미분방정식 모델링

(a) The volume of solution in the tank remains constant because the inflow and outflow are at the same rate. (b) Salt enters the tank with the inflow, which occurs at a rate of $3$ gallons per minute at a concentration of $20$ grams per gallon. Observing that $$20 \frac{\text{grams}}{\text{gallon}} \cdot 3 \frac{\text{gallons}}{\text{minute}} = 60 \frac{\text{grams}}{\text{minute}}\,,$$ we see that salt is entering the tank at a rate of $60$ grams per minute. (c) Letting $S(t)$ denotes the number of grams of salt in the tank at time $t$, since there are $100$ gallons of solution in the tank at any time $t$, the concentration of salt in the tank in grams per gallon is $$\frac{S(t)}{100} \frac{\text{grams}}{\text{gallon}}$$ at any time $t$. (d) With water leaving the tank at $3$ gallons per minute, it follows that salt is leaving the tank at a rate of $$\frac{S(t)}{100} \frac{\text{grams}}{\text{gallon}} \cdot 3 \frac{\text{gallons}}{\text{minute}} = \frac{3S(t)}{100} \frac{\text{grams}}{\text{minute}}\,.$$ (e) Taking into account the rates at which salt enters and leaves the tank as established in (b) and (d) and noting that the units on $\frac{dS}{dt}$ are grams per minute, we have shown that $$\frac{dS}{dt} = 60 - \frac{3}{100} S\,.$$ (f) Setting $60 - \frac{3}{100} S = 0$, we find that $S = 2000$ is the only equilibrium solution. This makes sense because the inflow is carrying $20$ grams of salt per gallon and thus when the tank reaches this concentration, there will be $20 \cdot 100$ total grams of salt. This equilibrium is stable (both intuitively and because $\frac{dS}{dt}$ is positive for $S < 2000$ and negative for $S > 2000$. (g) Assuming that $S(0) = 0$, we can solve the IVP with $\frac{dS}{dt} = 60 - \frac{3}{100} S$. For somewhat simpler computations, we write $60 - \frac{3}{100}S = -\frac{3}{100}(S-2000)$. Separating variables, $$\frac{1}{S-2000} \frac{dS}{dt} = -\frac{3}{100}$$ and integrating with respect to $t$, $$\int \frac{1}{S-2000} \, dS = -\int \frac{3}{100} \, dt$$ so that $$\ln|S-2000| = -\frac{3}{100}t + C\,.$$ Solving for $S$ in the usual way, $$S = 2000 + Ke^{-\frac{3}{100}t}$$ and using the initial condition that $S(0) = 0$, we find $K = -2000$ so that $$S(t) = 2000 - 2000e^{-\frac{3}{100}t}\,.$$ (h) As $t \to \infty$, we know $e^{-\frac{3}{100}t} \to 0$, and thus $S(t) \to 2000$. This makes sense because the inflow carries $20$ grams of salt per liter continuously. As time goes on, the concentration of solution in the tank should tend to that same concentration, and with $100$ gallons of solution, the amount of salt will tend to $20 \cdot 100 = 2000$ grams. This value also matches the stable equilibrium that we found in (f).