Question

Use power series to solve this problem: 4xy''+2y'-y=0

Answer 1

\[y = \sum_{r=0}^{\infty}a_{r}x^{r}\] \[y' = \sum_{r=1}^{\infty}r.a_{r}x^{r-1} = \sum_{r=0}^{\infty}(r+1)a_{r+1}x^{r} \Rightarrow 2y' = 2\sum_{r=0}^{\infty}(r+1)a_{r+1}x^{r}\] \[y'' = \sum_{r=2}^{\infty}r(r-1).a_{r}x^{r-2} = \sum_{r=1}^{\infty}(r+1)r.a_{r+1}x^{r-1} \Rightarrow 4xy'' = 4\sum_{r=1}^{\infty}(r+1)r.a_{r+1}x^{r}\]
Try substituting these in you differential equation

Answer 2

substituting we get
\[\sum_{r=1}^{\infty}(4(r+1)r.a_{r+1} + 2(r+1).a_{r+1}-a_{r})x^{r} + 2a_{1}+a_{0} = 0\]
thus \(2a_{1}+a_{0} = 0\) and \[a_{r+1} = \frac{a_{r}}{2(r+1)(2r+1)}, r \ge 1\]