How do you determine the series solution to $ y'=y $ or $ y''=-y$ around an arbitrary point, but I would love to see an example around the point 1. I know the solution is $ c_{1}e^{x} $ and $c_{1} \sin(x)+c_{2} \cos(x)$ but when I try to do it with power series I always get the same aproximation just moved to the right or left on the graph.
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Suppose that $y$ has a power series $$ y(x)=\sum_{n=0}^{\infty}c_n(x-1)^n $$ about the point $x=1$. Then the power series for $y'(x)$ is $$ y'(x)=\sum_{n=1}^{\infty}nc_n(x-1)^{n-1}=\sum_{n=0}^{\infty}(n+1)c_{n+1}(x-1)^n. $$ If $y=y'$, then we must have that all of the coefficients agree... so, from the coefficients of $(x-1)^n$ in both $y$ and $y'$, we find that for $n\geq 0$, $$ c_n=(n+1)c_{n+1}. $$ Rearranging, this says that $c_{n+1}=\frac{c_n}{n+1}$. Inductively, you can prove that this implies $$ c_n=\frac{c_0}{n!}, $$ so that $$ y(x)=\sum_{n=0}^{\infty}\frac{c_0}{n!}(x-1)^n=c_0e^{x-1}. $$ (Note that this isn't different from your answer, as we can absorb the extra $e^{-1}$ in to the constant.)
Nick Peterson
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