# Power Series Differential Equation

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!!

• That‘s just the $\sin$ function. – Frieder Jäckel Apr 8 '18 at 22:27
• How do you know that it is the sin function? My teacher did a really poor job at teaching us how to even recognize what function it is. – Chad Bowman Apr 8 '18 at 22:31
• You can differentiate power series termwise – Frieder Jäckel Apr 8 '18 at 22:34
• I did that and got ((2n+1)(-1)^n(x)^2n)/(2n+1)! but I am not sure where to start with my index. I have tested 0,1, and 2 but none of them work. – Chad Bowman Apr 8 '18 at 22:37

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 $$y''+y=0$$

• Thank you so much – Chad Bowman Apr 9 '18 at 1:23

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

$$y^{\prime\prime}+y=0$$

• Thank you for the help!! – Chad Bowman Apr 9 '18 at 1:24
• You are welcome. – John Wayland Bales Apr 9 '18 at 5:15