Let's break down each part and provide explanations: --- ### **Step 1: Setting up the System of Differential Equations** We need to model the change in solute in each tank. The general equation for each tank is: \[ \frac{dQ_i}{dt} = \text{(input rate of solute)} - \text{(output rate of solute)} \] #### **Tank 1:** - **Input:** From outside: \(10\, \text{L/min} \times 0.3\, \text{kg/L} = 3\, \text{kg/min}\) - **Input:** From Tank 2: \(10\, \text{L/min} \times \frac{Q_2}{1000}\) - **Output:** To Tank 2: \(10\, \text{L/min} \times \frac{Q_1}{1000}\) So, \[ Q_1' = 3 + \frac{10}{1000}Q_2 - \frac{10}{1000}Q_1 = 3 + \frac{1}{100}Q_2 - \frac{1}{100}Q_1 \] #### **Tank 2:** - **Input:** From Tank 1: \(10\, \text{L/min} \times \frac{Q_1}{1000}\) - **Output:** To Tank 1: \(10\, \text{L/min} \times \frac{Q_2}{1000}\) So, \[ Q_2' = \frac{1}{100}Q_1 - \frac{1}{100}Q_2 \] **Matrix Form:** \[ \begin{bmatrix} Q_1' \\ Q_2' \end{bmatrix} = \begin{bmatrix} -\frac{1}{100} & \frac{1}{100} \\ \frac{1}{100} & -\frac{1}{100} \end{bmatrix} \begin{bmatrix} Q_1 \\ Q_2 \end{bmatrix} + \begin{bmatrix} 3 \\ 0 \end{bmatrix} \] **Explanation:** This system represents the inflow and outflow of solute in each tank, accounting for concentration, flow rate, and external addition. --- ### **Step 2: Long-term Behavior as \( t \to \infty \)** At equilibrium (\( t \to \infty \)), the amounts of solute in each tank stop changing, so \( Q_1' = 0 \) and \( Q_2' = 0 \): \[ 0 = -\frac{1}{100}Q_1 + \frac{1}{100}Q_2 + 3 \] \[ 0 = \frac{1}{100}Q_1 - \frac{1}{100}Q_2 \] From the second equation: \(\frac{1}{100}Q_1 = \frac{1}{100}Q_2 \implies Q_1 = Q_2\) Plug into the first equation: \[ 0 = -\frac{1}{100}Q_1 + \frac{1}{100}Q_1 + 3 \implies 0 = 3 \] But this seems off! Let's double-check: - If \( Q_1 = Q_2 \), the system at equilibrium should have \( Q_1' = 3 \), which is nonzero. This means the system will never reach a finite equilibrium; the solute in both tanks increases without bound because we are continually adding solute to Tank 1 at \( 3\, \text{kg/min} \), but never removing it from the system. **Explanation:** As \( t \to \infty \), both tanks will accumulate more and more solute because the added solute is never removed. The system does not reach a finite equilibrium, so both \( Q_1 \) and \( Q_2 \) go to infinity. --- ### **Final Answer** - **Step 1:** The system of differential equations is: \[ \begin{bmatrix} Q_1' \\ Q_2' \end{bmatrix} = \begin{bmatrix} -\frac{1}{100} & \frac{1}{100} \\ \frac{1}{100} & -\frac{1}{100} \end{bmatrix} \begin{bmatrix} Q_1 \\ Q_2 \end{bmatrix} + \begin{bmatrix} 3 \\ 0 \end{bmatrix} \] (This describes how solute moves and is added to the tanks.) - **Step 2:** As \( t \to \infty \), the amount of solute in each tank goes to infinity because solute is continually added and never removed. **Final Answer:** In the long run, both Tank 1 and Tank 2 will have **infinite** amounts of solute.

**Step 1: Set up the system of differential equations** - **Inflow of solute to Tank 1:** External source with concentration 0.3 kg/L and flow rate 10 L/min: \[ \text{Inflow} = 10 \times 0.3 = 3\, \text{kg/min} \] - **Inflow to Tank 1 from Tank 2:** Flow rate 10 L/min, concentration \( \frac{Q_2}{1000} \) kg/L: \[ \text{From Tank 2} = 10 \times \frac{Q_2}{1000} = \frac{Q_2}{100}\, \text{kg/min} \] - **Outflow from Tank 1 to Tank 2:** Flow rate 10 L/min, concentration \( \frac{Q_1}{1000} \) kg/L: \[ \text{Outflow} = 10 \times \frac{Q_1}{1000} = \frac{Q_1}{100}\, \text{kg/min} \] - **Differential equation for \(Q_1(t)\):** \[ \frac{dQ_1}{dt} = \text{Inflow} + \text{From Tank 2} - \text{Outflow} = 3 + \frac{Q_2}{100} - \frac{Q_1}{100} \] - **Similarly for Tank 2:** Inflow from Tank 1: \(\frac{Q_1}{100}\) kg/min Outflow to Tank 1: \(\frac{Q_2}{100}\) kg/min \[ \frac{dQ_2}{dt} = \frac{Q_1}{100} - \frac{Q_2}{100} \] **Final system of equations:** \[ \boxed{ \begin{cases} Q_1' = 3 + \frac{Q_2}{100} - \frac{Q_1}{100} \\ Q_2' = \frac{Q_1}{100} - \frac{Q_2}{100} \end{cases} } \] --- **Step 2: Long-term behavior as \( t \to \infty \)** - At equilibrium, \(Q_1' = 0\) and \(Q_2' = 0\): \[ \begin{cases} 0 = 3 + \frac{Q_2}{100} - \frac{Q_1}{100} \\ 0 = \frac{Q_1}{100} - \frac{Q_2}{100} \end{cases} \] - From the second equation: \[ \frac{Q_1}{100} = \frac{Q_2}{100} \implies Q_1 = Q_2 \] - Substitute into the first: \[ 0 = 3 + \frac{Q_1}{100} - \frac{Q_1}{100} \implies 0 = 3 \] This is impossible, indicating the amounts of solute grow without bound. **Explanation:** Since there's a continuous inflow of solute (3 kg/min) into Tank 1, and the system transfers solute between tanks with no outflow removing solute from the system, the amount of solute in each tank will increase indefinitely over time. --- ### **Final answers:** - **In the long run (\( t \to \infty \)):** - **Tank 1:** The amount of solute tends to infinity (kg). - **Tank 2:** The amount of solute also tends to infinity (kg). **Summary:** Because the system continually adds solute and there's no mechanism for removal, the solute amounts in both tanks grow without bound over time.

**Step 1: Set up the differential equations** - **Flow rates and concentrations:** - External inflow to Tank 1: 10 L/min with concentration 0.3 kg/L Inflow rate of solute: \(10 \times 0.3 = 3\, \text{kg/min}\) - External inflow to Tank 2: 10 L/min with concentration 0 kg/L (no solute) Inflow rate of solute: \(10 \times 0 = 0\, \text{kg/min}\) - Flow from Tank 1 to Tank 2: 10 L/min, concentration in Tank 1: \(\frac{Q_1}{1000}\) kg/L Outflow of solute: \(10 \times \frac{Q_1}{1000} = \frac{Q_1}{100}\, \text{kg/min}\) - Flow from Tank 2 to Tank 1: 10 L/min, concentration in Tank 2: \(\frac{Q_2}{1000}\) kg/L Outflow of solute: \(10 \times \frac{Q_2}{1000} = \frac{Q_2}{100}\, \text{kg/min}\) - **Differential equations:** - **Tank 1:** \[ \frac{dQ_1}{dt} = \text{Inflow} + \text{from Tank 2} - \text{outflow to Tank 2} \] \[ = 3 + \frac{Q_2}{100} - \frac{Q_1}{100} \] - **Tank 2:** \[ \frac{dQ_2}{dt} = \text{Inflow from Tank 1} - \text{outflow to Tank 1} \] \[ = \frac{Q_1}{100} - \frac{Q_2}{100} \] **Answer:** \[ \boxed{ \begin{cases} Q_1' = 3 + \frac{Q_2}{100} - \frac{Q_1}{100} \\ Q_2' = \frac{Q_1}{100} - \frac{Q_2}{100} \end{cases} } \] --- **Step 2: Long-term behavior as \( t \to \infty \)** - **Set derivatives to zero for equilibrium:** \[ 0 = 3 + \frac{Q_2}{100} - \frac{Q_1}{100} \] \[ 0 = \frac{Q_1}{100} - \frac{Q_2}{100} \] - From the second equation: \[ \frac{Q_1}{100} = \frac{Q_2}{100} \implies Q_1 = Q_2 \] - Substitute into the first: \[ 0 = 3 + \frac{Q_1}{100} - \frac{Q_1}{100} \implies 0 = 3 \] This contradiction indicates the amounts of solute will grow without bound over time because: - The system continually adds 3 kg/min to Tank 1. - There is no removal of solute from the system. - The transfer between tanks redistributes but does not eliminate solute. **Conclusion:** - As \( t \to \infty \), both \( Q_1(t) \) and \( Q_2(t) \) tend to infinity. **Final answer:** - **Long-term, the amount of solute in Tank 1 is infinite kg.** - **Long-term, the amount of solute in Tank 2 is infinite kg.**

**Step 1: Set up the differential equations** Given data: - Flow rate between tanks: 10 L/min - Volumes: 1000 L (constant) - Inflow to Tank 1 (external): 10 L/min at 0.3 kg/L → solute inflow: \(10 \times 0.3 = 3\, \text{kg/min}\) - Inflow to Tank 2 (external): 10 L/min at 0 kg/L → solute inflow: 0 kg/min - All tanks are perfectly mixed. **Concentrations:** - In Tank 1: \( \frac{Q_1}{1000} \) kg/L - In Tank 2: \( \frac{Q_2}{1000} \) kg/L **Differential equations:** - **Tank 1:** \[ \frac{dQ_1}{dt} = \text{Inflow} + \text{from Tank 2} - \text{outflow to Tank 2} \] \[ = 3 + 10 \times \frac{Q_2}{1000} - 10 \times \frac{Q_1}{1000} \] \[ = 3 + \frac{Q_2}{100} - \frac{Q_1}{100} \] - **Tank 2:** \[ \frac{dQ_2}{dt} = \text{from Tank 1} - \text{to Tank 1} \] \[ = 10 \times \frac{Q_1}{1000} - 10 \times \frac{Q_2}{1000} \] \[ = \frac{Q_1}{100} - \frac{Q_2}{100} \] **Final system:** \[ \boxed{ \begin{cases} Q_1' = 3 + \frac{Q_2}{100} - \frac{Q_1}{100} \\ Q_2' = \frac{Q_1}{100} - \frac{Q_2}{100} \end{cases} } \] --- **Step 2: Long-term behavior as \( t \to \infty \)** Set derivatives to zero for steady state: \[ 0 = 3 + \frac{Q_2}{100} - \frac{Q_1}{100} \] \[ 0 = \frac{Q_1}{100} - \frac{Q_2}{100} \] From the second: \[ Q_1 = Q_2 \] Substitute into the first: \[ 0 = 3 + \frac{Q_1}{100} - \frac{Q_1}{100} \implies 0 = 3 \] which is impossible. This indicates the amounts do not reach a finite steady state; instead, they grow without bound because of continuous inflow. **Final conclusion:** - As \( t \to \infty \), **both \(Q_1\) and \(Q_2\) tend to infinity**. - **In the long run:** \[ \boxed{ \text{Tank 1:} \quad \text{infinite kg} \\ \text{Tank 2:} \quad \text{infinite kg} } \] **Why does this make sense?** Because solute is being continually added to Tank 1 at 3 kg/min, and with no removal mechanism, the amount of solute in both tanks will grow indefinitely over time.