Differential equation $y'(t) = -\Big(a+y(t+a)-y(t)\Big)_+$ I would like to solve the following differential equation on $[0,T]$ (or on $[0,+\infty]$ if it makes sense):
$$y'(t) = -\Big(a+y(t+a)-y(t)\Big)_+,$$
where $a$ is a positive constant and $y(t) = 0$ for all $t\geq T$ (or $y(+\infty) =0$).
 A: This delay-differential equation is linear.
By inspection, a solution of the inhomogeneous equation is
$$y(t)=-\frac a{a+1}t.$$
Then with $y(t):=z(t)-\dfrac a{a+1}t$,
$$z'(t)+z(t+a)-z(t)=0.$$
We try $e^{bt}$, giving the compatibility condition
$$b+e^{ba}-1=0.$$
Finally,
$$y(t)=ce^{bt}-\frac a{a+1}t$$ where $b$ is the solution of the above condition.
I do not guarantee that these are the only solutions.
A: Some points concerning the delayed differential equation.
Considering $t \ge  a$ and using the Laplace transform we have
$$
sY(s)+e^{-a s}Y(s) -Y(s) = y_0-a
$$
or
$$
Y(s) = \frac{y_0-a}{s+e^{-a s}-1}
$$
now looking at the poles for $Y(s)$ we have with $s = x + i y$
$$
s+e^{-a s}-1 = 0\Rightarrow \cases{x+e^{-a x}\cos(a y) - 1 = 0\\ y - e^{-a x}\sin(a y) = 0}
$$
Follows a plot showing at the intersections for blue and red paths, the pole locations, for $a = 1$. As we can observe, the behavior is rather complex. The pole locations depend on the $a$ values. From those considerations we can conclude also about stability. Note the existence of a double pole at the origin as well as infinite more inside the left complex semi-plane. 

