Stuck while using Laplace transform to solve delayed DE. I am studying delayed DE's, and while I've solved the following equation
$$y'(t) = y(t-1)$$ $$y[-1,0] = y_0$$ where $y_0$ is a constant by using the method of steps to get $$y(t) = \sum y_0\frac{t^n}{n!} + c$$ on $[n,n+1]$, I'm trying to now solve this using a Laplace transform. (Also, could someone confirm if the above is the correct answer? I remember following online notes, but it was a while ago, and I don't remember if this was supposed to be the final solution, since it's only defined on a specific interval?)
Anyway, until now, I've worked out the following by applying a Laplace transform to both sides of the equation. (I'll be skipping some more simple steps while writing this down though)
$$\int_0^\infty e^{-st}y'(t)dt = \int_0^\infty e^{-st} y(t-1)dt$$
$$sY(s) - Y(0) = \int_{-1}^\infty e^{-s(\tau+1)} y(\tau)d\tau$$
where $Y(s) = \mathcal{L}\{y(t)\}$, and $t-1 = \tau$.
I then continue and get
$$sY(s) - y_0 = e^{-s}\bigg[\int_{-1}^0 e^{-s\tau} y(\tau)d\tau +
\int_{0}^\infty e^{-s\tau} y(\tau)d\tau \bigg]$$
$$sY(s) - y_0 = e^{-s}\bigg[y(\tau)\frac{e^{-s\tau}}{-s}\bigg|^0_{-1} + 
\frac{1}{s}\int_{-1}^{0} y'(\tau)e^{-s\tau}d\tau +Y(s)\bigg]$$
where I used integration by parts to evaluate the first integral in the upper equation that's from $-1$ to $0$. This is where I'm stuck, because in the last equation, while I can evaluate the first part, and obviously simplify the rest of the equation, I'm not sure how to solve the middle integral. I tried letting it be equal to a random integral $I$ to use integration by parts again and get a value, but I ended up with a $0=0$ equation lol. I might be fuzzy on my calculus skills, so can someone help me evaluate that center integral? Or basically just this following integral overall $$\int_{-1}^0 e^{-s\tau} y(\tau)d\tau$$
I feel like it's probably something really simple I'm missing, but I would be super helpful if someone could help point it out. Thank you!!
 A: I have to make a necessary assumption in order to solve your differential equation using Laplace transform, that being $y(t)$ is a causal function, meaning we don't care about its history ($y=0$ if $t<0$). That being said, if we shift the function 1 unit to the right, it would still be zero for $t<1$, and the equation turns to:
$$y'(t)=y(t-1)H(t-1)$$
where $H(t)$ is the Heaviside or unit step function.
Taking Laplace transforms:
$$sY(s)-y_0 = Y(s)e^{-s}$$
$$sY(s)-Y(s)e^{-s}=y_0  $$
$$(s-e^{-s})Y(s)=y_0  $$
$$Y(s) = \frac{y_0}{s-e^{-s}} = \frac{y_0}{s} \; \frac{1}{1-\frac{e^{-s}}{s}} = \frac{y_0}{s} \sum_{n=0}^{+\infty} \left( \frac{e^{-s}}{s} \right)^n = y_0 \sum_{n=0}^{+\infty}  \frac{e^{-ns}}{s^{n+1}} $$
if $\left| \frac{e^{-s}}{s} \right|<1$
Now there's another assumption to make, and that is the inverse Laplace of a series is the series of the inverse Laplace (operator and series commute). It's not obvious, hard to prove, but widely used to solve for example, RLC circuits with periodic voltage sources.
Then,
$$\mathcal{L}^{-1} \left\{ \frac{e^{-ns}}{s^{n+1}} \right\} = \mathcal{L}^{-1} \left\{ \frac{1}{n!} \frac{n!}{s^{n+1}} e^{-ns} \right\} = \frac{1}{n!} \mathcal{L}^{-1} \left. \left\{ \frac{n!}{s^{n+1}} \right\}  \right|_{t-n} H(t-n) = \frac{(t-n)^{n}}{n!} H(t-n)$$
Resulting in
$$y(t) = y_0 \sum_{n=0}^{+\infty} \frac{(t-n)^{n}}{n!} H(t-n) = y_0 \left[ H(t) +(t-1)H(t-1) + \frac{1}{2} (t-2)^2 H(t-2) +...\right]$$
Where you can compute by hand or with GeoGebra that its derivative is a shifted version of itself. Here we've found a solution of the delayed differential equation restricting the solution to be causal, a requirement to use Laplace transforms in the first place.
