Solution to non-autonomous delay differential equation? If you define a special function called the Lambert W function, you can explicitly solve the classic delay differential equation $x'(t) = x(t - a)$ by supposing the solution is some $\exp(\lambda t)$ and solving for the eigenvalues explicitly and construct a solution as $c_0\exp \left( \frac{W(a)}{a} t \right)$, along with their superposition as $\sum c_k\exp \left( \frac{W_k(a)}{a} t \right)$ where $W_k$ is the $k$th branch of the $W$ function.
My question is: how can we solve non-autonomous cases, like $x'(t) = t \cdot x(t - a)$, or more generally, $x'(t) = f(t) x(t - a) + g(t),$ for continuous functions $f, x$ and $g$?
 A: If you are okay with series solutions then a good starting point is to use the theory of first order linear functional equations.
We start with the easiest case which is we assume $g(t) \ne 0$ then we wish to solve:
$$ x'(t) - f(t) x(t-a) = g(t) $$
We can write this as
$$ x'(t+a) - f(t+a) x(t) = g(t+a) $$
This is a first order linear functional equation and thus has two primitive solutions. We begin by turning our attention to the slightly more general problem (so I can even remember what to do in the first place):
Suppose $O_1$, $O_2$ are arbitrary linear operators then we want to characterize the solutions of:
$$ O_1 [F(x)] + O_2[F(x)] = g(x) $$
The two primitive solutions are:
$$ F = O_1^{-1}[g] - O_1^{-1} \left[O_2 \left[O_1^{-1} \left[g \right] \right] \right] + O_1^{-1} \left[ O_2 \left[ O_1^{-1} \left[ O_2 \left[ O_1^{-1} [g] \right] \right]  \right]  \right] - ...   $$
$$ F = O_2^{-1}[g] - O_2^{-1} \left[O_1 \left[O_2^{-1} \left[g \right] \right] \right] + O_2^{-1} \left[ O_1 \left[ O_2^{-1} \left[ O_1 \left[ O_2^{-1} [g] \right] \right]  \right]  \right] - ... $$
Here the raised index denotes operator composition.
So in your case $O_1[x] = x'(t+a)$ and $O_2[x] = -f(t+a)x$. And we have that:  $O_1^{-1}[x] = \int \left[ x(t+a) \right]$ and $O_2^{-1}[x] = -\frac{1}{f(t+a)}x$
So the two primitive solutions to the equation
$$ x'(t+a) - f(t+a)x(t) = g(t+a) $$
are:
$$ x = \left( \int  g(t+a) \right) + \left( \int f(t+a) \int g(t+a) \right)  + \left( \int f(t+a) \int f(t+a) \int g(t+a) \right) + ...   $$
$$ x = - \frac{g(t+a)}{f(t+a)} - \frac{1}{f(t+a)}\left(\frac{g(t+2a)}{f(t+2a)} \right)'  - \frac{1}{f(t+a)} \left( \frac{1}{f(t+2a)} \left(\frac{g(t+3a)}{f(t+3a)} \right)'\right)' - ...  $$
So there you have it! (Note with the integrals the bounds can be from some arbitrary constant $\alpha$ that you get to pick to a variable $t$ ex: $\int_{\alpha}^{t}$)
Now we turn our attention to the fierce dragon that is $g(x)=0$. Unfortunately in this situation these formulas SOMETIMES break completely (i.e. they'll just give the trivial solution of $x=0, \infty$). If that does happen to you, i'm sorry for your luck but stay tuned :)
So to tackle this we again look at the more general case again:
$$ O_1 [f(x)] + O_2[f(x)] = 0 $$
We can write this as
$$ O_2^{-1}[O_1[f(x)]] + f(x) = 0 $$
We re-label $-O_2^{-1}[O_1] = O_3$ to write
$$ O_3[f(x)] - f(x) = 0 $$, so now consider some arbitrary function $h(x)$ and consider the double sided infinite sum
$$ f = ... O_3^{-2}[h(x)] + O_3^{-1}[h(x)] + h(x) + O_3[h(x)] + O_3^2[h(x)] + ... $$
For any $h$ this formal solution technically solves our equation BUT it is the case that this expression either NEVER converges or converges to 0 unless that $h$ is selected very carefully.
Lets take a look at your equation
$$ x'(t) = t x(t-a) $$
This is the same as
$$ \frac{1}{t+a} x'(t) - x(t) = 0 $$
So the sequence we build is
$$ ... + \int (t+a) \int (t+a) h(t) +  \int (t+a) h(t) + h(t) + \frac{1}{t+a} h'(t) + \frac{1}{t+a} \left( \frac{1}{t+a} h'(t) \right)' + ... $$
The left side can be attacked with integration by parts and teh right side with the product rule to turn this into series which will strongly "suggest" what choice of $h$ leads to actual convergence. I will return to do that but anyways I hope this answers your question and piques your interest.
