Therefore we use the notation \(P(t)\) for the population as a function of time. It can no way reach c in reality, because dpdt\frac{{dp}}{{dt}}dtdp will get lesser and lesser, but the population reaches the carrying capacity as time t reaches infinity. Steady States 2003-2022 Chegg Inc. All rights reserved. Biologists have found that in many biological systems, the population grows until a certain steady-state population is reached. StudySmarter is commited to creating, free, high quality explainations, opening education to all. Draw the direction field for the differential equation from step \(1\), along with several solutions for different initial populations. To check when the population size is increasing at an increasing rate, we need to look for when \(\frac{d^{2}P}{dt^{2}}>0\). Published:April222013. \nonumber \]. xn. Step 1: Setting the right-hand side equal to zero leads to P = 0 and P = K as constant solutions. Such a phenomenon of starting with slow growth, expanding into exponential growth, and slowing down into logarithmic growth, is modeled by the logistic distribution. Sign up to read all wikis and quizzes in math, science, and engineering topics. xn+1=rxn(1xn){\displaystyle x_{n+1}=rx_{n}\left(1-x_{n}\right)} (1) where xnis a number between zero and one, that represents the ratio of existing population to the maximum possible population. \[ P(t)=\dfrac{1,072,764C_2e^{0.2311t}}{1+C_2e^{0.2311t}} \nonumber \], To determine the value of the constant, return to the equation, \[ \dfrac{P}{1,072,764P}=C_2e^{0.2311t}. The logistic differential equation dN/dt=rN(1-N/K) describes the situation where a population grows proportionally to its size, but stops growing when it reaches the size of K. In this context, the two parameters have physical interpretations: Both of these values should be predetermined in order to approximate the curve, but they are not necessarily easy to calculate; for example, the carrying capacity for humans is unknown (but suspected to be slightly less than 10 billion). Then \(\frac{P}{K}\) is small, possibly close to zero. Later on, much research and development is done, and the industry grows roughly exponentially. What do these solutions correspond to in the original population model (i.e., in a biological context)? There is a point in the middle of the graph where the graph switches concavity. Earn points, unlock badges and level up while studying. Step 1: Setting the right-hand side equal to zero gives P = 0 and P = 1, 072, 764. This analysis can be represented visually by way of a phase line. If and , then , and the equilibrium solutions are or . Will you pass the quiz? At the starting point of a research, the observed growth will be very little since the researcher attempts to gain acknowledgement. The first solution indicates that when there are no organisms present, the population will never grow. As long as \(P_0K\), the entire quantity before and including \(e^{rt}\)is nonzero, so we can divide it out: \[ e^{rt}=\dfrac{KP_0}{P_0} \nonumber \], \[ \ln e^{rt}=\ln \dfrac{KP_0}{P_0} \nonumber \], \[ rt=\ln \dfrac{KP_0}{P_0} \nonumber \], \[ t=\dfrac{1}{r}\ln \dfrac{KP_0}{P_0}. When \(P\) is between \(0\) and \(K\), the population increases over time. Use the solution to find the population size at at \(t=20\) and \(t=100\). The continuous version of the logistic model is described by the differential equation (1) where is the Malthusian parameter (rate of maximum population growth) and is the so-called carrying capacity (i.e., the maximum sustainable population). Multiply both sides of the equation by \(K\) and integrate: \[ \dfrac{K}{P(KP)}dP=rdt. If p = c, then the right-hand side becomes zero, and the population remains unchanged. If you're seeing this message, it means we're having trouble loading external resources on our website. \nonumber \], Substituting the values \(t=0\) and \(P=1,200,000,\) you get, \[ \begin{align*} C_2e^{0.2311(0)} =\dfrac{1,200,000}{1,072,7641,200,000} \\[4pt] C_2 =\dfrac{100,000}{10,603}9.431.\end{align*}\], \[ \begin{align*} P(t) =\dfrac{1,072,764C_2e^{0.2311t}}{1+C_2e^{0.2311t}} \\[4pt] =\dfrac{1,072,764 \left(\dfrac{100,000}{10,603}\right)e^{0.2311t}}{1+\left(\dfrac{100,000}{10,603}\right)e^{0.2311t}} \\[4pt] =\dfrac{107,276,400,000e^{0.2311t}}{100,000e^{0.2311t}10,603} \\[4pt] \dfrac{10,117,551e^{0.2311t}}{9.43129e^{0.2311t}1} \end{align*}\]. After a month, the rabbit population is observed to have increased by \(4%\). Solving it using either approach gives the solution f(x)=exex+Cf(x) = \frac{e^x}{e^x + C}f(x)=ex+Cex for a constant CCC. \end{align*} \nonumber \]. Finally, the research will come to an end undoubtedly. \nonumber \]. (Hint: use the slope field to see what happens for various initial populations, i.e., look for the horizontal asymptotes of your solutions.). Logistic Differential Equation Formula First we will discover how to recognize the formula for all logistic equations, sometimes referred to as the Verhulst model or logistic growth curve, according to Wolfram MathWorld. To solve this equation for \(P(t)\), first multiply both sides by \(KP\) and collect the terms containing \(P\) on the left-hand side of the equation: \[\begin{align*} P =C_1e^{rt}(KP) \\[4pt] =C_1Ke^{rt}C_1Pe^{rt} \\[4pt] P+C_1Pe^{rt} =C_1Ke^{rt}.\end{align*}\]. Step 1: Setting the right-hand side equal to zero leads to P = 0 P = 0 and P = K P = K as constant solutions. Sketch curves of the population \ ( N (t) \) for initial values \ ( N_ {0}=100 \) and \ ( N_ {0}=200 \). Referencing the general logistic differential equation, we can see that the proportionality constant \(k\) equals 0.08 and the carrying capacity \(M\) equals 1000. A differential equation that incorporates both the threshold population \(T\) and carrying capacity \(K\) is, \[ \dfrac{dP}{dt}=rP\left(1\dfrac{P}{K}\right)\left(1\dfrac{P}{T}\right) \nonumber \]. Therefore, any quantity experiencing this kind of growth can be modeled logistically; these fields range from medicine to machine learning to chemistry to linguistics. Give feedback. Powered by WOLFRAM TECHNOLOGIES Step 2: Rewrite the differential equation in the form, \[ \dfrac{dP}{dt}=\dfrac{rP(KP)}{K}. food, living space, etc. Suppose this is the deer density for the whole state (39,732 square miles). A logistic differential equation is an ordinary differential equation whose solution is a logistic function. dxdf = f (1f) dxdf f = f 2. An example of an exponential growth function is \(P(t)=P_0e^{rt}.\) In this function, \(P(t)\) represents the population at time \(t,P_0\) represents the initial population (population at time \(t=0\)), and the constant \(r>0\) is called the growth rate. \(\dfrac{dP}{dt}=rP\left(1\dfrac{P}{K}\right),\quad P(0)=P_0\), \(P(t)=\dfrac{P_0Ke^{rt}}{(KP_0)+P_0e^{rt}}\), \(\dfrac{dP}{dt}=rP\left(1\dfrac{P}{K}\right)\left(1\dfrac{P}{T}\right)\). For the derivation of the logistic differential equation solution, see the Deep Dive below. What is Logistic Differential Equation(LDE) Differential equations can be used to represent the size of population as it varies over time t. A logistic differential equation is an ordinary differential equation whose solution is a logistic function.. An exponential growth and decay model is the simplest model which fails to take into account such constraints that prevent indefinite growth but . Or, if provided with an initial condition, it can be graphed as a logistic sigmoid curve. With a basic knowledge of limits, we can see that no matter what the constant \(A\) is, \(\lim_{t \to \infty} P(t)=M\) as long as \(M\) and \(k\) are positive. So, \(P\) is increasing between 0 and 5,000. accessed April 9, 2015, www.americanscientist.org/issa-magic-number). Write the logistic differential equation and initial condition for this model. Logistic or population growth is not the same as natural growth because it has exponential growth with confined exponential function. What is another name for the carrying capacity in the logistic differential equation? All solutions to the logistic differential equation are of the form. Steady States We can verify that the function \(P(t)=P_0e^{rt}\) satisfies the initial-value problem. At the time the population was measured \((2004)\), it was close to carrying capacity, and the population was starting to level off. The carrying capacity of an organism in a given environment is defined to be the maximum population of that organism that the environment can sustain indefinitely. So, the value (1pc)>0\left( {1 - \frac{p}{c}} \right)>0(1cp)>0 becomes close to 1, and the right-hand side of this equation is almost equal to rp. The logistic differential equation is an autonomous differential equation, so we can use separation of variables to find the general solution, as we just did in Example \(\PageIndex{1}\). First determine the values of \(r,K,\) and \(P_0\). Then the right-hand side of Equation \ref{LogisticDiffEq} is negative, and the population decreases. Here \(P_0=100\) and \(r=0.03\). The carrying capacity in a given environment is the maximum population (of a creature) that the environment can sustain indefinitely without disturbing the environment. Source: http://monkeysuncle.stanford.edu/?p=933. Every intersection of the green line and the red parabola represents a value of . d x d t = r x ( 1 x K), i.e., a continuous-time dynamical system which gives you a function x ( t), t R, given an initial value x ( 0). \end{align*}\], \[ r^2P_0K(KP_0)e^{rt}((KP_0)P_0e^{rt})=0. At the beginning of the school day, one student starts a rumor. At the end, it reaches a limiting value, when a whole lot of other researchers started a similar research, and there are no more simplest enhancements that can be possible to make in this equipment. The solution is kind of hairy, but it's worth bearing with us! Logistic differential equations are useful in various other fields as well, as they often provide significantly more practical models than exponential ones. However, as more students hear the rumor, the rate at which the rumor spreads increases. Hence the right-hand side is still positive, but the value (1pc)\left( {1 - \frac{p}{c}} \right)(1cp) gets smaller, which results in reduced growth rate. The equation is used in the following manner. \end{align*}\]. The logistic differential equation is derived by taking the derivative of the logistic equation. This observation corresponds to a rate of increase \(r=\dfrac{\ln (2)}{3}=0.2311,\) so the approximate growth rate is 23.11% per year. \end{align*}\], Step 5: To determine the value of \(C_2\), it is actually easier to go back a couple of steps to where \(C_2\) was defined. The logistic difference equation was developed mainly to analyze population growth. The right-hand side is equal to a positive constant multiplied by the current population. c. Using this model we can predict the population in 3 years. Various factors limit the rate of growth of a particular population, including birth rate, death rate, food supply, predators, and so on. Finally, in the late stages when there are multiple competitors in the market and the "easiest" improvements have been performed, the innovation slows down significantly and approaches some limiting value. Log in. They are also useful in a variety of other contexts, including machine learning, chess ratings, cancer treatment (i.e. What is the logistic difference equation model? Here the function p ( t) represents the population of any creature as a function of time t. Let us consider the initial population is small with respect to carrying capacity. The LibreTexts libraries arePowered by NICE CXone Expertand are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. The logistic difference equation is given by where r is the so-called driving parameter. Logistic functions model bounded growth - standard exponential functions fail to take into account constraints that prevent indefinite growth, and logistic functions correct this error. The logistic differential equation incorporates the concept of a carrying capacity. A group of Australian researchers say they have determined the threshold population for any species to survive: \(5000\) adults. File Type: pdf. The variable \(P\) will represent population. . Let's go ahead and plug these values into the logistic differential general solution. The second solution indicates that when the population starts at the carrying capacity, it will never change. Imagine you go to school with 1,000 other students. Compute limxy(x)\displaystyle \lim_{x \rightarrow \infty}y(x)xlimy(x). How do these values compare? For example, we can use logistic differential equation to observe growth in a research on, say, some medical equipment. [more] A logistic differential equation is an ODE of the form f' (x) = r\left (1-\frac {f (x)} {K}\right)f (x) f (x) = r(1 K f (x))f (x) where r,K r,K are constants. \end{align*}\], \[ \begin{align*} P(t) =\dfrac{1,072,764 \left(\dfrac{25000}{4799}\right)e^{0.2311t}}{1+(250004799)e^{0.2311t}}\\[4pt] =\dfrac{1,072,764(25000)e^{0.2311t}}{4799+25000e^{0.2311t}.} We use the method of separation of variables to solve the logistic differential equation. The threshold population is defined to be the minimum population that is necessary for the species to survive. K is the carrying capacity. \nonumber \], We define \(C_1=e^c\) so that the equation becomes, \[ \dfrac{P}{KP}=C_1e^{rt}. The paramenters of the system determine what it does. Now, we'll find the second derivative and set it equal to 0. e = the natural logarithm base (or Euler's number) x 0 = the x-value of the sigmoid's midpoint. The carrying capacity \(K\) is 39,732 square miles times 27 deer per square mile, or 1,072,764 deer. Where, L = the maximum value of the curve. New user? Solve the initial-value problem for \(P(t)\). A green line intersects back and forth between the graphs of and , beginning at the point . We use the variable \(K\) to denote the carrying capacity. This phase line shows that when \(P\) is less than zero or greater than \(K\), the population decreases over time. Test your knowledge with gamified quizzes. Stop procrastinating with our study reminders. The graph of the logistic equation is pictured below. As the number of students who know the rumor begins to reach the total student population of 1,000, the rate at which the rumor spreads decreases as there are fewer students left to tell. The graph of the logistic differential equation is identical to the _________ graph at first. Create beautiful notes faster than ever before. Population growth is an environmental phenomenon which has restrictions, such as space and resources. \[P(100)=\frac{1000}{1+4e^{-0.08(100)}}\]. Then we will learn how to find the limiting capacity and maximum growth grate for logistic functions. Every solution to the logistic differential equation follows the form of this solution. Luckily, we can use the initial value from the problem to solve for \(A\)! The logistic equation takes the form The logistic equation takes the form (4.14) x [ n + 1 ] = r x [ n ] ( 1 x [ n ] ) The logistic differential equation has a general solutionP(t). Using an initial population of \(200\) and a growth rate of \(0.04\), with a carrying capacity of \(750\) rabbits. \end{align*}\]. As we saw in class, one possible model for the growth of a population is the logistic equation: Here the number is the initial density of the population, is the intrinsic growth rate of the population (for given, finite initial resources available) and is the carrying capacity, or maximum potential population density. d. After \(12\) months, the population will be \(P(12)278\) rabbits. In particular, use the equation, \[\dfrac{P}{1,072,764P}=C_2e^{0.2311t}. Let y(x)y(x)y(x) be a function satisfying dydx=8y2y2\dfrac{dy}{dx}=8y-2y^2dxdy=8y2y2 and y(2016)=1y(2016)=1y(2016)=1. Figure \(\PageIndex{1}\) shows a graph of \(P(t)=100e^{0.03t}\). Start with a fixed value of the driving parameter, r, and an initial value of x0. The logistic differential equation is given by \frac { {dp}} { {dt}} = rp\left ( {1 - \frac {p} {c}} \right) dtdp = rp(1 cp). The limit is like the carrying capacity of land: A certain region won't support extra growth because as a particular population grows, its resources get reduced. Support us and buy the Calculus workbook with all the packets in one nice spiral bound book. Sign up, Existing user? The raw residual is the difference between the actual response and the estimated probability from the model. For , solutions converge instead to . Like its continuous counterpart, it can be used to model the growth or decay of a process, population, or financial instrument. The formula for the raw . \end{align*}\]. "The Logistic Difference Equation" In other words, the population grows slower and slower as time passes until it reaches its carrying capacity \(M\). We use the variable \(T\) to represent the threshold population. Logistic curve. Working under the assumption that the population grows according to the logistic differential equation, this graph predicts that approximately \(20\) years earlier \((1984)\), the growth of the population was very close to exponential. What is the idea behind the carrying capacity in the logistic differential equation? Now exponentiate both sides of the equation to eliminate the natural logarithm: \[ e^{\ln \dfrac{P}{KP}}=e^{rt+C} \nonumber \], \[ \dfrac{P}{KP}=e^Ce^{rt}. This gives us small pc\frac{p}{c}cp, which we can assume to be close to zero. The logistic difference equation is given by x t + 1 = x t e r ( 1 - x t ) (3) It can be derived as a discrete time analogy to the logistic differential equation, which is given by dx dt = r x ( 1 - x ) (4) where x is the population density (scaled by its carrying capacity) and r is the maximal growth rate of the population at low values of x. of the users don't pass the Logistic Differential Equation quiz! If this r>0, then the population grows expeditiously, simulating exponential growth. \label{eq20a} \], The left-hand side of this equation can be integrated using partial fraction decomposition. A phase line describes the general behavior of a solution to an autonomous differential equation, depending on the initial condition. Interact on desktop, mobile and cloud with the free WolframPlayer or other Wolfram Language products. Whenever p>c, there will be decrease in the population. Practice math and science questions on the Brilliant iOS app. The logistic difference equation is given by <math> (1)\qquad x_{n+1} = r x_n (1-x_n) </math> where: <math>x_n</math> represents the population at generation n, and hence x0represents the initial population (at generation 0) ris a positive number, and represents the growth rate of the population. The resulting differential equation f(x)=r(1f(x)K)f(x)f'(x) = r\left(1-\frac{f(x)}{K}\right)f(x)f(x)=r(1Kf(x))f(x) can be viewed as the result of adding a correcting factor rf(x)2K-\frac{rf(x)^2}{K}Krf(x)2 to the model; without this factor, the differential equation would be f(x)=rf(x)f'(x)=rf(x)f(x)=rf(x), which has an exponential solution. The units of time can be hours, days, weeks, months, or even years. Sign up to highlight and take notes. Set individual study goals and earn points reaching them. \label{eq30a} \]. A logistic differential equation is an ODE of the form f(x)=r(1f(x)K)f(x)f'(x) = r\left(1-\frac{f(x)}{K}\right)f(x)f(x)=r(1Kf(x))f(x) where r,Kr,Kr,K are constants. The above shows the results of fitting the U.S. population from 1790-1930 to a logistic function. Before the hunting season of 2004, it estimated a population of 900,000 deer. Essentially, when the population reaches 2,500 people, the rate at which the population grows slows down. \[P(3)=\dfrac{1,072,764e^{0.2311(3)}}{0.19196+e^{0.2311(3)}}978,830\,deer \nonumber \]. Draw a slope field for this logistic differential equation, and sketch the solution corresponding to an initial population of \(200\) rabbits. The rumor spreads like wildfire throughout your school. | Find, read and cite all the research you need on . The left-hand side represents the rate at which the population increases (or decreases). Once all 1,000 students have heard the rumor, the rumor cannot spread any more. The logistic differential equation is used to model population growth that is proportional to the size of the population and takes into account that there are a limited number of resources necessary for survival. Step 1: Setting the right-hand side equal to zero gives \(P=0\) and \(P=1,072,764.\) This means that if the population starts at zero it will never change, and if it starts at the carrying capacity, it will never change. The graph of this solution is shown again in blue in Figure \(\PageIndex{6}\), superimposed over the graph of the exponential growth model with initial population \(900,000\) and growth rate \(0.2311\) (appearing in green). What is the formula for logistics differential equation? . This value is a limiting value on the population for any given environment. Put into words, as long as the carrying capacity and the constant of proportionality are positive, regardless of the size of the population at \(t=0\), the population will reach the carrying capacity as time passes. \end{align*}\]. For a constant of proportionality \(k\), a population size \(P\), and some carrying capacity \(M\), the logistic differential equation is, \[\frac{dP}{dt}=kP\left(1-\frac{P}{M}\right)\]. Finally, substitute the expression for \(C_1\) into Equation \ref{eq30a}: \[ P(t)=\dfrac{C_1Ke^{rt}}{1+C_1e^{rt}}=\dfrac{\dfrac{P_0}{KP_0}Ke^{rt}}{1+\dfrac{P_0}{KP_0}e^{rt}} \nonumber \]. Differential equations can be used to represent the size of a population as it varies over time. Snapshot 5: a solution that is chaotic and ultimately unpredictable; it can, however, be modeled as a simpler, three-cycle approximation, Snapshots 2, 4, and 6: the stairstep diagrams of snapshots 1, 3, and 5, respectively, Victor Hakim The general solution to the differential equation would remain the same. This happens because the population increases, and the logistic differential equation states that the growth rate decreases as the population increases. When little time has passed, and fff is small, this correcting term is effectively null and the population rises nearly exponentially, which is roughly consistent with empirical evidence. Later on, when the researcher has extended his research and derived some conclusions, the results of the research grow exponentially. where \(A\) is some constant that depends on the initial condition. As long as \(P>K\), the population decreases. Forgot password? Thus, the quantity in parentheses on the right-hand side of Equation \ref{LogisticDiffEq} is close to \(1\), and the right-hand side of this equation is close to \(rP\). The logistic equation is a discrete-time version of the logistic differential equation discussed in the previous section. Since the population varies over time, it is understood to be a function of time. This nonlinear difference equation is intended to capture two effects: reproductionwhere the population will increase at a rate proportionalto the current population when the population size is small. Next, factor \(P\) from the left-hand side and divide both sides by the other factor: \[\begin{align*} P(1+C_1e^{rt}) =C_1Ke^{rt} \\[4pt] P(t) =\dfrac{C_1Ke^{rt}}{1+C_1e^{rt}}. It is easy to see if the solution converges to a single point, oscillates in "square-like" fashion, or is completely unpredictable. Logistic comes from the Greek logistikos (computational). This differential equation can be coupled with the initial condition \(P(0)=P_0\) to form an initial-value problem for \(P(t).\). In this section, we study the logistic differential equation and see how it applies to the study of population dynamics in the context of biology. The harmonic oscillator is quite well behaved. http://demonstrations.wolfram.com/TheLogisticDifferenceEquation/. So, let's find out when both the first and second derivatives are greater than 0. Use the solution to predict the population after \(1\) year. will represent time. Then, the right-hand side of the equation becomes negative, resulting in reduction in the population. Since we already know the general solution for a logistic differential equation, all we need to do is extract the pertinent information from the differential equation and plug it into the solution. \[\int \frac{dP}{P(1- \frac{P}{M})}=\int k dt\]. When \(P>2500\), \(\frac{d^{2}P}{dt^{2}}<0\). Create flashcards in notes completely automatically. where \(r\) represents the growth rate, as before. calc_7.9_packet.pdf. The logistic difference equation (or logistic map) , a nonlinear first-order recurrence relation, is a time-discrete analogue of the logistic differential equation, . This is a Bernoulli differential equation (and also a separable differential equation). Legal. To find this point, set the second derivative equal to zero: P ( t ) = P 0 K e r t ( K P 0 ) + P 0 e r t P ( t ) = r P 0 K ( K P 0 ) e r t ( ( K P 0 ) + P 0 e r t ) 2 P ( t ) = r 2 P 0 K ( K P 0 ) 2 e r t r 2 P 0 2 K ( K P 0 ) e 2 r t ( ( K P 0 ) + P 0 e r t ) 3 = r 2 P 0 K ( K P 0 ) e r t ( ( K P 0 ) P 0 e r t ) ( ( K P 0 ) + P 0 e r t ) 3 . Best study tips and tricks for your exams. The equation of logistic function or logistic curve is a common "S" shaped curve defined by the below equation. Furthermore, it states that the constant of proportionality never changes. This makes sense from a practical perspective: large populations will necessarily compete for resources (e.g. As time goes on, however, the correcting term becomes more and more significant. The population size is growing when \(\frac{dP}{dt}>0\). To solve the logistic differential equation, we will integrate it with separation of variables. At first, the rumor spreads slowly as only a few people have heard it. This means that if the population starts at zero it will never change, and if it starts at the carrying capacity, it will never change. This equation is graphed in Figure \(\PageIndex{5}\). The logistic curve is also known as the sigmoid curve. Simply put, there is a restriction, which is carrying capacity for logistic growth, in the sense that in the beginning stages, population usually grows exponentially but becomes flat as the resources becomes limited. All we're missing is the constant \(A\). This possibility is not taken into account with exponential growth. More about Logistic Differential Equation, Derivatives of Inverse Trigonometric Functions, Initial Value Problem Differential Equations, Integration using Inverse Trigonometric Functions, Particular Solutions to Differential Equations, Frequency, Frequency Tables and Levels of Measurement, Absolute Value Equations and Inequalities, Addition and Subtraction of Rational Expressions, Addition, Subtraction, Multiplication and Division, Finding Maxima and Minima Using Derivatives, Multiplying and Dividing Rational Expressions, Solving Simultaneous Equations Using Matrices, Solving and Graphing Quadratic Inequalities, The Quadratic Formula and the Discriminant, Trigonometric Functions of General Angles, Confidence Interval for Slope of Regression Line, Hypothesis Test of Two Population Proportions, The logistic differential equation is used to model population growth that is proportional to the size of the population and takes into account that there are a limited number of resources necessary for survival - it describes a situation where the population will stop growing once it reaches a carrying capacity, For a constant of proportionality \(k\), a population size \(P\), and some carrying capacity \(M\), the logistic differential equation is \(\frac{dP}{dt}=kP(1-\frac{P}{M})\) and measures the growth of a population over time, All solutions to the logistic differential equation are of the form \(P(t)=\frac{M}{1+Ae^{-kt}}\) where \(A\) is some constant that depends on the initial condition, No matter what the constant \(A\) is, \(\lim_{t \to \infty} P(t)=M\) as long as \(M\) and \(k\) are positive; thus \(M\) can be thought of as an.
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