Question

A tank contains 100 gallons of water and 50 oz of salt. Water containing a salt concentration of \(\frac{1}{4}\left(1+\frac{1}{2} \sin t\right)\) oz/gal flows into the tank at a rate of 2 gal/min, and the mixture in the tank flows out at the same rate. $$ \begin{array}{l}{\text { (a) Find the amount of salt in the tank at any time. }} \\ {\text { (b) Plot the solution for a time period long enough so that you see the ultimate behavior }} \\ {\text { of the graph. }} \\ {\text { (c) The long-time behavior of the solution is an oscillation about a certain constant level, }} \\ {\text { What is this level? What is the amplitude of the oscillation? }}\end{array} $$

Short answer

(a) The amount of salt in the tank at any time is given by the function: $$S(t)=25(t-50\cos(t))+50-50e^{-\frac{t}{50}}$$ (b) The long-time behavior of the solution graph will be an oscillation centered around 50 oz, with a maximum amplitude of 25 oz. This is due to the sinusoidal term in the function. (c) The constant level of the long-time behavior is 50 oz, which represents the center of oscillation. The amplitude of oscillation is 25 oz, representing the maximum deviation from the constant level.

Step-by-step solution

(1. Set up the differential equation)

We can begin by creating a differential equation to represent the given problem. Let \(S(t)\) denote the amount of salt (in oz) in the tank at time \(t\), and let \(V(t)=100\) denote the volume (in gallons) of the tank's content. The rate of salt entering the tank is given by the salt concentration multiplied by the rate of flow into the tank, which is \(2 \cdot \frac{1}{4}\left(1+\frac{1}{2} \sin t\right)=\frac{1}{2}\left(1+\frac{1}{2} \sin t\right)\). The rate of salt leaving the tank is given by the salt concentration in the tank multiplied by the rate of flow out of the tank, which is \(\frac{S(t)}{100} \cdot 2=\frac{S(t)}{50}\). Therefore, the rate of change of salt in the tank is given by the following differential equation:$$\frac{dS}{dt}=\frac{1}{2}\left(1+\frac{1}{2} \sin t\right)-\frac{S(t)}{50}$$

(2. Solve the differential equation)

To solve the given differential equation, first notice that the particular solution is \(S_p=25(t-a\cos(t))+50\), where \(a=50\). To solve for the homogenous part, observe that the corresponding homogenous equation is \(\frac{dS}{dt}=-\frac{S(t)}{50}\). The homogeneous equation can be solved by separating variables:$$\frac{dS}{S(t)}=-\frac{1}{50}dt$$Integrate both sides to obtain:$$\ln|S(t)|=-\frac{t}{50}+C$$Exponentiation gives the general solution of the homogeneous equation:$$S_{h}(t)=De^{-\frac{t}{50}}$$where \(D\) is an integration constant. Combining the particular and homogeneous solutions gives the general solution for the original equation:$$S(t)=25(t-50\cos(t))+50+De^{-\frac{t}{50}}$$

(3. Determine the initial condition)

Since we know the initial amount of salt in the tank is 50 oz, we can use this information to find the value of \(D\):$$S(0)=25(0-50\cos(0))+50+De^{-\frac{0}{50}}=50$$Solving for \(D\), we get \(D=-50\).

(4. Find the general solution)

Now that we have the value for \(D\), we can write the general solution of the differential equation that gives the amount of salt in the tank at any time:$$S(t)=25(t-50\cos(t))+50-50e^{-\frac{t}{50}}$$

(5. Analyze long-time behavior)

After a long period of time, the exponential term will decay to zero, meaning that the long-time behavior of the solution will be dominated by the sinusoidal term. We can then write the long-time solution as:$$S(t) \approx 25(t-50\cos(t))+50$$The ultimate behavior of the graph will be an oscillation centered around 50 oz, with a maximum amplitude of 25 oz (corresponding to the coefficient of the sine term).

Key concepts

Initial Value Problem

An initial value problem in the context of differential equations requires finding a solution that not only satisfies the differential equation but also meets an initial condition. In our water tank example, we have an initial value problem where we start with an initial amount of salt in the tank, 50 oz. This initial condition is crucial as it allows us to determine the integration constant and therefore pinpoint the specific solution that fits our scenario.

To solve such problems, we first establish the differential equation describing the system. Then, by applying the initial condition, we calculate any constants that appear in the general solution. For instance, in the sample solution, the initial condition is used to find the constant, resulting in a particularized solution that applies to our situation.

This step ensures that the derived equation accurately reflects the state of the system at the start of the analysis, providing a complete snapshot of both the past and future of the system.

Homogeneous Equation

In differential equations, a homogeneous equation is a type where all terms are a function of the dependent variable and its derivatives. It does not include \(t\) or any other independent terms on its own. In the example exercise, the homogeneous equation is derived separately before the full solution. It is represented without the external forces or input terms.

For the problem at hand, the homogeneous equation is given by:
$$\frac{dS}{dt} = -\frac{S(t)}{50}.$$ Solving this involves separating variables, integrating, and finding a general form with an integration constant. The solution to the homogeneous equation provides insight into the natural dynamics of the system without any external influence, such as water entering the tank. Understanding this part helps us see how the system behaves independently.

• Separate variables to isolate differentials.
  
• Integrate both sides to find the natural log function.
  
• Solve for \(S(t)\) to find the general form.
  

These steps help us build the structure for the complete solution by later adding a particular solution, which reflects external influences.

Particular Solution

The particular solution reflects how the system behaves under specific external influences. In solving differential equations, it accounts for input functions or constants that are not present in the homogeneous solution. For our tank example, this particular solution caters to the steady addition of salt alongside the homogeneous behavior.

The particular solution here is given by:
$$S_p(t) = 25(t - 50\cos(t)) + 50.$$ This expression addresses the periodic input of salt concentration as described in the exercise. It structures the overall equation to include effects like periodic changes brought about by the sinusoidal input.

• Considers sinusoidal inflow due to its oscillatory nature.
  
• Modifies the system response accordingly.
  

By combining it with the homogeneous solution, a comprehensive picture of the system’s response emerges.

Oscillation Analysis

Oscillation analysis involves examining the repetitive variation, typically in time, within the solution to a differential equation. It's critical when time-dependent inputs, such as sinusoidal waves, affect the system. Understanding these oscillations informs us about the stability and long-term behavior of the system.In our example, the solution
$$S(t) \approx 25(t - 50\cos(t)) + 50$$ oscillates due to the sinusoidal term. Over a long period, the effect of any transitory components like exponential decay becomes negligible. The stable characteristic is the consistent oscillation around 50 oz (the constant level), with a clear amplitude of 25 oz defined by half the coefficient of the sine function.Oscillation analysis helps to predict:

• Long-term behavior of the system.
  
• The central constant level (here, 50 oz).
  
• The amplitude of variation around this level (here, 25 oz).
  

Therefore, oscillation analysis teaches us how external periodic inputs shape the system’s ongoing states.