To solve this equation we need to integrate the right hand side with respect to $t$. The equation is an ordinary, first-order, linear differential equation. The general form of such an equation is,
$$\frac {dy}{dx} + P(x)y = Q(x) $$and can be solved using and approach call the integrating factor method. Let's rearrange equation (1) to fit this general form and use an integrating factor to solve it. Equation (1) becomes,
$$\frac {dC}{dt} + \lambda C = aR \hspace{1cm} (2) $$So in our case we have $P(x) = \lambda$ and $Q(x) = aR$. We define the integrating factor as,
$$IF = e^{\int {P(x) dx}}$$and therefore in the case of equation (2), the integrating factor is,
$$ \begin{array}{rcl} IF &=& e^{\int {\lambda dt}} \\ &=& e^{\lambda t} \\ \end{array} $$The approach is to multiply both sides of the equation by the integrating factor,
$$\frac {dC}{dt}e^{\lambda t} + \lambda Ce^{\lambda t} = aRe^{\lambda t} $$The trick here is to notice that the left hand side can be simplified using the Product Rule
$$ \frac {d}{dt}(Ce^{\lambda t}) = \frac {dC}{dt}e^{\lambda t} + \lambda Ce^{\lambda t} $$and so we can say,
$$ \frac {d}{dt}(Ce^{\lambda t}) = aRe^{\lambda t} $$Now we can integrate,
$$ \begin{array}{rcl} Ce^{\lambda t} &=& \int {aRe^{\lambda t}dt} + c \\ &=& aR \int {e^{\lambda t} dt} + c \\ &=& \frac {aR}{\lambda} e^{\lambda t} + c \\ \end{array} $$where $c$ is an arbitrary constant of integration. Rearranging, to solve for $C$, gives
$$C(t) = \frac {aR}{\lambda} + ce^{-\lambda t} \hspace{1cm} (3)$$So now we need to find the value of the integration constant, $c$. Let's define our initial conditions. When $t = 0$, we'll say that the test concentration is $C_0$. When $t = 0$, $e^{-\lambda t} = 1$. Therefore
$$ \begin{array}{rcl} C_0 &=& \frac {aR}{\lambda} + c \\ c &=& C_0 - \frac {aR}{\lambda} \\ \end{array} $$Substituting for $c$ in equation (3) gives
$$ \begin{array}{rcl} C(t) &=& \frac {aR}{\lambda} + (C_0 - \frac {aR}{\lambda}) e^{-\lambda t} \\ &=& \frac {aR}{\lambda} + C_0e^{-\lambda t} - \frac {aR}{\lambda}e^{-\lambda t} \\ &=& \frac {aR}{\lambda} - \frac {aR}{\lambda}e^{-\lambda t} + C_0e^{-\lambda t} \\ &=& \frac {aR}{\lambda} [1 - e^{-\lambda t}] + C_0e^{-\lambda t} \hspace{1cm} (4) \\ \end{array} $$So this is the general solution to the equation (1). It comprises two terms,the first which describes the accumulation of tests produced during the time $t$ together with concommitant loss, while the second term describes the loss of the tests already present at $t_0$.
In the case that $C_0 = 0$, i.e. we start from a fresh sediment containing no tests, equation (4) reduces to,
$$C(t) = a \frac {R}{\lambda} \Big[1 - \exp(-\lambda t) \Big] \hspace{1cm} (2)$$Whereas in the case of no test production (e.g. $a = 0$ or $R = 0$), it reduces to the basic taphonomic decay equation,
$$C(t) = C_0e^{-\lambda t}$$
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