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 $1/\text{sec}$.

$$\frac{dy}{dt} + 2y = 50e^{-10t}$$

Here, $P(t) = 2$ and $Q(t) = 50e^{-10t}$.

$$e^{2t} \left( \frac{dy}{dt} + 2y \right) = e^{2t} (50e^{-10t})$$

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:

$$\frac{d}{dt}(y e^{2t}) = 50e^{-8t}$$

$$\int \frac{d}{dt}(y e^{2t}) dt = \int 50e^{-8t} dt$$ $$y e^{2t} = 50 \left( -\frac{1}{8} e^{-8t} \right) + C$$ $$y e^{2t} = -\frac{25}{4} e^{-8t} + C$$

$$y(t) = e^{-2t} \left( -\frac{25}{4} e^{-8t} + C \right)$$ $$y(t) = -\frac{25}{4} e^{-10t} + C e^{-2t}$$

Substitute $t=0$ and $y(0)=40$:

$$40 = -\frac{25}{4} e^{-10(0)} + C e^{-2(0)}$$ $$40 = -\frac{25}{4} (1) + C (1)$$ $$40 = -\frac{25}{4} + C$$ $$C = 40 + \frac{25}{4}$$ $$C = \frac{160}{4} + \frac{25}{4}$$ $$C = \frac{185}{4}$$

$$y(t) = -\frac{25}{4} e^{-10t} + \frac{185}{4} e^{-2t}$$

The mass $y(t)$ of RA$_2$ for $t \ge 0$ is:

$$\boxed{y(t) = \frac{185}{4} e^{-2t} - \frac{25}{4} e^{-10t} \text{ kg}}$$