I converted this ode into a linear matrix form like $y'=Ay$ and tried to solve this, but I couldn't find a fundamental solution which satisfies $\Phi (0)=I$, which is required in one of my assignment questions. Is it possible? Because I think there is no such fundamental matrix.
Btw, $f(t)=1$ if $0<=t<\pi$, 0 if $\pi <=t<2\pi$, and it is $2\pi$-periodic.
Note that the character of the solution depends on the sign of $\epsilon$.
Let $A_t = \begin{bmatrix} 0 & -t \\ 1 & 0 \end{bmatrix}$. You need only evaluate $\Phi(t,0)$ for $t \in [0, 2\pi)$.
Then if the initial condition is $y_0 \in \mathbb{R}^2$, the solution for $t \in [0,\pi)$ is given by $e^{A_\epsilon t} y_0$, and the value for $t \in [\pi,2 \pi)$ will be $e^{A_0 (t-\pi)} e^{A_\epsilon \pi} y_0$, etc, etc.
Hence $\Phi(t,0) = \begin{cases} e^{A_\epsilon t}, & t \in [0,\pi) \\ e^{A_0 (t-\pi)} e^{A_\epsilon \pi}, & t \in [\pi, 2 \pi) \end{cases}$.
It is easy to see that $\Phi(0,0) = I$.
If $t = 2 \pi n + \theta$, with $\theta \in [0,2 \pi)$, then we see $\Phi(t,0) = \Phi(\theta, 0) \Phi(0, 2 \pi)^n$.
Finally, since $\Phi(t,0) = \Phi(t, t_0) \Phi(t_0, 0)$, we have $\Phi(t, t_0) = \Phi(t,0) \Phi(t_0, 0)^{-1}$.