I am doing this problem $y'-2xy=0$. I am solving it using power series, and i am just questioning about one thing. I desire to know how to set the indices equal to each other to solve it using power series.
This is what I have set up \begin{equation}\sum_{n=1}^\infty nc_nx^{n-1}-2x\sum_{n=0}^\infty c_n x^{n}\end{equation}
That's good. $$\begin{equation}\sum_{n=1}^\infty nc_nx^{n-1}-2x\sum_{n=0}^\infty c_n x^{n}\end{equation}=0$$ $$\begin{equation}\sum_{n=1}^\infty nc_nx^{n-1}-2\sum_{n=0}^\infty c_n x^{n+1}\end{equation}=0$$ Change the indices: $$\begin{equation}\sum_{n=0}^\infty (n+1)c_{n+1}x^{n}-2\sum_{n=1}^\infty c_{n-1} x^{n}\end{equation}=0$$ $$c_1+\begin{equation}\sum_{n=1}^\infty (n+1)c_{n+1}x^{n}-2\sum_{n=1}^\infty c_{n-1} x^{n}\end{equation}=0$$ Now you have all the series starting at $n=1$ $$c_1+\begin{equation}\sum_{n=1}^\infty ((n+1)c_{n+1}-2 c_{n-1} )x^{n}\end{equation}=0$$ Hence , $$ \begin{cases} c_1=0 \\ (n+1)c_{n+1}=2 c_{n-1} \text { for } n \ge 1 \end{cases} $$ $$\implies c_{2n}=\dfrac {c_0}{n!}$$ $$\boxed {y(x)= \sum_{n=0}^\infty c_{2n}x^{2n}=\sum_{n=0}^\infty \dfrac {c_0x^{2n}}{n!}=c_0e^{x^2}}$$