How do I solve $L(L(y)) = 0$ where, $ L(y) = y^{(n)} + y$ for any $ y(t) \in C^{(n)}(\mathbb{R})$?
If I expand $$L(L(y)) = (y^{(n)}+y)^{(n)} + y^{(n)} + y = y^{(2n)} + 2y^{(n)} + y$$ then I don't even know if $y^{(n)}(t)$ is further differentiable.
If I try compute characteristic polynomial I get $\lambda^n=-1$, what has complex solutions, but I'm working on the polynomial of the regular expression, not the one calling itself.
The characteristic equation is $$ \bigl(\lambda^n+1\bigr)^2=0. $$ It has $n$ double roots, the $n$-th roots of $-1=e^{\pi i}$, which are $$ e^{\bigl(\tfrac{\pi}{n}+\tfrac{2k\pi}{n}\bigr)i}=\cos\Bigl(\frac{(2\,k+1)\pi}{n}\Bigr)+i\sin\Bigl(\frac{(2\,k+1)\pi}{n}\Bigr),\quad 0\le k\le n-1. $$ This gives $2\,n$ independent solutions for $0\le k\le n-1$: $$ e^{\cos\bigl(\tfrac{(2\,k+1)\pi}{n}\bigr)t}\cos\Bigl(\sin\Bigl(\frac{(2\,k+1)\pi}{n}\Bigr)\,t\Bigr),\quad e^{\cos\bigl(\tfrac{(2\,k+1)\pi}{n}\bigr)t}\sin\Bigl(\sin\Bigl(\frac{(2\,k+1)\pi}{n}\Bigr)\,t\Bigr),\\ t\,e^{\cos\bigl(\tfrac{(2\,k+1)\pi}{n}\bigr)t}\cos\Bigl(\sin\Bigl(\frac{(2\,k+1)\pi}{n}\Bigr)\,t\Bigr),\quad t\,e^{\cos\bigl(\tfrac{(2\,k+1)\pi}{n}\bigr)t}\sin\Bigl(\sin\Bigl(\frac{(2\,k+1)\pi}{n}\Bigr)\,t\Bigr). $$ The general solution is a linear combination of them.