{2x^5}{15}+...)+a_1(x-x^2+\frac{2x^3}{3}-\frac{2x^4}{3}+\frac{7x^5}{15}...)$ The independent solutions are: $y_1(x)=1-\frac{2x^3}{3}+\frac{x^4}{3}-\frac{2x^5}{15}+...\\ y_2(x)=x-x^2+\frac{2x^3}{3}-\frac{2x^4}{3}+\frac{7x^5}{15}...$ By then, we have: $a_{k+3}=-\frac{2}{k+3}a_{k+2}-\frac{4a^k}{(k+2)(k+3)}$ Consequently, $\lim \frac{a_{k+3}}{a_{k+2}}=0$ Therefore, the radius of convergence of the series solution is $R=\infty$