Let $T > 0$ and suppose $a$ and $b$ are continuous functions on $\mathbb{R}$ such that $a(t + T) \equiv a(t)$ and $b(t + T) \equiv b(t)$. Show that if $y$ is a solution to \begin{align} y' = a(t)y + b(t)\end{align} and $\bar{y}(t) \equiv y(t+T)$ for all $t\in\mathbb{R}$, then $\bar{y}$ is also a solution to this equation.
Let $y$ be a solution to the given equation and set $\bar{y}(t) \equiv y(t+T)$. Then,
$$ \bar{y}' = a(t)\bar{y} + b(t) \implies y(t+T) = a(t+T)y(t+T) + b(t+T) $$
but since $a$ and $b$ are periodic, then $$ y(t+T) = a(t)y(t+T) + b(t). $$
What must I connect from here?