Answer to Question #130284 in Differential Equations for Muhammad Tayyab

Question

solve the initial value problem using power series
(x^2 +2)y"-xy'+y=0
Y(0)=7 ,y'(0)=9

Answer 1

$(x^2+2)y''-xy'+y=0\newline y=C_0x^\sigma+C_1x^{\sigma+1}+C_2x^{\sigma+2}+...\\ (x^2+2)*(C_0\sigma(\sigma-1)x^{\sigma-2}+C_1(\sigma+1)\sigma x^{\sigma-1}+...)-x*(C_0\sigma x^{\sigma-1}+C_1(\sigma+1)x^{\sigma}+...)+C_0x^\sigma+C_1x^{\sigma+1}+C_2x^{\sigma+2}+...=0\\ Let\space coefficients \space near \space x^k(where \space k=\sigma-2, \ldots,\infin)\space be\, equal \space to \space zero. \space \\Then \space we \space have \space the \space next\space system:\\ \begin{cases} 2C_0\sigma(\sigma-1)=0\\ 2C_1\sigma(\sigma+1)=0\\ C_0\sigma(\sigma-1)+(\sigma+2)(\sigma+1)C_2-C_0\sigma+C_0=0\\ ........\\ C_k(\sigma+k-1)^2+2C_{k+2}(\sigma+k+2)(\sigma+k+1)=0 \end{cases}\\ Let\space\sigma=0\space then:\\ C_k=\dfrac{(-1)^{k/2}C_0}{2^{k/2}k!}, \space if \space k \space is\,divisible \space by\, 2, and\space C_k=\dfrac{(-1)^{(k-1)/2}C_1}{2^{(k-1)/2}k!}\space \space otherwise\space .\\ y=\underset{k=0,\space k=k+2}{\sum}\dfrac{(-1)^{k/2}C_0}{2^{k/2}k!}x^k +\underset{k=1,\space k=k+2}{\sum}\dfrac{(-1)^{(k-1)/2}C_1}{2^{(k-1)/2}k!}x^k\newline y(0)=7=>C_0=7\\ y'(0)=9=>C_1=9\\ y=7\underset{k=0,\space k=k+2}{\sum}\dfrac{(-1)^{k/2}}{2^{k/2}k!}x^k +9\underset{k=1,\space k=k+2}{\sum}\dfrac{(-1)^{(k-1)/2}}{2^{(k-1)/2}k!}x^k\newline$