Lipschitz constant for $F(t,y,z)=(z,f(y,z)\sin t)$ 
Let $f\in C^1(\mathbb R^2,\mathbb R)$.
  Prove that all solutions for $x''=f(x,x')\sin t$ such that $x(0)=x(2\pi)$ and $x'(0)=x'(2\pi)$ have period $2\pi$.

I'm in the process of solving the above differential equation.
To do this I define $F\colon \mathbb R\times \mathbb R^2\to \mathbb R^2,F(t,y,z)=(z,f(y,z)\sin t)$.
Then the system is equivalent to $$F(x'(t), f(x(t),x'(t)))=(x'(t),x''(t))$$
Now in order to later be sure there is only one solution, I need to prove that
$\Vert F(t,y_1,z_1)-F(t,y_2,z_2)\Vert\leq K\Vert(y_1,z_1)-(y_2,z_2)\Vert$ for some $K>0$.
$$\Vert F(t,y_1,z_1)-F(t,y_2,z_2)\Vert=\Vert(z_1,f(y_1,z_1)\sin t)-(z_2,f(y_2,z_2)\sin t)\Vert=\Vert (z_1-z_2,(f(y_1,z_1)-f(y_2,z_2))\sin t)\Vert$$ and now I don't know what to do. I don't see how could I possibly related the last expression with $\Vert (y_1,z_1)-(y_2,z_2)\Vert$.
If you're feeling particularly charityful I wouldn't mind having some hints on the rest of the problem.
 A: To obtain uniqueness, it suffices to check that the Lipschitz condition
$$\Vert F(t,y_1,z_1)-F(t,y_2,z_2)\Vert\leq K\Vert (y_1,z_1)-(y_2,z_2)\Vert \tag1 $$
hold locally: that is, every point $(t,y_0,z_0)$ has a neighborhood in which (1) holds with some $K$ (which can depend on the point $(t,y_0,z_0)$ and on the size of neighborhood). 
Every continuously diffentiable function is locally Lipschitz. Indeed, because the derivatives in $y$ and $z$ are continuous, they are bounded in some rectangular neighborhood of $(t,y_0,z_0)$. Let $M$ be an upper bound for $|F_y|$ and $|F_z|$. The mean value theorem yields
$$\Vert F(t,y_1,z_1)-F(t,y_2,z_1)\Vert\leq M |y_1-y_2| \tag2 $$
$$\Vert F(t,y_2,z_1)-F(t,y_2,z_2)\Vert\leq M |z_1-z_2| \tag3 $$
From (2) and (3), 
$$\Vert F(t,y_1,z_1)-F(t,y_2,z_2)\Vert\leq M (|y_1-y_2|+|z_1-z_2|) \le 2M\|(y_1,z_1)-(y_2,z_2)\| \tag4 $$
which is the desired Lipschitz estimate. 
The particular form of $F$ does not matter here; all you have to check is that it has continuous  partial derivatives   in $y$ and in $z$.
