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Do a qualitative analysis of the family of solutions to each of the differential equations below. Then, in another color pen or pencil, highlight the graphs of the solutions corresponding to the given initial conditions. (If the solution is asymptotic to a horizontal line, draw and label that line.) \(\begin{array}{llll}\text { (a) } \frac{d y}{d t}=4 y-8 ; & y(0)=0, & y(0)=-1, & y(0)=3\end{array}\) (b) \(\frac{d y}{d t}=y^{2}-4\) \(y(0)=-1, \quad y(0)=-3, \quad y(0)=4\) (c) \(\frac{d y}{d t}=(y-1)(y-2)(y+1) ; \quad y(0)=0, \quad y(0)=3\) \(\begin{array}{llll}\text { (d) } \frac{d y}{d t}=y^{2}+5 y-6 ; & y(0)=-5, & y(0)=-7, & y(0)=2\end{array}\)

Short Answer

Expert verified
Qualitative analysis reveals the general direction and segment forms of solution curves for the respective differential equations, based on their equilibrium solution scenarios and initial conditions. Evaluating the differential equations, one can observe direction changes in solutions at these equilibrium points, which then guide the graphing of solution curves for the required initial conditions. Thus several patterns of slopes are produced to complete the qualitative analysis.

Step by step solution

01

Solving part (a)

For the differential equation \(\frac{dy}{dt} = 4y - 8\), an equilibrium is given when \(\frac{dy}{dt} = 0\), or \(y = 2\). A upward sloping line represents solutions where \(y > 2\) since \(\frac{dy}{dt} > 0\) in that condition, and a downward sloping represents solutions where \(y < 2\) as \(\frac{dy}{dt} < 0\) in that condition. With the initial conditions \(y(0) = 0, -1, 3\), we sketch and highlight a solution curving upward from \(0\), one curving downward from \(3\), and one going upwards from \(-1\), all towards \(y = 2\).
02

Solving part (b)

For the differential equation \(\frac{dy}{dt} = y^{2} - 4\), equilibrium solutions are at \(y = -2\) and \(y = 2\). Since \(\frac{dy}{dt} > 0\) when \(|y| > 2\), solutions will slope away from the equilibrium lines. With initial conditions \(y(0) = -1, -3, 4\), we sketch and highlight three separate solution curves accordingly.
03

Solving part (c)

For the differential equation \(\frac{dy}{dt} = (y - 1)(y - 2)(y + 1)\), there are three equilibrium solutions where \(y = -1, 1, 2\). Solutions will be directed away from \(y = 1\) and \(y = -1\) and towards \(y = 2\) when the conditions are met. Initial conditions \(y(0) = 0,3\) are provided, hence two solution curves are drawn and highlighted.
04

Solving part (d)

For the differential equation \(\frac{dy}{dt} = y^{2} + 5y - 6\), the equilibriums are where \(y = -6\) or \(y = 1\). Solution curves are directed away from the equilibrium \(y = -6\) when \(y < -6\) or \(y > 1\), and towards it when \(-6 < y < 1\). With initial conditions \(y(0) = -5, -7, 2\), three separate solution curves are sketched and highlighted.

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Key Concepts

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

Qualitative Analysis of Differential Equations
Qualitative analysis deals with the behavior of solutions to differential equations without solving them explicitly. It provides insight into how solutions evolve over time by examining the structure of the differential equation itself.
For example, in the differential equation \(\frac{dy}{dt} = 4y - 8\), understanding that solutions above or below a certain point (the equilibrium) behave differently is pivotal. This point often indicates stability or lack thereof in solutions. Here, the equilibrium is \(y = 2\).
This method allows us to predict solution behaviors, such as whether they will increase or decrease over time, approach certain values, or eventually become unbounded. By sketching solution curves, we can visually interpret these dynamics, making the abstract math more tangible.
Equilibrium Solutions in Differential Equations
Equilibrium solutions are specific solutions to a differential equation where change does not happen over time; mathematically, this is when \(\frac{dy}{dt} = 0\). They highlight where solutions can settle or stabilize if they start nearby.
  • For \(\frac{dy}{dt} = 4y - 8\), we find the equilibrium by solving \(4y - 8 = 0\) obtaining \(y = 2\).
  • In \(\frac{dy}{dt} = y^2 - 4\), setting \(y^2 - 4 = 0\) yields equilibria at \(y = \pm 2\).
These solutions are crucial as they guide whether other solutions will grow towards them or diverge away, depending on the initial conditions. Understanding equilibria helps categorize the stability of these points and how other solutions behave relative to them.
Initial Conditions and Their Role
Initial conditions specify the state of the system at the beginning of observation and profoundly influence the resulting solution curves. Each unique initial condition can lead to drastically different behavior.
For instance, in \(\frac{dy}{dt} = 4y - 8\) with \(y(0) = 0\), the solution begins below the equilibrium \(y = 2\) and will rise towards it. In contrast, an initial condition of \(y(0) = 3\) starts above \(y = 2\) and will move downward, reaching the equilibrium from above.
Initial conditions are essential in plotting solution curves; they provide the starting point from which we extend our qualitative guesses into quantitative solutions.
Understanding Solution Curves
Solution curves are graphical representations showing how solutions to a differential equation change over time. They help visualize the behavior of solutions provided by the differential equation, taking into account initial conditions and equilibria.
  • Consider \(\frac{dy}{dt} = y^2 - 4\). Solution curves reveal how trajectories begin from initial conditions like \(y(0) = -1\), allowing us to see this solution will move towards equilibrium points \(y = -2\) or \(y = 2\).
  • The nature of these paths helps us understand phenomena such as stability, convergence, and divergence relative to equilibrium points.
Such analysis is crucial in applications where understanding the dynamics over time can predict the eventual outcome, and is commonly used in fields like biology, physics, and engineering.

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Most popular questions from this chapter

We can construct a model for the spread of a disease by assuming that people are being infected at a rate proportional to the product of the number of people who have already been infected and the number of those who have not. Let \(P(t)\) denote the number of infected people at time \(t\) and \(N\) denote the total population affected by the epidemic. Assume \(N\) is xed throughout the time period we are considering. We are assuming that every member of the population is susceptible to the disease and the disease is long in duration (there are no recoveries during the time period we are analyzing) but not fatal (no deaths during this period). The assumption that people are being infected at a rate proportional to the product of those who are infected and those who are not could re ect a contagious disease where the sick are not isolated. Write a differential equation whose solution is \(P(t)\).

The population of wildebeest in the Serengeti was decimated by a rinderpest plague in the \(1950 \mathrm{~s}\). In 1961 the Serengeti supported a population of a quarter of a million wildebeest. By 1978 the wildebeest population was \(1.5\) million and by 1991 it had reached 2 million. (Craig Packer Into Africa, Chicago, The University of Chicago Press, 1996 p. \(250 .\) ) Given this data, would you be more inclined to model the growth of the wildebeest population using an exponential growth model or using a logistic growth model? Explain your reasoning.

(a) You plan to save money starting today at a rate of \(\$ 4000\) per year over the next 30 years. You will deposit this money at a nearly continuous rate (a constant amount each day) into a bank account that earns \(5 \%\) interest compounded continuously. Let \(B(t)\) be the balance of money in the account \(t\) years from now, where \(0 \leq t \leq 30\) i. Write a differential equation whose solution is \(B(t)\). ii. Write an integral that is equal to \(B(30)\), the amount in the account at the end of 30 years. (b) Now assume that instead of making deposits continuously, you decide to make a deposit of \(\$ 4000\) once a year, starting today and continuing until you have made a total of 30 deposits. Suppose the bank account pays \(5 \%\) interest compounded annually. i. Write a geometric sum equal to the balance immediately after the final deposit. ii. Find a closed form expression (no \(+\cdots+\), no summation notation) for this sum.

Sketch a representative family of solutions for each of the following differential equations. $$ \frac{d x}{d t}=(1-x)(x+2)(x-3) $$

Sketch a representative family of solutions for each of the following differential equations. (a) \(\frac{d y}{d t}=\sin t\) (b) \(\frac{d y}{d t}=\sin y\)

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