Why can you use the one-sided laplace transform to solve differential equation describing a causal LTI-system?

In an example, an equation describing a causal LTI-system is

$$(D^2 + 5D + 6) y(t) = (D+1) x(t)$$

where $$y(t) = y_{zs}(t) + y_{zi}(t)$$ and the initial conditions are $$y(0^-) = 2, \dot{y}(0^-) = 1$$.
$$x(t) = e^{-4t}u(t)$$ and we want to calculate $$y(t)$$.

My teacher said in an example of a solution that because $$x(t) = 0, \quad t < 0$$ we can take the one-sided/unilateral laplace transform of the RHS (since then the unilateral and bilateral laplace transform are the same, I guess ). Further, the explanation continued with that because the system is causal, the impulse response $$h(t) = 0$$ for $$t < 0$$ and therefore $$y_{zs}(t) = (x*h)(t) = 0$$ for $$t < 0$$. Therefore, supposedly, $$y(t<0)=0$$ and we can take the one-sided laplace transform of the LHS too.

I am questioning the part in boldface, because what about the zero-input response $$y_{zi}(t)$$, how do we know its value for $$t<0$$? I would like to belive it is not equal to $$0$$ for $$t<0$$ due to the initial conditions given at $$0^-$$ not being zero and $$0^- < 0$$. If it is not equal to $$0$$ for $$t<0$$ how can we know we can take the one-sided laplace transform of the LHS?

Note: I also read this question in which it is stated that non-zero initial conditions make the system non-linear and time-varying, which also makes me think the equation at hand is inconsistent with the fact that it describes an LTI-system?

Edit 1 in response to the comment about Lathi's linear systems and signals:
I did not remember that this example was indeed from the book, but I have read the example and the section "Comments on initial conditions at $$0^-$$ and at $$0^+$$ " before. The section explains that we cannot expect the total response $$y(t)$$ to satisfy the initial conditions given at $$0^-$$ at $$0$$, which makes perfect sense I think since $$0^- \neq 0$$. It goes on to say that there is another version of the laplace transform, $$L_+$$ that is not as convenient to work with. The author's discussion might have gone a bit over my head, because unfortunately I can't understand how it answers my questions, that is how we can assume $$y(t)$$ to be causal (in order to use the unilateral laplace transform on both sides of the equation) and the note about the non-zero initial conditions implying the system to not be LTI.

The reason I have not accepted the answer given so far is that it is a bit to advanced for me to judge its correctness and therefore I wanted to wait a bit in case there would be more input for my question. But eventually I will just assume it is correct and accept it.

• That example can be found in Lathi's book Linear Systems and Signals. Have you read the section "Comments on Initial Conditions at $0^−$ and at $0^+$"? There your question is actually answered, at least as far as I can see. If not, then please explain any remaining doubts by editing your current question. Dec 5 '20 at 11:43
• @MattL. Thanks for your comment, I have edited my question. Dec 5 '20 at 12:44
• Do you have doubts about the interval $t\in (-\infty,0^-]$ or about the infinitesimal interval $t\in [0^-,0]$? Dec 5 '20 at 13:07
• @MattL. I have doubts about the interval $t \in (-\infty, 0^-]$, I think the other interval would not matter since it is infinitesimal. Dec 5 '20 at 13:11

Unfortunately, the way the problem is framed it is very difficult to do in a rigorous manner. You can, by remembering that $$0^-$$ is shorthand for $$-\epsilon$$ as $$\epsilon \to 0$$, and taking limits everywhere, and having expressions where you're looking at intervals of $$t$$ from $$-\infty$$ to $$-2\epsilon$$ and other maddening things.
The easy way to do it is to separate the problem: solve the problem for the $$x(t)$$ that's given. Then augment the problem by finding an $$x(t)$$ that is zero everywhere except in $$0^- < t < 0^+$$ that will result in $$y(0^+) = 2$$ (note the $$0^+$$ instead of $$0^-$$) and $$\dot y(0^+) = 1$$. This involves setting $$x(t)$$ to a linear combination of $$\delta(t)$$ and $$\delta^2(t)$$ (as if $$\delta(t)$$ weren't wacky enough, $$\delta^2(t) = d/dt\ \delta(t)$$).
Then add the two solutions together. It may not be what your prof had in mind, but it takes the least amount of hand-waving, and if you really want rigor you can break $$\epsilon$$ and put it to work.
By doing the above trick of expressing $$x(t)$$ as a linear combination of the specified $$x(t)$$ plus whatever sum of the Dirac impulse and it's derivatives, you cast the problem into one where the system is linear -- you're just using physically impossible values for $$x(t)$$.