= Solution
The Euler–Lagrange equations are
$$
\ddot x+\omega\cos(\omega t)=0,\qquad \ddot y+y=0.
$$
For $\omega\ne0$ their general solution is
$$
x=c_1t+c_2+\frac{\cos(\omega t)}\omega,\qquad
y=c_3\cos t+c_4\sin t,
$$
while for $\omega=0$, $x=c_1t+c_2$.
The action has continuous translation symmetry in $x$; it is also invariant up to endpoint terms under adding homogeneous Jacobi solutions $at+b$ to $x$ and $a\cos t+b\sin t$ to $y$. For $\omega=0$ it additionally has continuous time-translation symmetry; for nonzero $\omega$ only the corresponding discrete period remains.
A complete set of four independent first <integrals> is
$$
C_1=\dot x+\sin(\omega t),
$$
$$
C_2=x-tC_1-\frac{\cos(\omega t)}\omega\quad(\omega\ne0),
\qquad C_2=x-t\dot x\quad(\omega=0),
$$
$$
C_3=y\cos t-\dot y\sin t,\qquad C_4=y\sin t+\dot y\cos t.
$$
In particular $(\dot y^2+y^2)/2=(C_3^2+C_4^2)/2$ is conserved; when $\omega=0$, the usual total energy is conserved as well.
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