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Chapter 8: Problem 41

A differential equation of the form \(y^{\prime}(t)=f(y)\) is said to be autonomous (the function \(f\) depends only on \(y\) ). The constant function \(y=y_{0}\) is an equilibrium solution of the equation provided \(f\left(y_{0}\right)=0\) (because then \(y^{\prime}(t)=0\) and the solution remains constant for all \(t\) ). Note that equilibrium solutions correspond to horizontal lines in the direction field. Note also that for autonomous equations, the direction field is independent of t. Carry out the following analysis on the given equations. a. Find the equilibrium solutions. b. Sketch the direction field, for \(t \geq 0\). c. Sketch the solution curve that corresponds to the initial condition \(y(0)=1\). $$y^{\prime}(t)=y(y-3)$$

Short Answer

Expert verified
Answer: The equilibrium solutions are \(y = 0\) and \(y = 3\). The solution curve for the initial condition \(y(0) = 1\) starts at \((0,1)\) and follows the direction field without touching the equilibrium lines.

Step by step solution

01

Find the equilibrium solutions

To find the equilibrium solutions, we will set \(y^{\prime}(t)=0\) and solve for \(y\): $$y(y-3) = 0$$ This equation is true when either \(y = 0\) or \(y = 3\). Therefore, the equilibrium solutions are \(y = 0\) and \(y = 3\).
02

Sketch the direction field

To sketch the direction field of the given differential equation \(y^{\prime}(t) = y(y-3)\), we need to compute the derivative for a range of \(y\) values and plot the corresponding slopes. Following the slope of the direction field, we can visualize how the function behaves: - For \(y < 0\), we have \(y^{\prime}(t) > 0\) (positive slope). - For \(0 < y < 3\), we have \(y^{\prime}(t) < 0\) (negative slope). - For \(y > 3\), we have \(y^{\prime}(t) > 0\) (positive slope). Now we can sketch the direction field, keeping in mind that equilibrium solutions are horizontal lines (constant \(y\) values), and the slopes vary about the \(y\) values.
03

Sketch the solution curve for given initial condition

Now let's sketch the solution curve for the initial condition \(y(0) = 1\). Locate the point \((0, 1)\) on the \(ty\)-plane, and note that it is between the equilibrium solutions found previously (\(y = 0\) and \(y = 3\)). Following along the direction field from the initial condition \((0,1)\), we can see that the solution curve will pass through points with negative slopes on the \(ty\)-plane. The curve will begin at \((0,1)\), following the direction of the negative slope, without touching the equilibrium lines (as those are constant solutions). In conclusion: - The equilibrium solutions are \(y = 0\) and \(y = 3\). - The direction field can be sketched using the slopes computed depending on \(y\). - The solution curve for the initial condition \(y(0) = 1\) starts at \((0,1)\) and follows the direction field without touching the equilibrium lines.

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Key Concepts

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

Equilibrium Solutions
An equilibrium solution for a differential equation represents a state where the system remains unchanging. For an autonomous differential equation expressed as \( y'(t) = f(y) \), an equilibrium solution occurs at constant values of \( y \) where the derivative equals zero. This implies the system is in balance.

To find these points, we solve \( f(y) = 0 \). For the given problem \( y'(t) = y(y - 3) \), setting this equal to zero gives us the points \( y = 0 \) and \( y = 3 \). At these points, \( y'(t) = 0 \), indicating no change in \( y \) over time. These are the constant solutions where the differential equation remains static.

Understanding equilibrium solutions is crucial as they often represent stable states in a system, like an object's rest position under the influence of forces. In a graph, these solutions are represented by horizontal lines, which help in visualizing the behavior of the system at various points.
Direction Field
A direction field, sometimes called a slope field, is a visual tool used to represent the behavior of solutions to a differential equation. For an autonomous differential equation like \( y'(t) = y(y-3) \), the direction field shows how the slope of solutions changes with respect to \( y \) at each point.

To construct a direction field, we first determine the slope \( y'(t) \) at various values of \( y \):
  • For \( y < 0 \), the slope is positive, implying that the solution curves rise as \( t \) increases.
  • For \( 0 < y < 3 \), the slope is negative, so the solution curves fall as \( t \) increases.
  • For \( y > 3 \), the slope becomes positive again, indicating rising curves.
These patterns help us sketch the direction field by drawing short line segments on the \( ty \)-plane indicating these slopes. Equilibrium solutions appear as horizontal lines in this field, marking points where \( y'(t) = 0 \), thus no slope or change.

Using a direction field allows us to get an intuitive grasp of how solutions behave over time and aids in predicting long-term behavior without solving the equation explicitly.
Solution Curve
A solution curve represents a specific solution to a differential equation given an initial condition. In this context, the curve is traced in the \(ty\)-plane starting from an initial condition, such as \( y(0) = 1 \).

For the differential equation \( y'(t) = y(y-3) \), we start at the point \( (0,1) \) on the \(ty\)-plane. This point falls between the equilibrium solutions \( y = 0 \) and \( y = 3 \), where the slope \( y'(t) \) is negative. This indicates that the solution curve will tend downwards, reflecting a decrease in \( y \) over time.

As we follow the direction field from \( (0,1) \), the curve follows the path dictated by the slope, moving in a direction determined by the interplay of forces described by the differential equation. Solution curves are crucial for visualizing and understanding the behavior of a system under given conditions, providing insights into how variables evolve over time relative to starting conditions.

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Most popular questions from this chapter

For each of the following stirred tank reactions, carry out the following analysis. a. Write an initial value problem for the mass of the substance. b. Solve the initial value problem and graph the solution to be sure that \(m(0)\) and \(\lim _{t \rightarrow \infty} m(t)\) are correct. A one-million-liter pond is contaminated and has a concentration of \(20 \mathrm{g} / \mathrm{L}\) of a chemical pollutant. The source of the pollutant is removed and pure water is allowed to flow into the pond at a rate of \(1200 \mathrm{L} / \mathrm{hr} .\) Assuming that the pond is thoroughly mixed and drained at a rate of \(1200 \mathrm{L} / \mathrm{hr}\), how long does it take to reduce the concentration of the solution in the pond to \(10 \%\) of the initial value?

Determine whether the following equations are separable. If so, solve the initial value problem. $$y^{\prime}(t)=\frac{e^{t}}{2 y}, y(\ln 2)=1$$

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 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 if \(A<-b / a,\) then the solution increases for \(t \geq 0\).

Solve the differential equation for Newton's Law of Cooling to find the temperature in the following cases. Then answer any additional questions. A pot of boiling soup \(\left(100^{\circ} \mathrm{C}\right)\) is put in a cellar with a temperature of \(10^{\circ} \mathrm{C}\). After 30 minutes, the soup has cooled to \(80^{\circ} \mathrm{C}\). When will the temperature of the soup reach \(30^{\circ} \mathrm{C} ?\)

a. Show that for general positive values of \(R, V, C_{i},\) and \(m_{0},\) the solution of the initial value problem $$m^{\prime}(t)=-\frac{R}{V} m(t)+C_{i} R, \quad m(0)=m_{0}$$ is \(m(t)=\left(m_{0}-C_{i} V\right) e^{-R t / V}+C_{i} V\) b. Verify that \(m(0)=m_{0}\) c. Evaluate \(\lim m(t)\) and give a physical interpretation of the result. d. Suppose \(^{t} \vec{m}_{0}^{\infty}\) and \(V\) are fixed. Describe the effect of increasing \(R\) on the graph of the solution.
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