# Find the general solution of $y''+y=0$

This is brand new material that I'm trying to learn and it says:

To find a power series solution about the point $x = 0$, we write $$y(x)=\sum_{n=0}^{\infty}a_nx^n$$

So then I differentiate:

$$y'(x)=\sum_{n=1}^{\infty}na_nx^{n-1}=0$$

$$y''(x)=\sum_{n=2}^{\infty}n(n-1)a_nx^{n-2}=0$$

Then I equate them to $y''+y=0$:

$$y''+y=0 \Rightarrow \sum_{n=2}^{\infty}n(n-1)a_nx^{n-2} +\sum_{n=0}^{\infty}a_nx^n =0$$

But here is where I get confused, the example says that the following two are equivalent:

$$\sum_{n=2}^{\infty}n(n-1)a_nx^{n-2}= \sum_{n=0}^{\infty}(n+2)(n+1)a_{n+2}x^n$$

But no amount of algebra I do gets me there. Can you explain to me how they are equivalent?

Once I get to that point I'm supposed to be able to then combine the two terms to get:

$$\sum_{n=0}^{\infty}\left((n+2)(n+1)a_{n+2}+a_n\right)x^n=0$$

• For the left hand side of the equivalency, try changing the n=2 to n=0. – MITjanitor Mar 30 '16 at 0:50
• When $n = 0$, you get $2$ and $1$, respectively, for the leading coefficients. Likewise, when $n = 2$ you get $2$ and $2 - 1 = 1$ for the two leading coefficients. – Jared Mar 30 '16 at 0:51
• If I may suggest, life is easier if you do not change the index in any derivative. So, if $y=\sum_{n=0}^{\infty}a_nx^n$, $y'=\sum_{n=0}^{\infty}na_nx^{n-1}$, $y''=\sum_{n=0}^{\infty}n(n-1)a_nx^{n-2}$ – Claude Leibovici Mar 30 '16 at 7:43

All you're doing is changing the index. Try writing out the first few terms of the series and notice the equivalence.

Try finding an explicit form for the coefficients. You should notice that you will get something along the lines of

$$\frac{(-1)^{n}}{(2n+1)!}$$ for the odd terms and

$$\frac{(-1)^n}{(2n)!}$$ for the even terms.

An easy way to convince yourself of this fact is to write out the first few terms of each series:

$$\sum_{n=2}^\infty n(n-1)a_nx^{n-2}=2(2-1)a_2x^0+3(3-1)a_3x^1+4(4-1)a_4x^2+...$$ $$\sum_{n=0}^\infty (n+2)(n+1)a_{n+2}x^n=(0+2)(0+1)a_2x^0+(1+2)(1+1)a_3x^1+(2+2)(2+1)a_4x^2+...$$

Now it is not hard to see why they are equal. Another way to see this is to replace $n$ with $n+2$ in the first one.

To be rigorous, you do a change of indexes:

$$\sum_2^\infty n(n - 1)a_nx^{n - 2} = \sum_{m = m(2) = 0}^\infty n(m)(n(m) - 1)a_{n(m)}x^{n(m) - 2}$$

It should be clear that:

$$m = n - 2 \rightarrow n = m + 2$$

Such that:

$$\sum_{m = 0}^\infty (m + 2)(m + 2 - 1)a_{m + 2}x^{m + 2 - 2} = \sum_0^\infty (m + 2)(m + 1)a_{m + 2}x^m$$

$$\sum_{n=2}^{\infty}n(n-1)a_nx^{n-2}= \sum_{n=0}^{\infty}(n+2)(n+1)a_{n+2}x^n$$

It is just an index transformation, not unlike a substitution in integration: $n' = n - 2 \iff n = n' + 2$:

\begin{align} \sum_{n=2}^{\infty}n(n-1)a_nx^{n-2} &= \sum_{n'+2=2}^{\infty}(n' + 2) (n' + 2 - 1) a_{n'+ 2} x^{n'+ 2 -2} \\ &= \sum_{n'=0}^{\infty-2}(n' + 2)(n'+1)a_{n'+2}x^{n'} \\ &= \sum_{n'=0}^{\infty}(n' + 2)(n'+1)a_{n'+2}x^{n'} \end{align} The final step would be to rename $n'$ into $n$.

Let $$A_n=n(n-1)a_nx^{n-2}$$ so that $$A_{n+2}=(n+2)(n+1)a_{n+2}x^n;$$ you want to know why $$\sum_{n=2}^\infty A_n=\sum_{n=0}^\infty A_{n+2}.$$ Well, $$\sum_{n=2}^\infty A_n=A_2+A_3+A_4+A_5+A_6+\cdots$$ while $$\sum_{n=0}^\infty A_{n+2}=A_{0+2}+A_{1+2}+A_{2+2}+A_{3+2}+A_{4+2}+\cdots=A_2+A_3+A_4+A_5+A_6+\cdots.$$