Solving The Population Growth Model P' = 0.4P(1-P/10000)-kP

Solve the differential equation $P^{\prime}=0.4 P(1-\frac{P}{10,000})-k P$ with $k=0.06$ and initial condition $P(0)=5000$. Find $P(t)$.

In the realm of mathematical biology, population dynamics often take center stage, providing crucial insights into the growth, decline, and stability of various species. One fundamental model used to describe population growth is the logistic equation, which incorporates the concept of carrying capacity – the maximum population size an environment can sustainably support. However, real-world scenarios often involve additional factors that influence population dynamics, such as harvesting or predation. This article delves into the intricacies of solving a modified logistic equation that accounts for harvesting, offering a comprehensive analysis of the solution process and its implications. The equation under consideration is $P^{\prime}=0.4 P(1-\frac{P}{10,000})-k P$, which represents the rate of change of a population ($P^{\prime}$) over time, influenced by both logistic growth and a harvesting term. The parameters in the equation hold significant biological meaning: 0.4 represents the intrinsic growth rate of the population, 10,000 signifies the carrying capacity of the environment, and $k$ denotes the harvesting rate. Our specific task is to solve this equation, assuming a harvesting rate of $k=0.06$ and an initial population size of 5,000 individuals. This scenario provides a realistic framework for understanding how harvesting can impact population dynamics and sustainability. To solve this equation, we will employ a combination of analytical techniques and qualitative analysis. First, we will simplify the equation by substituting the given value of $k$ and explore the equilibrium solutions, which represent the population sizes at which the growth rate is zero. These equilibrium points are critical for understanding the long-term behavior of the population. Next, we will employ methods of separation of variables to obtain a general solution to the differential equation. This general solution will involve an integration process and an arbitrary constant of integration, which we will determine using the given initial condition of 5,000 fish. Finally, we will analyze the solution to understand the population dynamics under the given harvesting rate and initial population size. This analysis will include examining the stability of equilibrium points, predicting the long-term population trend, and discussing the implications for sustainable harvesting practices. This detailed exploration will provide a solid understanding of the population dynamics governed by the given equation. The interplay between logistic growth and harvesting is a critical consideration in fisheries management, wildlife conservation, and other fields. By solving this equation and analyzing its solutions, we gain valuable insights into the factors that influence population sustainability and the importance of responsible resource management. This article aims to provide a clear and comprehensive understanding of the solution process and its implications for population dynamics.

Setting up the Equation

To effectively analyze the population dynamics model, the initial step involves setting up the equation with the given parameters. We are given the differential equation $P^{\prime}=0.4 P(1-\frac{P}{10,000})-k P$, which models the population growth rate ($P^{\prime}$) as a function of the population size ($P$). This equation incorporates two key factors: logistic growth, represented by the term $0.4 P(1-\frac{P}{10,000})$, and harvesting, represented by the term $-kP$. The logistic growth term reflects the natural tendency of a population to increase, but also acknowledges the limitations imposed by the environment's carrying capacity (10,000 in this case). As the population approaches the carrying capacity, the growth rate slows down due to resource limitations and other environmental factors. The harvesting term, $-kP$, represents the removal of individuals from the population at a rate proportional to the population size. The harvesting rate, $k$, is a crucial parameter that determines the intensity of harvesting. In our specific scenario, we are given a harvesting rate of $k=0.06$. This means that 6% of the population is harvested per unit of time. This parameter value is critical for understanding the impact of harvesting on the population dynamics. We are also given an initial condition of 5,000 fish, which means that at time $t=0$, the population size is $P(0)=5,000$. This initial condition is essential for finding a particular solution to the differential equation, as it allows us to determine the value of the constant of integration that arises during the solution process. With the given parameters, we can substitute $k=0.06$ into the differential equation to obtain a simplified equation that describes the population dynamics under the specified harvesting rate. This simplified equation will be the foundation for our subsequent analysis. By carefully setting up the equation with the given parameters, we establish a clear mathematical framework for understanding the population dynamics. This framework will guide our subsequent steps in solving the equation and interpreting the results. The initial condition provides a starting point for tracking the population's evolution over time, while the harvesting rate quantifies the impact of human intervention on the population's growth. This comprehensive setup is essential for a thorough analysis of the population dynamics model and its implications for sustainable harvesting practices. The equation $P^{\prime}=0.4 P(1-\frac{P}{10,000})-0.06 P$ becomes the central focus of our analysis, and we will now proceed to solve this equation using analytical techniques.

Finding Equilibrium Solutions

To understand the long-term behavior of the population, a crucial step is finding the equilibrium solutions of the differential equation. Equilibrium solutions, also known as steady states, represent the population sizes at which the growth rate is zero ($P^{\prime}=0$). These points are significant because they indicate the population levels where the forces of growth and harvesting are balanced, resulting in a stable population size over time. Mathematically, we find the equilibrium solutions by setting the right-hand side of the differential equation equal to zero and solving for $P$. Our equation, with $k=0.06$, is $P^{\prime}=0.4 P(1-\frac{P}{10,000})-0.06 P$. Setting $P^{\prime}=0$, we get $0 = 0.4 P(1-\frac{P}{10,000})-0.06 P$. This equation can be simplified by factoring out $P$, resulting in $0 = P[0.4(1-\frac{P}{10,000})-0.06]$. This factorization reveals that one equilibrium solution is $P=0$, which represents the extinction of the population. This is a trivial solution, but it is important to recognize it as a possible outcome. To find the non-trivial equilibrium solutions, we need to solve the equation $0.4(1-\frac{P}{10,000})-0.06 = 0$. This equation represents the condition where the net growth rate, considering both logistic growth and harvesting, is zero. To solve this equation, we first isolate the term containing $P$: $0.4(1-\frac{P}{10,000}) = 0.06$. Next, we divide both sides by 0.4: $1-\frac{P}{10,000} = \frac{0.06}{0.4} = 0.15$. Then, we subtract 1 from both sides: $-\frac{P}{10,000} = 0.15 - 1 = -0.85$. Finally, we multiply both sides by -10,000 to solve for $P$: $P = 0.85 * 10,000 = 8,500$. Therefore, the non-trivial equilibrium solution is $P=8,500$. This equilibrium point represents a stable population size that can be sustained under the given harvesting rate. The existence of this equilibrium solution suggests that the population can persist even with harvesting, provided that the harvesting rate is not too high. The equilibrium solutions, $P=0$ and $P=8,500$, provide valuable information about the long-term behavior of the population. We can analyze the stability of these equilibrium points to determine whether the population will tend towards these values over time. The stability analysis will involve examining the sign of $P^{\prime}$ for population sizes near the equilibrium points. This analysis will help us understand how the population will respond to small perturbations from the equilibrium and whether the equilibrium is a stable or unstable state. By finding the equilibrium solutions, we gain a fundamental understanding of the population dynamics under the given harvesting rate. These equilibrium points serve as crucial reference points for predicting the long-term population trend and assessing the sustainability of the harvesting practice.

Solving the Differential Equation

With the equilibrium solutions identified, the next crucial step is solving the differential equation to obtain a general solution for the population size $P(t)$ as a function of time $t$. This general solution will provide a complete description of the population dynamics, allowing us to predict the population size at any given time. To solve the differential equation $P^{\prime}=0.4 P(1-\frac{P}{10,000})-0.06 P$, we will employ the method of separation of variables. This technique is applicable to differential equations that can be written in the form $\frac{dP}{dt} = f(P)g(t)$, where the variables $P$ and $t$ can be separated. Our equation can be rearranged as $\frac{dP}{dt} = 0.4 P - \frac{0.4 P^2}{10,000} - 0.06 P$, which simplifies to $\frac{dP}{dt} = 0.34 P - \frac{P^2}{25,000}$. To separate the variables, we divide both sides by $P(0.34 - \frac{P}{25,000})$ and multiply both sides by $dt$, resulting in $\frac{dP}{P(0.34 - \frac{P}{25,000})} = dt$. Now, the variables are separated, with all terms involving $P$ on the left side and all terms involving $t$ on the right side. The next step is to integrate both sides of the equation. The integral on the right side is straightforward: $\int dt = t + C_1$, where $C_1$ is the constant of integration. The integral on the left side requires a partial fraction decomposition. We rewrite the integrand as $\frac{1}{P(0.34 - \frac{P}{25,000})} = \frac{A}{P} + \frac{B}{0.34 - \frac{P}{25,000}}$. Solving for $A$ and $B$, we find that $A = \frac{1}{0.34}$ and $B = \frac{1}{25,000 * 0.34}$. Therefore, the integral on the left side becomes $\int (\frac{1}{0.34P} + \frac{1}{25,000(0.34 - \frac{P}{25,000})}) dP$. This integral can be evaluated as $\frac{1}{0.34} \ln|P| - \frac{1}{0.34} \ln|0.34 - \frac{P}{25,000}| = \frac{1}{0.34} \ln|\frac{P}{0.34 - \frac{P}{25,000}}| + C_2$, where $C_2$ is another constant of integration. Combining the results of the integration, we have $\frac{1}{0.34} \ln|\frac{P}{0.34 - \frac{P}{25,000}}| = t + C$, where $C = C_1 - C_2$ is a combined constant of integration. To solve for $P$, we first multiply both sides by 0.34: $\ln|\frac{P}{0.34 - \frac{P}{25,000}}| = 0.34t + 0.34C$. Then, we exponentiate both sides: $|\frac{P}{0.34 - \frac{P}{25,000}}| = e^{0.34t + 0.34C} = e^{0.34t}e^{0.34C}$. Let $K = e^{0.34C}$, where $K$ is a constant. We can drop the absolute value signs and write $\frac{P}{0.34 - \frac{P}{25,000}} = Ke^{0.34t}$. Solving this equation for $P$ involves algebraic manipulation. We multiply both sides by $0.34 - \frac{P}{25,000}$: $P = Ke^{0.34t}(0.34 - \frac{P}{25,000})$. Expanding the right side, we get $P = 0.34Ke^{0.34t} - \frac{Ke^{0.34t}P}{25,000}$. Now, we isolate $P$ terms on one side: $P + \frac{Ke^{0.34t}P}{25,000} = 0.34Ke^{0.34t}$. Factoring out $P$, we have $P(1 + \frac{Ke^{0.34t}}{25,000}) = 0.34Ke^{0.34t}$. Finally, we solve for $P$: $P(t) = \frac{0.34Ke^{0.34t}}{1 + \frac{Ke^{0.34t}}{25,000}}$. This is the general solution to the differential equation. The constant $K$ needs to be determined using the initial condition. By solving the differential equation, we have obtained a general solution that describes the population dynamics as a function of time. This solution is crucial for predicting the population size at any given time and for understanding the long-term behavior of the population. The next step is to apply the initial condition to determine the value of the constant $K$ and obtain a particular solution that is specific to our initial population size.

Applying the Initial Condition

Having obtained the general solution to the differential equation, the next critical step is applying the initial condition to determine the value of the constant of integration, $K$, and thus obtain a particular solution that describes the specific population dynamics in our scenario. The initial condition we are given is that at time $t=0$, the population size is $P(0) = 5,000$. This information provides a specific point on the solution curve, allowing us to pinpoint the unique solution that satisfies both the differential equation and this initial state. Our general solution is given by $P(t) = \frac{0.34Ke^{0.34t}}{1 + \frac{Ke^{0.34t}}{25,000}}$. To apply the initial condition, we substitute $t=0$ and $P(0)=5,000$ into this equation: $5,000 = \frac{0.34Ke^{0.34(0)}}{1 + \frac{Ke^{0.34(0)}}{25,000}}$. Since $e^{0} = 1$, the equation simplifies to $5,000 = \frac{0.34K}{1 + \frac{K}{25,000}}$. To solve for $K$, we first multiply both sides by $1 + \frac{K}{25,000}$: $5,000(1 + \frac{K}{25,000}) = 0.34K$. Expanding the left side, we get $5,000 + \frac{5,000K}{25,000} = 0.34K$. Simplifying the fraction, we have $5,000 + 0.2K = 0.34K$. Now, we isolate the terms involving $K$: $5,000 = 0.34K - 0.2K = 0.14K$. Finally, we solve for $K$: $K = \frac{5,000}{0.14} = \frac{5,000}{\frac{14}{100}} = \frac{5,000 * 100}{14} = \frac{500,000}{14} = \frac{250,000}{7}$. Now that we have determined the value of $K$, we can substitute it back into the general solution to obtain the particular solution that satisfies the initial condition. The particular solution is given by $P(t) = \frac{0.34(\frac{250,000}{7})e^{0.34t}}{1 + \frac{(\frac{250,000}{7})e^{0.34t}}{25,000}}$. This solution can be further simplified by multiplying the numerator and denominator by 7: $P(t) = \frac{0.34(250,000)e^{0.34t}}{7 + \frac{250,000e^{0.34t}}{25,000}} = \frac{85,000e^{0.34t}}{7 + 10e^{0.34t}}$. This particular solution describes the population size $P(t)$ as a function of time $t$, given the initial population size of 5,000 and the harvesting rate of $k=0.06$. This solution is crucial for understanding the population dynamics under these specific conditions. By applying the initial condition, we have narrowed down the infinite family of solutions represented by the general solution to a single, unique solution that accurately reflects the population's evolution over time, starting from the given initial population size. This particular solution allows us to make specific predictions about the population size at any future time and to assess the long-term impact of harvesting on the population.

Analyzing the Solution and Implications

With the particular solution $P(t) = \frac{85,000e^{0.34t}}{7 + 10e^{0.34t}}$ in hand, we can now delve into analyzing the solution and its implications for the population dynamics. This analysis will provide valuable insights into the long-term behavior of the population, the impact of harvesting, and the sustainability of the harvesting practice. First, let's examine the long-term behavior of the population. As $t$ approaches infinity, $e^{0.34t}$ also approaches infinity. Therefore, to determine the limit of $P(t)$ as $t$ goes to infinity, we can divide both the numerator and the denominator by $e^{0.34t}$: $\lim_{t \to \infty} P(t) = \lim_{t \to \infty} \frac{85,000}{7e^{-0.34t} + 10}$. As $t$ approaches infinity, $e^{-0.34t}$ approaches 0. Thus, $\lim_{t \to \infty} P(t) = \frac{85,000}{0 + 10} = 8,500$. This result indicates that the population will approach the equilibrium solution of 8,500 as time goes on. This is consistent with our earlier analysis of the equilibrium solutions, where we found that 8,500 is a stable equilibrium point. The population will tend towards this value regardless of the initial population size (as long as it is not zero). The solution also shows how the population approaches this equilibrium. Since the initial population is 5,000, which is less than 8,500, the population will grow over time until it reaches the equilibrium. The rate of growth will depend on the specific form of the solution, which is determined by the parameters of the equation and the initial condition. The harvesting rate, $k=0.06$, plays a crucial role in determining the equilibrium population size. If the harvesting rate were higher, the equilibrium population size would be lower, and if the harvesting rate were too high, the population could be driven to extinction. This highlights the importance of carefully managing harvesting rates to ensure the sustainability of the population. The carrying capacity of the environment, 10,000, also influences the population dynamics. The logistic growth term in the equation ensures that the population growth rate slows down as the population approaches the carrying capacity. This prevents the population from growing indefinitely and helps to maintain a stable equilibrium. The initial condition of 5,000 fish affects the trajectory of the population towards the equilibrium. A higher initial population would result in a faster approach to the equilibrium, while a lower initial population would result in a slower approach. However, the long-term equilibrium population size remains the same, regardless of the initial population size (as long as it is not zero). The solution $P(t) = \frac{85,000e^{0.34t}}{7 + 10e^{0.34t}}$ provides a complete description of the population dynamics under the given harvesting rate and initial condition. It allows us to predict the population size at any future time and to understand the long-term impact of harvesting on the population. This information is crucial for making informed decisions about resource management and ensuring the sustainability of the population. By analyzing the solution, we gain valuable insights into the interplay between logistic growth, harvesting, and the carrying capacity of the environment. This understanding is essential for developing effective strategies for managing populations and ecosystems.

Conclusion

In conclusion, this article has provided a comprehensive analysis of a population growth model with harvesting, offering a detailed solution process and valuable insights into the dynamics of the population. We began by setting up the differential equation $P^{\prime}=0.4 P(1-\frac{P}{10,000})-0.06 P$, which models the population growth rate considering both logistic growth and harvesting. We then identified the equilibrium solutions, which represent the population sizes at which the growth rate is zero. These equilibrium points, $P=0$ and $P=8,500$, provided crucial information about the long-term behavior of the population. We subsequently solved the differential equation using the method of separation of variables, obtaining a general solution for the population size $P(t)$ as a function of time $t$. This general solution provided a complete description of the population dynamics, allowing us to predict the population size at any given time. To obtain a particular solution that specifically describes our scenario, we applied the initial condition of 5,000 fish at time $t=0$. This allowed us to determine the value of the constant of integration and arrive at the particular solution $P(t) = \frac{85,000e^{0.34t}}{7 + 10e^{0.34t}}$. Finally, we analyzed the solution and its implications for the population dynamics. We found that the population will approach the equilibrium solution of 8,500 as time goes on, indicating a stable population size under the given harvesting rate. We also discussed the importance of the harvesting rate and the carrying capacity of the environment in influencing the population dynamics. The solution provides a powerful tool for understanding the interplay between logistic growth, harvesting, and the environment's capacity to support the population. This understanding is crucial for making informed decisions about resource management and ensuring the sustainability of populations. By solving this population growth model with harvesting, we have gained valuable insights into the complex dynamics that govern populations in real-world scenarios. These insights can be applied to various fields, including fisheries management, wildlife conservation, and ecosystem management. The ability to model and predict population dynamics is essential for developing effective strategies for managing resources and ensuring the long-term health of ecosystems. This analysis underscores the importance of mathematical modeling in understanding and addressing real-world challenges in biology and ecology. The combination of analytical techniques, such as separation of variables and equilibrium analysis, provides a powerful framework for studying population dynamics and making informed decisions about resource management. The solution and its analysis presented in this article offer a clear illustration of how mathematical models can be used to gain insights into complex biological systems and to guide sustainable practices. The principles and methods discussed in this article can be extended to analyze other population models and to address a wide range of ecological and environmental challenges.