For the Delayed Differential Equation(DDE) of neutral type
$y'(t) = y'(t - \tau) +a(y(t) - y(t - \tau))$
I am trying to obtain the characteristic equation and then solve it so that I can write the solution in terms of real solutions but I am having trouble finding some of the real solutions in one of the factors of the characteristic equation.
So far I have assumed that solution is of the form $y(t) = e^{\lambda t}$ and substituted this into the equation so that I can simplify it using these steps
$\lambda e^{\lambda t} = \lambda e^{\lambda (t - \tau)} +a(e^{\lambda t} - e^{\lambda(t - \tau)})$
$\lambda e^{\lambda t} = \lambda e^{\lambda t}e^{-\lambda \tau} +a(e^{\lambda t} - e^{\lambda t}e^{- \lambda \tau})$
Dividing both sides by $e^{\lambda t}$ and subtraction the LHS from the RHS I get
$0 = \lambda (e^{-\lambda \tau} - 1) + a(1 - e^{-\lambda \tau})$
Simplifying I then get the characteristic equation $(1 - e^{-\lambda \tau})(a - \lambda) = 0$
I understand one of the possible values of $\lambda$ that satisfy this equation is $\lambda = a$ but I am having trouble understanding what are the real solutions which satisfy the other factor of the equation $1 = e^{-\lambda \tau}$. In a simpler version of my problem when $1 = e^{- \lambda}$ the roots are $2\pi i n$ but I am trying to find only real solutions which satisfy $1 = e^{- \lambda \tau}$(assuming there are indeed real solutions other than the trivial case of $\lambda = 0)$. Any help would be much appreciated, thanks.
Note that $u(t)=y(t)−y(t−τ)$ satisfies $u'=au$ and thus has the solution $u(t)=Ce^{at}$. $$y(t)=y(t−τ)+Ce^{at}=y(t−2τ)+Ce^{a(t−τ)}+Ce^{at}=...=y(t−nτ)+Ce^{at}\frac{1-e^{-anτ}}{1-e^{-aτ}}.$$
Thus you can select any differentiable function for the restriction of $y$ to $[0,τ]$, determine $C=e^{-aτ}(y(τ)-y(0))$ and then for any $t=nτ+s$, $s\in[0,τ)$, the value of the solution function is $$y(t)=y(s)+\frac{e^{at}-e^{-as}}{e^{aτ}-1}(y(τ)-y(0)).$$