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I have a general question about this following problem \begin{equation} y''+y=0\end{equation} The required method to solve this problem is based on generating the power series solution, using the power series method. I began my problem by setting up the summations. \begin{align}\sum_{n=2}^\infty(n)(n-1)c_nx^{n-2}+\sum_{n=0}^\infty x^nc_n&=0\end{align} Then I did two substitutions, and then got to the following equation: \begin{equation}\sum_{k=0}^\infty[(k+2)(k+1)c_{k+2}+c_k]x^k=0\end{equation} Then I set the part that I know could zero out which was the inner portion of the sum: \begin{equation}(k+2)(k+1)c_{k+2}+c_k=0 \end{equation} Then I get the following equation: \begin{equation} c_{k+2}=-\frac{c_k}{(k+2)(k+1)}\end{equation} After that I decided to do the following and break it apart into a table: \begin{array}{|c|c|}k=0&k=1\\c_2=-\frac{c_0}{2\cdot1}&c_3=-\frac{c_1}{3\cdot2\cdot1}\\ \hline k=2 & k=3 \\ c_4 = \frac{c_o}{4\cdot 3\cdot 2 \cdot 1} & c_5 = \frac{c_1}{5\cdot 4\cdot 3\cdot 2\cdot 1}\end{array} Based off the pattern I that sgn changes, and that there is factorial in the denominator is my attempt right thus far, and how to collaborate them into a power series solution?

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  • $\begingroup$ Are you just trying to find a formula for the general $c_k$ term? You seem to have a good idea of the formula, so why not use induction? $\endgroup$
    – user771918
    Apr 21, 2020 at 18:00
  • $\begingroup$ In my note book I figured out its sine and cosine but I get lost with how to isolate them into those two solutions $\endgroup$ Apr 21, 2020 at 18:01
  • $\begingroup$ You need to use Taylor series of sine and cosine functions +1 $\endgroup$ Apr 21, 2020 at 18:18

2 Answers 2

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From your work it follows inductively that for $n\geq0$, $$c_{2n}=\frac{(-1)^n c_0}{(2n)!}, \qquad c_{2n+1}=\frac{(-1)^n c_1}{(2n+1)!} .$$ So we find that $$\begin{align}y(x)&=\sum_{n=0}^\infty c_nx^n =\sum_{n=0}^\infty c_{2n}x^{2n}+\sum_{n=0}^\infty c_{2n+1}x^{2n+1}\\ &=c_0\sum_{n=0}^\infty \frac{(-1)^nx^{2n}}{(2n)!}+c_1\sum_{n=0}^\infty \frac{(-1)^nx^{2n+1}}{(2n+1)!}=c_0\cos(x)+c_1\sin(x).\end{align}$$

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  • $\begingroup$ Yeah its at ordinary points so your answer is right I am just waiting. $\endgroup$ Apr 21, 2020 at 18:05
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Your attempt looks correct to me. You can use the Taylor series definition of: $$\cos (x)=\sum_{n=0}^\infty (-1)^n\dfrac {x^{2n}}{(2n)!}$$ $$\sin (x)=\sum_{n=0}^\infty (-1)^n\dfrac {x^{2n+1}}{(2n+1)!}$$

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