Step 1: Identify the given differential equation and parameters. The rate of change of mass $y(t)$ of RA$_2$ is given by the differential equation: $$ \frac{dy}{dt} = 50e^{-10t} - ky $$ We are given $k = 2 \text{ kg/sec}$ and the initial condition $y(0) = 40 \text{ kg}$. For the equation to be dimensionally consistent, if $y$ is in kg and $dy/dt$ is in kg/sec, then $k$ must be in $1/\text{sec}$. We will assume that the unit "kg/sec" for $k$ is a typo and use $k=2$ with units of $1/\text{sec}$. Step 2: Substitute the value of $k$ into the differential equation. $$ \frac{dy}{dt} = 50e^{-10t} - 2y $$ Step 3: Rearrange the equation into the standard form of a first-order linear differential equation, $\frac{dy}{dt} + P(t)y = Q(t)$. $$ \frac{dy}{dt} + 2y = 50e^{-10t} $$ Here, $P(t) = 2$ and $Q(t) = 50e^{-10t}$. Step 4: Calculate the integrating factor, $I(t) = e^{\int P(t) dt}$. $$ I(t) = e^{\int 2 dt} = e^{2t} $$ Step 5: Multiply the differential equation by the integrating factor. $$ e^{2t} \left( \frac{dy}{dt} + 2y \right) = e^{2t} (50e^{-10t}) $$ The left side is the derivative of the product $y(t)I(t)$: $\frac{d}{dt}(y e^{2t})$. The right side simplifies to $50e^{2t - 10t} = 50e^{-8t}$. So, the equation becomes: $$ \frac{d}{dt}(y e^{2t}) = 50e^{-8t} $$ Step 6: Integrate both sides with respect to $t$. $$ \int \frac{d}{dt}(y e^{2t}) dt = \int 50e^{-8t} dt $$ $$ y e^{2t} = 50 \left( -\frac{1}{8} e^{-8t} \right) + C $$ $$ y e^{2t} = -\frac{25}{4} e^{-8t} + C $$ Step 7: Solve for $y(t)$ by dividing by $e^{2t}$. $$ y(t) = e^{-2t} \left( -\frac{25}{4} e^{-8t} + C \right) $$ $$ y(t) = -\frac{25}{4} e^{-10t} + C e^{-2t} $$ Step 8: Use the initial condition $y(0) = 40 \text{ kg}$ to find the constant $C$. Substitute $t=0$ and $y(0)=40$: $$ 40 = -\frac{25}{4} e^{-10(0)} + C e^{-2(0)} $$ $$ 40 = -\frac{25}{4} (1) + C (1) $$ $$ 40 = -\frac{25}{4} + C $$ $$ C = 40 + \frac{25}{4} $$ $$ C = \frac{160}{4} + \frac{25}{4} $$ $$ C = \frac{185}{4} $$ Step 9: Substitute the value of $C$ back into the equation for $y(t)$. $$ y(t) = -\frac{25}{4} e^{-10t} + \frac{185}{4} e^{-2t} $$ The mass $y(t)$ of RA$_2$ for $t \ge 0$ is: $$ \boxed{y(t) = \frac{185}{4} e^{-2t} - \frac{25}{4} e^{-10t} \text{ kg}} $$