# Eigenvalue of one for the inhomogeneous equation

I am trying to solve the problem from Gerald Teschl's book "Ordinary Differential Equations and Dynamical Systems" on p. 97:

\begin{equation*} x^{\prime}(t) = A(t)x(t) + g(t) \end{equation*}

where both A(t), g(t) periodic of period T. Show that this equation has a unique period of period T if and only if 1 is not an eigenvalue of the monodromy matrix $M(t_0)$.

(Hint: Note that x(t) is periodic if and only if x(T)=x(0) and use the variation of constants formula)

A strict application of the formula gives me:

\begin{equation*} x(T)= M(t_0) x_0 + \int_0^T \Pi(T, s) g(s) ds = x_0 \end{equation*}

I will be tempted to conclude that if the integral is 0 then $det(M(t_0) - I)=0$ hence 1 should not be an eigenvalue of $M(t_0)$.

I do not see why the integral is zero and if this is the right approach.

Why should the integral be $0$? If $1$ is not an eigenvalue, then $I-M(t_0)$ is invertible and hence $$x_0 = (I-M(t_0))^{-1}\int_0^T \Pi(T, s) g(s) ds$$ is the solution you are looking for.