This problem and the following questions are motivated by a simple problem in the theory of visco-elasticity.
Assume that $\sigma:\mathbb{R}-\{0\}\to\mathbb{R}$ is a function defined as \begin{align*} \sigma(t)= \begin{cases} 0 & t\lt0 \\ \sigma_0 & t\gt0 \end{cases} \tag{1} \end{align*} with $\sigma_0\in\mathbb{R}$. Suppose that we are looking for a function $\varepsilon$ such that \begin{align*} &q_1 \dot{\varepsilon}(t)=\sigma(t)+p_1\dot{\sigma}(t),\quad \forall t \in \mathbb{R}-\{0\}, \\ &\varepsilon(0^-):=\lim_{t\to0^-}\varepsilon(t)=0, \tag{2} \end{align*} where the over dot denotes differentiation with respect to $t$ and $q_1\in\mathbb{R}-\{0\},\,p_1\in\mathbb{R}$.
Question 1. Does Eq.$(2)$ uniquely determines $\varepsilon(t)$ over $\mathbb{R}-\{0\}$?
I suppose that the answer to Question 1 is yes. I am not sure about this, please correct me if I am wrong. Bearing this in mind, let us find this function.
Direct Method
If $t<0$ then the ODE in $(2)$ implies that $\varepsilon(t)=A$. But the initial condition forces $A=0$ and we have \begin{align*} \varepsilon(t)=0,\quad t<0. \tag{3} \end{align*} If $t>0$ then the ODE reduces to $q_1\dot{\varepsilon}(t)=\sigma_0$ which has the solution \begin{align*} \varepsilon(t)=\frac{\sigma_0}{q_1}t+B,\quad t>0 \tag{4} \end{align*}
Question 2. How can I find the constant $B$?
In my text, the final answer is given as \begin{align*} \varepsilon(t)= \begin{cases} 0 & t<0 \\ \frac{\sigma_0}{q_1}t+\frac{p_1\sigma_0}{q_1} & t>0 \end{cases} \tag{5} \end{align*}
Laplace Transform Method
I checked the answer in $(5)$ via the Laplace transform method. If we define our Laplace transform as $$\mathcal{L}(f)(s):=\int_{0^-}^{+\infty}f(t)\exp(-st)dt\tag{6}$$ (note $0^-$ in the lower limit). Then we can easily obtain the above answer. Denoting that $\bar f(s):=\mathcal{L}(f)(s)$, using $\bar{\dot{f}}(s)=-f(0^-)+s\bar f(s)$ and $\bar\sigma(s)=\frac{\sigma_0}{s}$ we obtain \begin{align*} q_1\big(-\varepsilon(0^-)+s\bar\varepsilon(s)\big)&=\bar\sigma(s)+p_1\big(-\sigma(0^-)+s\bar\sigma(s)\big) \\ q_1s\bar\varepsilon(s)&=(1+p_1s)\bar\sigma(s)\\ \bar\varepsilon(s)&=\frac{(1+p_1s)}{q_1s^2}\sigma_0\\ \bar\varepsilon(s)&=\frac{1}{q_1}\Big(\frac{1}{s^2}+\frac{p_1}{s}\Big)\sigma_0\\ \varepsilon(t)&=\frac{1}{q_1}\big(t+p_1\big)\sigma_0\\ \varepsilon(t)&=\frac{\sigma_0}{q_1}t+\frac{p_1\sigma_0}{q_1} \tag{7} \end{align*}
which is the desired answer. But, I think that some part of this calculation is not allowed as I am sure that the direct method is completely OK!
Read the comments below Dylan's answer for an answer to this discrepancy.
Further Observations
The answer to Question 1 is no in the sense of classical analysis and yes in the sense of distribution theory. In conclusion, by just using classical analysis we cannot get anything more than Eqs. $(3)$ and $(4)$. Then it is natural to ask
Question 3. How can we obtain the solution $(5)$ by just using classical analysis?
Indeed, the main problem we are interested to solve is some other problem. The limit of the solution of this problem coincides with $(5)$. Suppose that \begin{align*} \sigma_\tau(t)= \begin{cases} 0 & t\leq-\frac{\tau}{2} \\ \frac{\sigma_0}{\tau}\big(t+\frac{\tau}{2}\big) & -\frac{\tau}{2} \leq t \leq \frac{\tau}{2} \\ \sigma_0 & t\ge\frac{\tau}{2} \end{cases} \tag{8} \end{align*} is given. We are looking for a continuous function $\varepsilon_\tau(t)$ such that \begin{align*} &q_1 \dot{\varepsilon}_\tau(t)=\sigma_\tau(t)+p_1\dot{\sigma}_\tau(t),\quad \forall t \in \mathbb{R}, \\ &\varepsilon_\tau(-\frac{\tau}{2})=0, \tag{9} \end{align*} Then, by just using classical analysis, we can prove that $$\lim_{\tau\to0^+}\varepsilon_\tau(t)=\varepsilon(t),\quad \forall t\in\mathbb{R}-\{0\}, \tag{10}$$ where $\varepsilon(t)$ is the function given by $(5)$. However, this method is painful and lengthy, specially for more complicated problems. Distribution theory is a nice tool which can give us $(5)$ much much faster!
Note that the following arithmetic should be interpreted in the sense of distribution theory.
Bearing in mind that $\dot\sigma(t)=\sigma_0\delta(t)$ with $\delta(t)$ being the Dirac delta function (or more precisely Dirac delta distribution), integrate the entire equation like this
\begin{align*} q_1\int_{-\tau}^\tau \dot\varepsilon(t)\ dt &= \int_{-\tau}^\tau\sigma(t)\ dt + p_1 \int_{-\tau}^\tau \dot\sigma(t)\ dt \\ q_1\big(\varepsilon(\tau)-\varepsilon(-\tau)\big)&=\int_{-\tau}^\tau\sigma(t)\ dt +p_1\sigma_0\int_{-\tau}^\tau\delta(t)dt \end{align*}
Then take the limit as $\tau \to 0^+$. You'll get
$$ q_1\big[\varepsilon(0^+) - \varepsilon(0^-)\big] = p_1\sigma_0 $$
Since $\varepsilon(0^-) = 0$, we have $\varepsilon(0^+) = \frac{p_1\sigma_0}{q_1}$ which is also the value of $B$.