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Chapter 9: Problem 17

Find the equilibrium solution of the following equations, make a sketch of the direction field, for \(t \geq 0,\) and determine whether the equilibrium solution is stable. The direction field needs to indicate only whether solutions are increasing or decreasing on either side of the equilibrium solution. $$y^{\prime}(t)=12 y-18$$

Short Answer

Expert verified
Answer: The equilibrium solution of the given differential equation is \(y_{eq} = \frac{3}{2}\), and it is stable as the solutions approach \(y_{eq}\) on either side.

Step by step solution

01

Find the equilibrium solution

To find the equilibrium solution, we need to set the derivative equal to zero and solve for y: $$y^{\prime}(t)=12y-18=0$$ Solving for y gives us the equilibrium solution: $$y_{eq}=\frac{18}{12}=\frac{3}{2}$$
02

Determine the direction of solutions on either side

To understand the stability of the equilibrium solution, we need to analyze the behavior of the solutions on either side of \(y_{eq}=\frac{3}{2}\). We will consider the sign of y'(t) for values of y greater and smaller than \(\frac{3}{2}\). When \(y > \frac{3}{2}\): $$y^{\prime}(t)=12y-18 > 12 \left(\frac{3}{2}\right) -18 = 0$$ So, the solutions are increasing in this region. When \(y < \frac{3}{2}\): $$y^{\prime}(t)=12y-18 < 12 \left(\frac{3}{2}\right) -18 = 0$$ So, the solutions are decreasing in this region.
03

Analyze the equilibrium solution's stability

From the previous step, we can see that when y is greater than \(\frac{3}{2}\), the solutions are increasing, whereas when y is smaller than \(\frac{3}{2}\), the solutions are decreasing. This behavior indicates that the equilibrium solution at \(y_{eq}=\frac{3}{2}\) is stable since the solutions approach \(y_{eq}\) on either side. To make a sketch of the direction field: - Mark \(y_{eq}=\frac{3}{2}\) as a horizontal line. - For \(y > \frac{3}{2}\), sketch arrows pointing to the right and upwards, indicating the solutions are increasing. - For \(y < \frac{3}{2}\), sketch arrows pointing to the right and downwards, indicating the solutions are decreasing.

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Key Concepts

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

Direction Field
A direction field, also known as a slope field, is a visual representation that helps us understand the behavior of a differential equation's solutions without actually solving the equation. Imagine it as a field of tiny arrows drawn on a graph, each indicating the slope or direction of the solution at that particular point. For our exercise, we have the differential equation \(y'(t) = 12y - 18.\)By evaluating this equation at various points, we can sketch a direction field:
  • Determine a series of y-values, and calculate \(y'(t)\) for each.
  • If \(y'(t) > 0\), draw an arrow pointing upwards (indicating a positive slope).
  • If \(y'(t) < 0\), draw an arrow pointing downwards (indicating a negative slope).
In this way, the direction field visualizes which way the solutions are trending, either increasing or decreasing, helping us see patterns of stability and equilibrium.
Stability Analysis
Stability analysis is a method used to determine the behavior of solutions around an equilibrium point. In simple terms, we check whether solutions approach or diverge from equilibrium over time. Consider the equilibrium solution\(y_{eq} = \frac{3}{2}\). To assess stability:
  • For \(y > \frac{3}{2}\), \(y'(t) = 12y - 18 > 0\), indicating solutions are moving upwards.
  • For \(y < \frac{3}{2}\), \(y'(t) = 12y - 18 < 0\), indicating solutions are moving downwards.
Because solutions head towards the equilibrium from both sides, our equilibrium solution is considered stable. This means that no matter where the system starts, it will eventually settle at \(y_{eq} = \frac{3}{2}\). This kind of stability is also referred to as asymptotic stability.
Differential Equations
A differential equation is a mathematical equation that relates a function with its derivatives. In our context, \(y'(t) = 12y - 18\) is a first-order linear differential equation. The aim is to understand how function \(y\) changes with respect to \(t\).

To solve a differential equation, we often look for equilibrium solutions, where the rate of change is zero (\(y'(t) = 0\)). These solutions help in understanding the long-term behavior of systems modeled by differential equations.
  • First-order indicates the equation involves the first derivative of \(y\).
  • Linear refers to the highest power of \(y\) and its derivatives being one.
Differential equations are critical tools for modeling real-world situations where change over time is involved, providing us with powerful insights into the behavior of dynamic systems.

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Most popular questions from this chapter

Find the equilibrium solution of the following equations, make a sketch of the direction field, for \(t \geq 0,\) and determine whether the equilibrium solution is stable. The direction field needs to indicate only whether solutions are increasing or decreasing on either side of the equilibrium solution. $$y^{\prime}(t)=-6 y+12$$

The amount of drug in the blood of a patient (in milligrams) administered via an intravenous line is governed by the initial value problem \(y^{\prime}(t)=-0.02 y+3\) \(y(0)=0,\) where \(t\) is measured in hours. a. Find and graph the solution of the initial value problem. b. What is the steady-state level of the drug? c. When does the drug level reach \(90 \%\) of the steady-state value?

Direction field analysis Consider the first-order initial value problem \(y^{\prime}(t)=a y+b, y(0)=A,\) for \(t \geq 0,\) where \(a, b,\) and \(A\) are real numbers. a. Explain why \(y=-b / a\) is an equilibrium solution and corresponds to a horizontal line in the direction field. b. Draw a representative direction field in the case that \(a>0\) Show that if \(A>-b / a\), then the solution increases for \(t \geq 0,\) and that if \(A<-b / a,\) then the solution decreases for \(t \geq 0\). c. Draw a representative direction field in the case that \(a<0\) Show that if \(A>-b / a\), then the solution decreases for \(t \geq 0,\) and that if \(A<-b / a,\) then the solution increases for \(t \geq 0\).

Explain how a stirred tank reaction works.

A physiological model A common assumption in modeling drug assimilation is that the blood volume in a person is a single compartment that behaves like a stirred tank. Suppose the blood volume is a four-liter tank that initially has a zero concentration of a particular drug. At time \(t=0,\) an intravenous line is inserted into a vein (into the tank) that carries a drug solution with a concentration of \(500 \mathrm{mg} / \mathrm{L} .\) The inflow rate is \(0.06 \mathrm{L} / \mathrm{min} .\) Assume the drug is quickly mixed thoroughly in the blood and that the volume of blood remains constant. a. Write an initial value problem that models the mass of the drug in the blood, for \(t \geq 0\) b. Solve the initial value problem, and graph both the mass of the drug and the concentration of the drug. c. What is the steady-state mass of the drug in the blood? d. After how many minutes does the drug mass reach \(90 \%\) of its steady-state level?
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