Consider the initial value problem $$xu''+\sin(x)u=0 \ \ \ \ u(0)=0, u'(0)=2$$ What can be said about the radius of convergence of this series?
I have determined that the first four nonzero terms (about the point $x=0$) of the power series are $$2x,-x^2,\frac{1}{6}x^3,-\frac{1}{72}x^4$$ How can I test for convergence? Intuitively it looks not to converge, but how can I show this?
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The solution will have the from $$u(x)=\sum_{k=0}^{\infty}A_kx^k.$$ Differentiating, \begin{align} \sum_{k=2}^{\infty} k(k-1)A_kx^{k-1}+\sin(x)\sum_{k=0}^{\infty}A_kx^k&=0 \\ \sum_{k=2}^{\infty} k(k-1)A_kx^{k-1}+\left(\sum_{k=0}^{\infty} \frac{(-1)^k x^{2k+1}}{(2k+1)!}\right)\sum_{k=0}^{\infty}A_kx^k&=0 \\ (2A_2x+6A_3x^2+12A_4x^3+20A_5x^4+30A_6x^5+42A_7x^6+..)+\left(2x^2+A_2x^3+\left(A_3-\frac{1}{3}\right)x^4+\left(A_4-\frac{A_2}{6!}\right)x^5+\left(A_5-\frac{A_3}{3!}+\frac{2}{5!}\right)x^6+..\right)=0 \end{align} Equating coefficients, $x^{2k}=0$ for $k\in\mathbb{Z^+}$. $$6A_3+2=0\implies A_3=-\frac{1}{3}$$ $$20A_5+A_3-\frac{1}{3}=0\implies A_5=\frac{1}{30}$$ $$42A_7+A_5-\frac{A_3}{3!}+\frac{2}{5!}=0\implies A_7=-\frac{19}{7560}$$ Hence $$u=2x-\frac{1}{3}x^3+\frac{1}{30}x^5-\frac{19}{7560}x^7+..$$