Let's break down the problem step by step and set up the differential equations for the amount of salt in each tank. ### **Given Data** - Tank 1: \(V_1 = 100\) gal, initially 50 lbs salt - Tank 2: \(V_2 = 200\) gal, initially 50 lbs salt - Flow rates: All **10 gal/min** - Pure water flows into Tank 1 - Tanks are well-mixed (uniform salt concentration) Let: - \(x(t)\): Amount of salt (lbs) in Tank 1 at time \(t\) (min) - \(y(t)\): Amount of salt (lbs) in Tank 2 at time \(t\) (min) ### **Step 1: Write the Salt Balance for Each Tank** #### **Tank 1:** - **Inflow:** Pure water, so no salt in - **Outflow:** Brine leaves Tank 1 to Tank 2 at 10 gal/min, carrying salt at concentration \( \frac{x}{100} \) lbs/gal \[ \frac{dx}{dt} = \text{Salt In} - \text{Salt Out} = 0 - 10 \cdot \frac{x}{100} = -0.1x \] #### **Tank 2:** - **Inflow:** Brine enters from Tank 1 at 10 gal/min, salt in at concentration \( \frac{x}{100} \) lbs/gal - **Outflow:** Brine leaves Tank 2 at 10 gal/min, salt out at concentration \( \frac{y}{200} \) lbs/gal \[ \frac{dy}{dt} = \text{Salt In} - \text{Salt Out} = 10\cdot\frac{x}{100} - 10\cdot\frac{y}{200} = 0.1x - 0.05y \] ### **Step 2: Set Up the System of ODEs** \[ \boxed{ \begin{align*} \frac{dx}{dt} &= -0.1x \qquad \text{with} \quad x(0) = 50 \\ \frac{dy}{dt} &= 0.1x - 0.05y \qquad \text{with} \quad y(0) = 50 \\ \end{align*} } \] --- ### **Step 3: Solve for \( x(t) \)** This is a simple first-order linear ODE: \[ \frac{dx}{dt} = -0.1x \] \[ \Rightarrow x(t) = x(0) e^{-0.1t} = 50e^{-0.1t} \] --- ### **Step 4: Solve for \( y(t) \)** Now, substitute the solution for \(x(t)\): \[ \frac{dy}{dt} + 0.05y = 0.1x(t) = 0.1 \cdot 50 e^{-0.1t} = 5e^{-0.1t} \] This is a first-order linear ODE. The integrating factor is \(e^{0.05t}\): Multiply both sides by integrating factor: \[ e^{0.05t} \frac{dy}{dt} + 0.05 e^{0.05t} y = 5e^{-0.1t} e^{0.05t} \] \[ \frac{d}{dt} \left( e^{0.05t} y \right) = 5 e^{-0.05t} \] Integrate both sides: \[ e^{0.05t} y = 5 \int e^{-0.05t} dt + C \] \[ = 5 \left( -\frac{1}{0.05} e^{-0.05t} \right) + C \] \[ = -100 e^{-0.05t} + C \] So, \[ y(t) = -100 e^{-0.05t} e^{-0.05t} + Ce^{-0.05t} \] \[ y(t) = -100 e^{-0.05t} + C e^{-0.05t} \] Wait, check: the integrating factor math: - The left side after integrating factor is \( \frac{d}{dt}(e^{0.05t}y) \) - The right is \( 5e^{-0.05t} \) - Integrate \( 5e^{-0.05t} \) gives \( -100e^{-0.05t} \) So \[ e^{0.05t} y = -100 e^{-0.05t} + C \] \[ y = -100 e^{-0.05t} + C e^{-0.05t} \] But let's check initial condition \(y(0) = 50\): \[ 50 = -100 \cdot 1 + C \cdot 1 \] \[ C = 150 \] So, \[ \boxed{ y(t) = -100 e^{-0.05t} + 150 e^{-0.05t} = 50 e^{-0.05t} + 100 (e^{-0.05t} - e^{-0.1t}) } \] Wait, let's check the integration step again, as the form seems off: Actually, after integrating factor: \[ e^{0.05t} y = 5 \int e^{-0.05t} dt + C = 5 \left(-\frac{1}{0.05} e^{-0.05t}\right) + C \] \[ = -100 e^{-0.05t} + C \] So, \[ y(t) = e^{-0.05t}\left(-100 e^{-0.05t} + C\right) \] \[ = -100 e^{-0.1t} + C e^{-0.05t} \] Apply initial condition \(y(0) = 50\): \[ 50 = -100 e^{0} + C e^{0} \implies C = 150 \] So, \[ \boxed{ y(t) = -100 e^{-0.1t} + 150 e^{-0.05t} } \] --- ### **Final Answers:** \[ \boxed{ \begin{align*} x(t) &= 50 e^{-0.1t} \\ y(t) &= -100 e^{-0.1t} + 150 e^{-0.05t} \end{align*} } \] --- **Explanation:** - \(x(t)\): Amount of salt in Tank 1 decreases exponentially as pure water washes out the salt. - \(y(t)\): Amount of salt in Tank 2 changes due to salt inflow from Tank 1 and outflow to the environment, resulting in a sum of exponentials. If you need to find the amount of salt after a specific time, just plug in \(t\) (in minutes) into these equations.

Let's provide a detailed, step-by-step solution with all calculations and reasoning. --- ### **Problem Recap:** - **Tank 1:** - Initial volume \(V_1 = 100\, \text{gal}\) - Initial salt \(x(0) = 50\, \text{lbs}\) - **Tank 2:** - Initial volume \(V_2 = 200\, \text{gal}\) - Initial salt \(y(0) = 50\, \text{lbs}\) - **Flow rates:** - Inflow into Tank 1: pure water at \(10\, \text{gal/min}\) - Outflow from Tank 1 to Tank 2: \(10\, \text{gal/min}\) - Outflow from Tank 2 to environment: \(10\, \text{gal/min}\) - **Assumptions:** - Tanks are well mixed, so concentration is uniform. - Salt density is uniform in each tank at any instant. --- ### **Step 1: Set up variables** Let: - \(x(t)\): lbs of salt in Tank 1 at time \(t\) - \(y(t)\): lbs of salt in Tank 2 at time \(t\) --- ### **Step 2: Write differential equations** **Tank 1:** - **Inflow:** pure water, no salt → 0 lbs/min - **Outflow:** to Tank 2 at 10 gal/min, carrying salt at concentration \(x(t)/V_1(t)\) Since water inflow is pure water, volume in Tank 1 decreases due to outflow: \[ V_1(t) = V_{1,0} - 10t = 100 - 10t \] The rate of change of salt: \[ \frac{dx}{dt} = \text{In} - \text{Out} \] Inflow of salt: \[ 0 \] Outflow of salt: \[ \text{Flow} \times \text{Concentration} = 10 \times \frac{x(t)}{V_1(t)} = 10 \times \frac{x(t)}{100 - 10t} \] Hence: \[ \boxed{ \frac{dx}{dt} = - \frac{10x(t)}{100 - 10t} } \] --- ### **Step 3: Solve for \(x(t)\)** Rewrite: \[ \frac{dx}{dt} = - \frac{10 x}{100 - 10 t} \] Separate variables: \[ \frac{dx}{x} = - \frac{10}{100 - 10 t} dt \] Note: \[ 100 - 10 t \Rightarrow \text{denominator} \] Change variable: \[ u = 100 - 10t \Rightarrow du = -10 dt \Rightarrow dt = -\frac{du}{10} \] Substitute: \[ \frac{dx}{x} = - \frac{10}{u} \times \left(- \frac{du}{10}\right) = \frac{1}{u} du \] Integrate both sides: \[ \int \frac{1}{x} dx = \int \frac{1}{u} du \] \[ \ln |x| = \ln |u| + C \] Back to original variables: \[ x(t) = K \times u = K (100 - 10 t) \] Use initial condition at \(t=0\): \[ x(0) = 50 = K \times 100 \Rightarrow K = \frac{50}{100} = 0.5 \] **Solution for \(x(t)\):** \[ \boxed{ x(t) = 0.5 (100 - 10 t) = 50 - 5 t } \] *This indicates that the amount of salt in Tank 1 decreases linearly over time, which makes sense because pure water flows in and salt leaves at a rate proportional to salt concentration.* --- ### **Step 4: Write the differential equation for Tank 2** - **Volume in Tank 2:** \[ V_2(t) = V_{2,0} + \text{Inflow} - \text{Outflow} = 200 + 10 t - 10 t = 200\, \text{gal} \] *Since inflow and outflow are equal, volume remains constant at 200 gal.* - **Salt in:** \(y(t)\) - **Inflow:** from Tank 1 at 10 gal/min, carrying salt at concentration \(x(t)/V_1(t)\) - **Outflow:** to environment at 10 gal/min, carrying salt at concentration \(y(t)/200\) Rate of change: \[ \frac{dy}{dt} = \text{Salt in} - \text{Salt out} \] Salt in: \[ 10 \times \frac{x(t)}{V_1(t)} = 10 \times \frac{50 - 5 t}{100 - 10 t} \] Salt out: \[ 10 \times \frac{y(t)}{200} = \frac{10 y(t)}{200} = \frac{y(t)}{20} \] So the differential equation: \[ \boxed{ \frac{dy}{dt} = 10 \times \frac{50 - 5 t}{100 - 10 t} - \frac{y}{20} } \] Simplify numerator: \[ 10 \times \frac{50 - 5 t}{100 - 10 t} \] Note numerator: \[ 10 \times (50 - 5 t) = 500 - 50 t \] and denominator: \[ 100 - 10 t \] Thus: \[ \frac{dy}{dt} + \frac{y}{20} = \frac{500 - 50 t}{100 - 10 t} \] --- ### **Step 5: Solve for \(y(t)\)** This is a linear first-order ODE: \[ \frac{dy}{dt} + P(t) y = Q(t) \] where: \[ P(t) = \frac{1}{20} \] \[ Q(t) = \frac{500 - 50 t}{100 - 10 t} \] **Integrating factor:** \[ \mu(t) = e^{\int P(t) dt} = e^{\int \frac{1}{20} dt} = e^{t/20} \] Multiply the entire differential equation by \(\mu(t)\): \[ e^{t/20} \frac{dy}{dt} + e^{t/20} \times \frac{1}{20} y = e^{t/20} \times \frac{500 - 50 t}{100 - 10 t} \] Left side is derivative of \(y e^{t/20}\): \[ \frac{d}{dt} \left( y e^{t/20} \right) = e^{t/20} \times \frac{500 - 50 t}{100 - 10 t} \] --- ### **Step 6: Integrate both sides** \[ y(t) e^{t/20} = \int e^{t/20} \times \frac{500 - 50 t}{100 - 10 t} dt + C \] Now, simplify the integral: Express numerator as: \[ 500 - 50 t \] Observe that: \[ 100 - 10 t \Rightarrow \text{denominator} \] Note that: \[ \frac{500 - 50 t}{100 - 10 t} = \frac{50(10 - t)}{10(10 - t)} = \frac{50 \times (10 - t)}{10 \times (10 - t)} = 5 \] **Wait, check this algebra:** \[ \frac{500 - 50 t}{100 - 10 t} = \frac{50(10 - t)}{10(10 - t)} = \frac{50}{10} = 5 \] Since numerator and denominator are proportional for all \(t\), the ratio simplifies to **5**! **This is a crucial insight.** **Therefore:** \[ Q(t) = 5 \] and the differential equation reduces to: \[ \frac{dy}{dt} + \frac{1}{20} y = 5 \] --- ### **Step 7: Solve the simplified ODE** Multiply through by the integrating factor \(e^{t/20}\): \[ \frac{d}{dt} \left( y e^{t/20} \right) = 5 e^{t/20} \] Integrate both sides: \[ y e^{t/20} = 5 \int e^{t/20} dt + C \] Integral: \[ \int e^{t/20} dt = 20 e^{t/20} + C' \] So, \[ y e^{t/20} = 5 \times 20 e^{t/20} + C = 100 e^{t/20} + C \] Divide both sides by \(e^{t/20}\): \[ y(t) = 100 + C e^{-t/20} \] Apply initial condition \(t=0\): \[ y(0) = 50 = 100 + C \Rightarrow C = -50 \] --- ### **### **Final solutions:** \[ \boxed{ \begin{aligned} x(t) &= 50 - 5 t \\ y(t) &= 100 - 50 e^{-t/20} \end{aligned} } \] --- ### **Summary:** | Quantity | Expression | Explanation | |------------|--------------|--------------| | Salt in Tank 1 | \(\boxed{x(t) = 50 - 5 t}\) | Decreases linearly from 50 lbs at 0 min, reaching zero at \(t=10\, \text{min}\). | | Salt in Tank 2 | \(\boxed{y(t) = 100 - 50 e^{-t/20}}\) | Starts at 50 lbs, approaches 100 lbs as \(t \to \infty\). | --- ### **Final note:** - At \(t=0\), initial conditions check out: - \(x(0) = 50\) - \(y(0) = 50\) - The salt in Tank 1 diminishes to zero at \(t=10\, \text{min}\): \[ 50 - 5 t = 0 \Rightarrow t=10\, \text{min} \] - The salt in Tank 2 approaches 100 lbs over time. --- **This completes the detailed solution with calculations and reasoning.**

The problem models the dynamic exchange of salt in a cascade of two well-mixed tanks with inflow and outflow, leading to coupled first-order differential equations solved via integrating factors. The solutions describe how salt concentrations evolve over time, approaching steady-state values determined by flow rates and initial conditions.