Answer to Question #289762 in Differential Equations for Ramya

Question

Find a power series solution of xy'=y

Answer 1

$\text{Let}\ y=\sum_{n=0}^{\infty}a_nx^n, \\ \text{Then}, y\prime=\sum_{n=1}^{\infty}na_nx^{n-1}, \\ \text{Therefore}, xy\prime=y\ \Rightarrow x\sum_{n=1}^{\infty}na_nx^{n-1}=\sum_{n=0}^{\infty}a_nx^n\\ \Rightarrow \sum_{n=1}^{\infty}na_nx^n=\sum_{n=0}^{\infty}a_nx^n\\ \Rightarrow\sum_{n=0}^{\infty}a_nx^n-\sum_{n=1}^{\infty}na_nx^n=0\\ \Rightarrow a_0+\sum_{n=1}^{\infty}a_nx^n-\sum_{n=1}^{\infty}na_nx^n=0\\ \Rightarrow a_0+\sum_{n=1}^{\infty}a_n(1-n)x^n=0\\ \Rightarrow \sum_{n=0}^{\infty}a_n(1-n)x^n=0\\ \Rightarrow \sum_{n=0}^{\infty}a_n(1-n)x^n=\sum_{n=0}^{\infty}(0 \times x^n)\\ \Rightarrow a_n(1-n)=0, \text{for}\ n\in\N\\ \Rightarrow a_n=\frac{0}{(1-n)}=0, \text{if}\ n\neq1\\ \Rightarrow a_1\in\R\\ \text{Hence, }y=\sum_{n=0}^{\infty}a_nx^n=a_0+a_1x+a_2x^2+\cdots=a_1x\\ \text{Therefore, the series solution to the given equation is:}\\ y=a_1x, \text{where } a_1\in \R.$