My Calc 2 teacher did not really teach us how to do these and I have a test on it tomorrow. Any help you can provide to help me understand this type of problem would be much appreciated.
The question:
Show that the given power series is a solution to the differential equation $$y''+y=0$$ $$y=\sum_{n=0}^{\infty}(-1)^n\frac { x^{2n+1}}{(2n+1)!}$$ Thank you in advance!!
On
Suppose you just write out the first few terms
\begin{eqnarray} y&=&x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots\\ y^\prime&=&1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\cdots\\ y^{\prime\prime}&=&-x+\frac{x^3}{3!}-\frac{x^5}{5!}+\frac{x^7}{7!}-\\ \end{eqnarray}
Then notice that
$$ y^{\prime\prime}+y=0$$
Hint
$$y=\sum_{n=0}^{\infty}(-1)^n\frac { x^{2n+1}}{(2n+1)!}$$ $$y'=\sum_{n=0}^{\infty}(-1)^n\frac { x^{2n}}{(2n)!}$$ $$y''=\sum_{n=1}^{\infty}(-1)^n\frac { x^{2n-1}}{(2n-1)!}$$ Substitute $n=m+1$ $$y''=\sum_{m=0}^{\infty}(-1)^{m+1}\frac { x^{2m+1}}{(2m+1)!}$$ $$y''=-\sum_{m=0}^{\infty}(-1)^{m}\frac { x^{2m+1}}{(2m+1)!}$$ Therefore $$y''=-y$$ And $$y''+y=0$$