How prove this ordinary equations $x''+2ax'+x=f(t)$ has a unique $2\pi$-periodic solution if and only if $a\neq 0$ let $f(t)$ be a continuous and $2\pi$-periodic function,and let $a$ be a constant,show that:
the ordinary equation $$x''+2ax'+x=f(t)$$ has a unique $2\pi$-periodic solution if and only if $a\neq 0$
My idea: let  $$f(t+2\pi)=f(t),t\in R$$
if $a\ge 1 $,then 
$$r^2+2ar+1=0,r_{1}=\sqrt{a^2-1}-a,r_{2}=-\sqrt{a^2-1}-a$$
and then
$$x''+2ax'+x=x''-(r_{1}+r_{2})x'+r_{1}r_{2}x=x''-r_{1}x'-r_{2}(x'-r_{1}x)=f(t)$$
so
$$(x'-r_{1}x)'-r_{2}(x'-r_{1}x)=f(t)$$
let $z=x'-r_{1}x$,then
$$z*=e^{r_{2}t}\int f(t)e^{-r_{2}t}dt$$
then straightforward calculation,we have
$$x*=\dfrac{1}{r_{2}-r_{1}}\left[e^{r_{2}t}\int f(t)e^{-r_{2}t}dt-e^{r_{1}t}\int f(t)e^{-r_{1}t}dt\right]$$
then I can't
Thank you for you help!
 A: I am assuming that $a$ is real.
Write the system as $\dot{x} = A x + bf$, where $A=\begin{bmatrix} -2 a & -1 \\ 1 & 0\end{bmatrix}$ and $b=
 \begin{bmatrix} 1 \\ 0 \end{bmatrix}$.
The general solution starting from an initial condition $x_0$ (at $t=0$) is given by
$x(t) = e^{At} \left(  x_0 + \int_0^t e^{- A \tau} f(\tau) dt \,b\right)$, and we see that there is a unique $2 \pi$-periodic solution iff there is a unique solution (that is, a unique $x_0$ such that the following equation holds) to the equation $x(2 \pi) = x_0$.
In other words, a unique solution exists iff there exists a unique $x_0$ satisfying $(I-e^{2 \pi A}) x_0 = \int_0^{2 \pi} e^{- A \tau} f(\tau) dt \,b$.
The eigenvalues of $A$ are $\lambda_\pm = -a \pm \sqrt{a^2-1}$.
Aside: Note that the only solution to $\lambda_\pm = ni$ ($n$ an integer) is $a=0, n=1$. This follows from squaring both sides of the equation $a+ni = \pm \sqrt{a^2-1}$, which reduces to $n(n-2 ai) = 1$.
First deal with $a=0$, which results in a harmonic oscillator. We have
$e^{At} = \begin{bmatrix} \cos t & - \sin t \\ \sin t & \cos t \end{bmatrix}$, and so $e^{2 \pi A} = I$.
The equation $x(2 \pi) = x_0$ reduces to $J=\int_0^{2 \pi} e^{- A \tau} f(\tau) dt \,b = 0$, so either $J \neq 0$ in which case there is no periodic solution, or $J = 0$ in which case any $x_0$ will result in a periodic solution. Hence there is no unique periodic solution (be careful with the wording here).
If $a \neq 0$, then we note that the eigenvalues of $e^{2 \pi A}$ are $e^{2 \pi \lambda_\pm}$, and from the aside above, we note that $e^{2 \pi \lambda_\pm} \neq 1$, and so $(I-e^{2 \pi A})$ is invertible. In particular, the unique solution to the equation $x(2 \pi) = x_0$ is given by
$x_0 = (I-e^{2 \pi A})^{-1} \int_0^{2 \pi} e^{- A \tau} f(\tau) dt \,b$.
A: Under the assumption that $f(t)$ and $x(t)$ are $2\pi$-periodic and sufficiently smooth, represent them as
\begin{align}
f(t)&=\sum_{k\in\Bbb Z} \hat f_k\,e^{ikt}\\
x(t)&=\sum_{k\in\Bbb Z} \hat x_k\,e^{ikt}\\
x'(t)&=\sum_{k\in\Bbb Z} ik\hat x_k\,e^{ikt}\\
x''(t)&=\sum_{k\in\Bbb Z} (-k^2)\hat x_k\,e^{ikt},
\end{align}
insert in differential equation and compare coefficients of the same exponential, then
$$
(-k^2+2ika+1)\hat x_k=\hat f_k
$$
which for $k=\pm1$ is only solvable for $\hat x_k$ if the factor before it is different from zero, i.e., if $a\ne 0$.
