Problem
The velocity for a body moving in a viscous fluid has the following diff. equation:
$$a v'(t) + b v(t) = c(t)$$ $$v(0) = 0$$
Where $a$ is the mass of the body, $b$ is the frictional force from the fluid, and $c(t)$ is a pulsive force applied to the body which is, at first, stationery in the water. The force $c(t)$ can be approximated with any arbitrary impulse function, but it has to meet the following conditions:
$$c(t) \ne 0 \text{, for a very short time interval around $t = 0$}$$
$$\int_{-\infty}^{\infty} c(t) \text{ } dt = d$$
The body is finally at rest for $t < 0$.
$$\text{a, b, d > 0, and they are shall be treated as known parameters}$$
Task: Find $v(t)$, for $t$ $\geq$ 0
My attempt
I first formulated a solution for $v(t)$ using Laplace transform for the diff. equation.
$$\mathscr{L}(a v'(t) + b v(t)) = \mathscr{L}(c(t)) \implies V(s) = \frac{C(s)}{as+b}$$
And then I tried to find an arbitrary impulse function that would meet the conditions.
We know that: $$\int_{-\infty}^{\infty} c(t) \text{ } dt = d$$
And we can also know that if we integrate the standard Dirac-pulse distribution/function $\mathcal{ D}(t)$ we get
$$\int_{-\infty}^{\infty} \mathcal{D}(t) \text{ } dt = 1$$
due to $d$ just being a real constant we get
$$d\int_{-\infty}^{\infty}\mathcal{D}(t) \text{ } dt = d$$ $$\iff$$ $$\int_{-\infty}^{\infty} d\mathcal{D}(t) \text{ } dt = d$$
And also
$$\mathcal{D}(t) = {\infty} \text{ for t = 0}$$
$$\mathcal{D}(t) = {0} \text{ for } t\neq 0$$ $$\implies$$ $$c(t) = d\mathcal{D}(t)$$ $$\implies$$ $$C(s) = d$$
So now we can solve for $V(s)$ $$V(s) = \frac{d}{as+b} = \frac{d}{a}\frac{1}{s+\frac{b}{a}}$$ Which gives us our inverse $$v(t) = \frac{d}{a}e^{-\frac{b}{a}t}$$
The issue
Now as from the description of the problem, we know that the function $v(t)$ is at rest for $t < 0$, which I interpret as $$v(t) = 0 \text{ for } t < 0$$
But in the function $v(t)$ we have no restriction for the domain of the function. So I'm thinking v(t) should've had a Heaviside step function multiplied to it. But also, the function depends fully on the parameters, {a,b,d}, for different values. Yes, we do get an exponential function, so it will always be positive, and $$\lim_{t \rightarrow \infty} v(t) = 0$$ But I don't like the fact that with a force applied to the body, the function would slowly descend towards 0.
Sorry for the messy post, I had to learn latex to rewrite all the mathematical bits by coding it in latex.
Summary of the issue is that we don't have any restrictions in our $v(t)$, so the function is defined for $t < 0$, I would like to see a Heaviside step function in our $v(t)$ but we have none, and I don't know right now how to proceed.
I don't really have anyone to verify my solution with, but after working a bit more with the problem I realized that my solution must be the right one. Because when I read the problem again, I could just interpret $v(t)$ as two cases. $$v(t) = \frac{d}{a}e^{-\frac{b}{a}t} \text{ for } t \geq 0$$ $$v(t) = 0 \text{ for } t < 0$$
Another addition would be that we define the Laplace transform of a function for the interval $t = [0,\infty)$
Also as we do have an engineering/real-life applicational problem, it's not reasonable to even think of the function for negative intervals for $t$. Because we want to evaluate the transient response.
I feel so stupid, I am now 99.99% sure that I have the right solution. But if anyone here would like to add anything or maybe prove me wrong, please do.