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In Problems 25-28 consider the two-compartment model for two tanks with respective volumes \(V_{1}\) and \(V_{2}\). $$ \begin{array}{l} \frac{d C_{1}}{d t}=\frac{q}{V_{1}}\left(C_{\infty}-C_{1}\right) \\ \frac{d C_{2}}{d t}=\frac{q}{V_{2}}\left(C_{1}-C_{2}\right) \end{array} $$ where \(C_{1}(t)\) is the concentration in the first tank and \(C_{2}(t)\) is the concentration in the second tank, and \(q\) is the volume of water flowing between the two tanks in one unit of time. Let \(C_{\infty}=0\), so that the fresh water is pumped into \(\tan \mathrm{k} 1\) and flushes solute from \(\operatorname{tank} 1\) into tank 2 . Now assume that \(C_{1}(0)=1\) and \(C_{2}(0)=0 .\) If \(q=1\) and \(V_{1}=V_{2}=1\), solve the pair of differential equations to find \(C_{1}(t)\) and \(C_{2}(t)\), and sketch both functions of time.

Step by step solution

01

Identify and Simplify the Problem

We are given two differential equations to solve, assuming the initial conditions: \(C_{1}(0) = 1\) and \(C_{2}(0) = 0\), along with \(q = 1\), \(V_1 = 1\), and \(V_2 = 1\). Also, \(C_{\infty} = 0\). The equations simplify to: \(\frac{d C_{1}}{d t} = -(C_{1})\) and \(\frac{d C_{2}}{d t} = (C_{1} - C_{2})\).

02

Solve the First Differential Equation

The equation \(\frac{d C_{1}}{d t} = -C_{1}\) can be solved by separation of variables. Integrating both sides gives \(\ln|C_{1}| = -t + C\). Solving for \(C_{1}\), we get \(C_{1}(t) = Ae^{-t}\), where \(A\) is a constant. Using the initial condition \(C_{1}(0) = 1\), we find \(A = 1\), thus \(C_{1}(t) = e^{-t}\).

03

Solve the Second Differential Equation

Substitute \(C_{1}(t) = e^{-t}\) into the second equation \(\frac{d C_{2}}{d t} = (e^{-t} - C_{2})\). This can be solved using the integrating factor method. Let \(\mu(t) = e^{\int 1 \, dt} = e^{t}\). Multiply through by the integrating factor: \(e^{t} \frac{d C_{2}}{d t} + e^{t}C_{2} = 1\). This simplifies to \(\frac{d}{dt}(e^{t}C_{2}) = e^{t}\).

04

Integrate to Find \(C_{2}(t)\)

Integrate both sides with respect to \(t\): \(e^{t}C_{2} = e^{t} + B\), where \(B\) is a constant. Solve for \(C_{2}\), which gives \(C_{2}(t) = 1 + B e^{-t}\). Using the initial condition \(C_{2}(0) = 0\), we solve for \(B\): \(1 + B = 0\), thus \(B = -1\). Therefore, \(C_{2}(t) = 1 - e^{-t}\).

05

Sketch the Functions

The function \(C_{1}(t) = e^{-t}\) is an exponential decay starting at 1 and approaching 0 as \(t\) increases. The function \(C_{2}(t) = 1 - e^{-t}\) is an exponential growth starting at 0 and approaching 1 as \(t\) increases. Plot these functions against time to visualize the changes in concentration over time.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Differential Equations

In mathematics, a differential equation is an equation that involves the derivatives of a function. These equations describe how a particular quantity changes over time, which makes them ideal for modeling dynamic systems. In this exercise, we dealt with a system of two differential equations:

• \(\frac{d C_{1}}{d t} = -(C_{1})\)
• \(\frac{d C_{2}}{d t} = (C_{1} - C_{2})\)

These equations are characteristic of a two-compartment model, where each compartment represents a tank with a certain concentration of solute. The rates of change in concentration in each tank are affected by the flow of the solution. Recognizing the form and structure of these differential equations allows us to choose appropriate methods to solve them, highlighting their importance in understanding how systems evolve over time.

Initial Value Problem

An initial value problem in the context of differential equations is a problem where a function is determined by specifying its value at a particular point, usually the initial condition. For this exercise:

• \(C_{1}(0) = 1\)
• \(C_{2}(0) = 0\)

These initial conditions tell us that at time \(t=0\), the concentration in the first tank is at its maximum, while the second tank starts with no concentration. Initial conditions are crucial because they provide the necessary information to determine the unique solution to a differential equation. By setting these values, you ensure that the solution path aligns exactly with the physical situation you're modeling.

Integrating Factor

The integrating factor is a technique used to solve linear first-order ordinary differential equations of the form \(\frac{dy}{dt} + P(t) y = Q(t)\). By multiplying through by an appropriate function, called the integrating factor, the differential equation can be transformed into an exact equation, making it easier to solve. For instance, with our differential equation for \(C_{2}\):

• Equation: \(\frac{d C_{2}}{d t} = e^{-t} - C_{2}\)
• Integrating factor: \(e^{t}\)

By multiplying through by \(e^{t}\), the equation simplifies to the derivative \(\frac{d}{dt}(e^{t}C_{2}) = e^{t}\). This transformation simplifies finding \(C_{2}(t)\) because it allows us to integrate directly, leading to a clear solution path.

Exponential Functions

Exponential functions are fundamental in many mathematical models, particularly when describing growth or decay processes. In this problem, both \(C_{1}(t)\) and \(C_{2}(t)\) are expressed in terms of exponential functions:

• \(C_{1}(t) = e^{-t}\)
• \(C_{2}(t) = 1 - e^{-t}\)

These functions describe the concentrations in our two-tank system over time. The term \(e^{-t}\) in \(C_{1}(t)\) indicates that the concentration in the first tank decreases exponentially, reflecting decay. On the other hand, \(C_{2}(t) = 1 - e^{-t}\) shows an exponential increase towards a stable concentration level in the second tank. This behavior captures the essence of how each compartment evolves, with tank one losing and tank two gaining concentration over time.