Question

Suppose that tank 1 initially contains 100 gallons of solution, with 50 pounds of dissolved salt, and tank 2 contains \(200 \mathrm{gal}\) lons, with 150 pounds of dissolved salt. Pure water flows into tank 1 at 2 gallons per minute, the well-mixed solution flows out and into tank 2 at the same rate, and finally, the solution in tank 2 drains away also at the same rate. Let \(x(t)\) and \(y(t)\) denote the amounts of salt in tanks 1 and 2 , respectively, at time \(t\). Find \(y(t)\). Hint: First find \(x(t)\) and use it in setting up the differential equation for tank \(2 .\)

Short answer

The function \(y(t) = 50e^{-t/100} + 50\) describes the salt amount in Tank 2 over time.

Step-by-step solution

Understand the system flow

The system involves two tanks. Tank 1 has a solution initially with 50 pounds of salt in 100 gallons, and Tank 2 has 150 pounds in 200 gallons. Pure water enters Tank 1 at 2 gallons per minute, and this well-mixed solution flows to Tank 2 at the same rate. From Tank 2, the solution drains at the same rate.

Write the differential equation for Tank 1

Since pure water is added to Tank 1, the concentration of salt decreases. The change in salt in Tank 1, \(x(t)\), is described by the differential equation: \(\frac{dx}{dt} = -\frac{2}{100}x(t)\) because when 2 gallons of the mixture leave Tank 1, 2/100 of the salt also leaves.

Solve the differential equation for Tank 1

The differential equation \(\frac{dx}{dt} = -\frac{1}{50}x(t)\) is separable. Integrating gives \(x(t) = C e^{-t/50}\). Given the initial condition \(x(0) = 50\), solve for \(C\): \(C = 50\). Thus, \(x(t) = 50e^{-t/50}\).

Write the differential equation for Tank 2

In Tank 2, saline enters at the rate of \(\frac{2}{100}x(t)\) from Tank 1 and leaves at the same rate \(\frac{2}{200}y(t)\). Hence, the differential equation is \(\frac{dy}{dt} = \frac{2}{100}x(t) - \frac{2}{200}y(t)\).

Substitute \(x(t)\) into the differential equation for Tank 2

Substituting \(x(t) = 50e^{-t/50}\) into the differential equation for \(y(t)\), we have \(\frac{dy}{dt} = 1 e^{-t/50} - \frac{1}{100}y(t)\).

Solve the differential equation for Tank 2

This is a first-order linear differential equation. Using the integrating factor method, solve: \(\mu(t) = e^{\int \frac{1}{100}dt} = e^{t/100}\). Then, \(y(t)e^{t/100} = \int e^{-t/50}e^{t/100} dt\). Simplifying, \(\int e^{-t/100} dt = -100e^{-t/100} + C_2\). Thus, \(y(t) = -100 + C_2 e^{-t/100}\). Applying initial condition \(y(0) = 150\), solve for \(C_2\) to find the complete solution for \(y(t)\): \(y(t) = 50e^{-t/100} + 50\).

Key concepts

Systems of Differential Equations

In many real-world situations, we encounter systems involving more than one differential equation, which are known as systems of differential equations. When working with systems of differential equations, each equation involves two or more functions and their derivatives. These systems often model complex interactions between variables over time.
A key part of solving these problems is understanding the flow within each element of the system. For instance, in our tank exercise, we have two interconnected tanks that affect each other because the combination influences how salt moves from Tank 1 to Tank 2. Analyzing this system involves writing down the differential equations for each tank based on their interactions.
To solve systems like these, you'll normally follow a few steps:

• Create equations that capture how each variable changes given the rates of flow.
• Solve one equation first, often the simpler one, and use its solution in the second differential equation.
• Apply all given initial conditions to find any remaining constants.

By following these steps, you can often uncover the behavior of all elements in such systems.

Exponential Growth and Decay

Exponential growth and decay describe processes where a quantity increases or decreases at a rate proportional to its current value. These processes are commonly modeled using the equation \( \frac{dN}{dt} = rN \), where \( N(t) \) is the quantity of interest and \( r \) is the rate.
In our exercise, the salt concentration in Tank 1 exemplifies exponential decay. As pure water enters Tank 1, the amount of salt decreases over time. This can be mathematically represented as \( \frac{dx}{dt} = -\frac{1}{50}x(t) \), where the negative sign indicates that the salt amount is decreasing. Solving this differential equation, you find \( x(t) = 50e^{-t/50} \). Here, the exponent captures the decay of salt concentration in Tank 1 over time.
Processes involving exponential decay or growth are ubiquitous in science and engineering, modeling phenomena such as radioactive decay or population growth.

Separable Differential Equations

Separable differential equations are a type of differential equation that can be reorganized into two separate integrals. This means you can isolate each variable on one side of the equation. This property makes them straightforward to solve compared to other types.
For example, consider the equation \( \frac{dx}{dt} = -\frac{1}{50}x(t) \. \) This is separable as you can rewrite it as \( \frac{1}{x}dx = -\frac{1}{50}dt \. \) By integrating both sides, you obtain the general solution of the form \( x(t) = Ce^{-t/50} \, \) where \( C \) is a constant determined by initial conditions.
In our tank problem, separable differential equations help find the time-dependent concentration of salt in the tanks. Their intuitive nature and clear solution path make them a favorite subject in many first courses on differential equations.