Periodic solutions of a planar ODE Linearizing the equation of the two body problem at a circular solution I came accross the following planar second order system of differential equations
$$
\begin{cases}
2\frac{1}{\omega^2} \ddot{u}=3u\cos2\omega t+3v\sin2\omega t+u\\
2\frac{1}{\omega^2}\ddot{v}=3u\sin2\omega t-3v\cos2\omega t+v
\end{cases}
$$
It is easy to see that $(u,v)=(-\sin\omega t,\cos\omega t)$ is a $\frac{2\pi}{\omega}$-periodic solution. I'm wondering if there are other independent ones.
 Do you have any suggestion to find them?
 A: If we define $\eta = u + i v$, then the set of equations can be rewritten as 
$$
0 = - \frac{2}{\omega^2} \ddot{\eta} + 3 e^{2 i \omega t} \bar{\eta} + \eta.
$$
If $\eta$ is periodic with period $2 \pi/\omega$, then it will be expressible as a power series of the form
$$
\eta = \sum_{m = - \infty}^\infty a_m e^{i m \omega t}.
$$
Putting this ansatz into the ODE and manipulating the series appropriately, we find that if $\eta$ is of this form and satisfies the above ODE, then we must have
$$
(1 + 2 m^2) a_m + 3 a^*_{2-m} = 0
$$
for all $m$.
This "recursion" relation relates the coefficients $a_m$ in pairs;  it relates $a_1$ to itself, $a_0$ to $a_2$, $a_{-1}$ to $a_3$, and so forth.  Note that all of these pairs of coefficients are determined independently:  the value of $\{a_0, a_2\}$ are unaffected by the values of $\{a_{-1}, a_3\}$ or any other pair.  Moreover, if we set $m \to 2-m$ in the above recursion relation and conjugate it, we obtain
$$
(1 + 2(2-m)^2) a^*_{2-m} + 3 a_m = 0,
$$
and combining the above two equations yields
$$
\left[ (1 + 2m^2) (1 + 2(2-m)^2) - 9 \right] a_m = 0
$$
for all $m$.  Assuming we want $a_m \neq 0$, this implies the quantity in brackets must vanish.  But the only roots of this polynomial are $m = 0, 1, 2$ (with 1 a double root).  Thus, the only $a_m$ coefficients that can be non-zero are $a_0$, $a_1$, and $a_2$.  Taking these cases separately:


*

*For $a_1 \neq 0$, the recursion relation is $3 a_1 + 3 a^*_1 = 0$, which implies that $a_1$ is pure imaginary.  Thus, the solution is $\eta = A i e^{i \omega t}$ for $A \in \mathbb{R}$, which (taking the real and imaginary parts) yields the solution you found:  $u(t) = - A \sin(\omega t)$, $v(t) = A \cos (\omega t)$. 

*For the pair $\{a_0, a_2\}$, we have $$a_0 = - 3 a_2^*.$$  Thus, we have a general solution of the form
$$
\eta(t) = a_0 - \frac{1}{3} a_0^* e^{2 i \omega t}
$$
for an arbitrary complex number $a_0$.  Expressing this in terms of two real coefficients $a_0 = A + iB$, the solution then becomes
$$
u(t) = A\left( 1 - \frac{1}{3} \cos (2 \omega t)\right) - \frac{B}{3} \sin (2 \omega t), \\
v(t) = B\left( 1 + \frac{1}{3} \cos (2 \omega t) \right) - \frac{A}{3} \sin (2 \omega t). \\
$$
The solutions found by Robert Israel correspond to $B = -3, A = 0$ and $A = -3, B = 0$ respectively.
These three independent solutions, and linear combinations thereof, are the only possible solutions with a period of $2\pi/\omega$.
A: Try $$u(t) = \sin(2 \omega t),\ v(t) = -\cos(2 \omega t) - 3$$
and
$$ u(t) = \cos(2 \omega t)-3,\ v(t) = \sin(2\omega t)$$
A fourth fundamental solution is non-periodic:
$$ u(t) = 3 t \omega \sin(\omega t) - 2 \cos(\omega t),\ v(t) = 3 t \omega \cos(\omega t) - 2 \sin(\omega t)$$
