We consider the linear differential equation $y''+y=0$.
I want to show that for any non trivial solution of the equation $y$, for any point $x_0\in \mathbb{R}$ $$y\sim c(x-x_0)^n, \ x\rightarrow x_0$$ for non zero real constant $c$ and non negative integer constant $n$.
Then I want to construct the particular solution of the equation that satisfies the conditions $y(0)=0$ and $y'(0)=1$ as a taylor series at the point $0$.
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For this to we search for the solution $y(x)$ that is represented as a power series?
Or can we not just use this and we have to prove this?
If we can use this we have that $$y(x)=\sum_{n=0}^\infty a_n(x-x_0)^n$$ Differentiating this twice and replacing the result into the DE we get a relation for the $a_n$ : $$a_{n+2}=-\frac{a_n}{(n+2)(n+1)}, \ n=0, 1, 2, \ldots $$ Then we get that $$y(x)=a_0\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n)!}(x-x_0)^{2n}+a_1\frac{(-1)^n}{(2n+1)!}(x-x_0)^{2n+1}$$ Is this correct?
Do we prove in that way the desired statement?
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For the seond part do we do the same just use the conditions?
From the given conditions we get $a_0=0$ and $a_1=1$, right?
Therefore we get from $$y(x)=a_0\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n)!}(x-x_0)^{2n}+a_1\frac{(-1)^n}{(2n+1)!}(x-x_0)^{2n+1}$$ the solution $$y(x)=\frac{(-1)^n}{(2n+1)!}(x-x_0)^{2n+1}$$ Is that correct?