Section 7.2 Reading Assignment: Exponential Change and Separable Differential Equations

Answer Only Exercise 1, 2, and 3 by using a screenshot provided Calculus Pearson textbook. Make sure you read these three questions very carefully and see on what it is asking for and what is really about.

References: Thomas' Calculus: Early Transcendentals | Calculus | Calculus | Mathematics | Store | Pearson+

Feedback: #1: Consider the computation in this example and examine the algebraic position of where the C ends up here. That is, in terms of the algebraic solution, where does the C end up in the expression? (0/2).

Section 7.2 Reading Assignment: Exponential Change and Separable Differential Equations Instructions. Read through this assignment and complete the three exercises below by reading the appropriate passages of the textbook. Section 9.1 gives a better introduction to differential equations compared to this earlier section in the book, but this section will give us the first main strategy for solving differential equations. One thing to watch out for is what happens to the constant of integration when solving a differential equation. Exercise 1. Read the solution of initial-value problem (1), i.e. the differential equation (1a) with initial condition (1b) on p. 436. After the integration step, there is a constant of integration (+C). Explain where this constant ends up in the final answer. It's important to remember this exercise, since solving differential equations will result in a different location for that constant of integration. This example also highlights the method for solving certain differential equations. The next exercises cover what these equations are and the strategy for solving these equations. Exercise 2. Read the introduction to the subsection "Separable Differential Equations" (p. 436 - 437). Explain what it means for a differential equation to be separable. Exercise 3. Read Example 1 (p. 437) and Example 2 (p. 438). Summarize how to solve a separable differential equation.The textbook leaves the solutions in Example 1 and 2 implicit rather than solving for y explicitly. This is an unfortunate choice in my opinion. We often prefer having explicit solutions, but also the constant of integration is quite important and can get up to some algebraic mischief as shown in Exercise 1. Notice that the equation in Example 2 is not completely obvious that the differential equation is separable. It can be important to do some preliminary algebra to determine when an equation is separable. We will be skipping the rest of the section. Hopefully, you have seen this material before when discussing exponential functions in your algebra or precalculus classes. It is worth noting that the reason why we considered these exponential models in those classes is based on being a solution to a basic differential equation.Chapter 7 Integrals and Transcendental Functions 436 Chapter 7 Integrals and Transcendental Functions If the amount present at time / = 0 is called y, then we can find y as a function of by solving the following initial value problem: Differential equation: dy = ky (la) Initial condition: y = jb when 7 = 0. (1b) If y is positive and increasing, then & is positive, and we use Equation (la) to say that the rate of growth is proportional to what has already been accumulated. If y is positive and decreasing, then & is negative, and we use Equation (la) to say that the rate of decay is proportional to the amount still left. We see right away that the constant function y = 0 is a solution of Equation (la) if vo - 0. To find the nonzero solutions, we divide Equation (la) by y: 1 dy 1 dy Integrate with respect to r In ly = ki + C Exponentiale. y = tele" If ly| = r, then y = 1r. y = Ak A is a shorter name for tel. By allowing A to take on the value 0 in addition to all possible values to , we can include the solution y = 0 in the formula. We find the value of A for the initial value problem by solving for A when y = > and 1 = 0:30 = Ac30 = A. * = 1.3 The solution of the initial value problem dy - = ky, W(0) = 3 * = 0.6 is (2) (a) Quantities changing in this way are said to undergo exponential growth if k > 0 and exponential decay if & 0) or decay dx (k (x) defined on an interval of x-values (perhaps infinite) such that 7 )(x) = fix, y(x)) on that interval. That is, when y(x) and its derivative y'(x) are substituted into the different tial equation, the resulting equation is true for all x in the solution interval. The general solution is a solution y(x) that contains all possible solutions and it always contains an arbitrary constant. Equation (3) is separable if f can be expressed as a product of a function of x and a function of y. The differential equation then has the form dy = g(x) H(Y). is a function of .. H is a function of y. Then collect all y terms with dy and all x terms with dr: H(y)" dy = g(x) dx. Now we simply integrate both sides of this equation: (4) After completing the integrations, we obtain the solution y defined implicitly as a function of x. The justification that we can integrate both sides in Equation (4) in this way is based on the Substitution Rule (Section 5.5):1 dy H(y(x)) dx = HOU)) -H((x) g(x)dx =/ g(x) dx. EXAMPLE 1 Solve the differential equation dx "=atye. y>-1. Solution Since 1 + y is never zero for y > -1, we can solve the equation by separat- ing the variables. dy = (1 + y)e' dy = (1 + y)e' dx Treat dudx as a quotient of differentials and multiply both sides by ax. dy = edx Divide by (1 + y)- dy Ity e' dx Integrate both sides. In (1 + y) = e+ C C represents the combined constants of integration. The last equation gives y as an implicit function of x.Chapter 7 Integrals and Transcendental Functions 438 Chapter 7 Integrals and Transcendental Functions dy EXAMPLE 2 Solve the equation y(x + 1) = = x(y + 1). Solution We change to differential form, separate the variables, and integrate: y(x + 1)dy = x(y + 1)dx y dy x dx y+l xtl Divide chy r + 1. "In (1 + y ) = x - In|x + 1| + C. The last equation gives the solution y as an implicit function of r. The initial value problem ady di = Ky.involves a separable differential equation, and the solution y = joe expresses exponen- tial change. We now present several examples of such change. 500 400 300 Unlimited Population Growth Yeast biomass (mp) 200 Strictly speaking, the number of individuals in a population (of people, plants, animals, or 100 bacteria, for example) is a discontinuous function of time because it takes on discrete val- ues, However, when the number of individuals becomes large enough, the population can 10 be approximated by a continuous function. Differentiability of the approximationg function Time (hr) is another reasonable hypothesis in many settings, allowing for the use of calculus to model and predict population sizes. FIGURE 7.6 Graph of the growth of a If we assume that the proportion of reproducing individuals remains constant and yeast population over a 10-hour period. assume a constant fertility, then at any instant / the birth rate is proportional to the based on the data in Table 7.3. number y(0) of individuals present. Let's assume, too, that the death rate of the popula- tion is stable and proportional to y(). If. further, we neglect departures and arrivals, the TABLE 7.3 Population of yeast growth rate dy/dr is the birth rate minus the death rate, which is the difference of the Yeast biomass two proportionalities under our assumptions. In other words, dy/dr = ky so that Time (hr) (mg) y = me", where y is the size of the population at time f = 0. As with all kinds of growth, there may be limitations imposed by the surrounding environment, but we will 9.6 not go into these here. (We treat one model imposing such limitations in Section 9.4.) 18.3 When & is positive, the proportionality dy/di - ky models unlimited population 29.0 growth. (See Figure 7.6.) 47.2 71.1 EXAMPLE 3 The biomass of a yeast culture in an experiment is initially 29 grams. 119.1 After 30 minutes the mass is 37 grams. Assuming that the equation for unlimited popula- 174.6 tion growth gives a good model for the growth of the yeast when the mass is below 257.3 100 grams, how long will its take for the mass to double from its initial value? 350.7 Solution Let y() be the yeast biomass after / minutes. We use the exponential growth 441.0 model dy/ di = ky for unlimited population growth, with solution y = joe". 10 513.3 We have y = y(0) = 29. We are also told that, 3(30) = 292 30) = 37