Consider the linear system $$x_1'=-\nu x_1 + a(t) x_1, ~~~~x_2'=-\nu x_2 + k a(t) x_1,$$ where $a(t+T)=a(t)$ for all $t$ and $\nu$ and $k$ are positive constants.
Find the characteristic exponents and show that there is a periodic solution of period $T$ if and only if $\int_{0}^{T} a(t)~dt=\nu T.$
My approach: Writing the system in the form $x'(t)=A(t)x(t),$ with \begin{bmatrix} -v+a(t) & 0 \\ k a(t) & -v \end{bmatrix}
Let $X(t)$ be the fundamental matrix solution of the given system (so $X(0)=E,$ the Identity matrix). i.e., $$X'(t)=A(t)X(t).$$ Let us define a constant matrix by $X(T+t)=X(T)B$ for all $t.$ Then by the choice of $X(t),~B=X(T).$ Now the characteristic multipliers, $\rho$ are the eigen values of $B.$ And the characterisitc exponents, $\mu$ are defined by $\rho=e^{\mu T}.$
I solved the two ODEs and got $$x_1=C e^{-\nu t} e^{\int_{0}^{t} a(s)~ds},~~~~x_2=e^{-\nu t} k a(t)C e^{\int_{0}^{t} a(s)~ds}+D,$$ where $C, D$ are arbitrary constants. I'm stuck here. Any help in continuing from here on ward and completing the problem is much appreciated.
I guess it is clear to you that if $$\int_{0}^{T} a(t)\,dt=\nu T,$$ then there is a periodic solution, right? (Simply compute $x(t+T)$ and use this identity.)
For the other direction, write the condition $x(t+T)=x(t)$ already using your solution and simplify. For example, the first equation gives $$ e^{-\nu (t+T)} e^{\int_{0}^{t+T} a(s)\,ds}=e^{-\nu t} e^{\int_{0}^{t} a(s)\,ds}, $$ which is the same as $$ e^{-\nu T} e^{\int_{t}^{t+T} a(s)\,ds}= 1. $$ Now use the fact that $a$ is $T$-periodic.