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Magazine Chapter 11: Problem 21
Give the solution to the logistic differential equation with initial condition. $$\frac{d P}{d t}=0.8 P\left(1-\frac{P}{8500}\right) \text { with } P_{0}=500$$
Short Answer
Expert verified
The solution is \( P(t) = \frac{8500}{1 + 16e^{-0.8t}} \).
Step by step solution
01
Identify the Logistic Differential Equation
The given differential equation is \( \frac{dP}{dt} = 0.8P \left( 1 - \frac{P}{8500} \right) \). This is a logistic differential equation, where the growth rate is proportional to both the current population \( P \) and the difference between the carrying capacity \( K = 8500 \) and the current population.
02
Write the General Solution Form
The general solution of a logistic differential equation \( \frac{dP}{dt} = rP \left( 1 - \frac{P}{K} \right) \) is given by \(P(t) = \frac{K}{1 + Ae^{-rt}}.\) Here, \( r = 0.8 \) and \( K = 8500 \). Rewrite the solution as \( P(t) = \frac{8500}{1 + Ae^{-0.8t}}. \)
03
Determine A Using Initial Condition
We use the initial condition \( P(0) = 500 \) to find the constant \( A \). Substituting these values into the general solution, we get:\[500 = \frac{8500}{1 + A}\]Solving for \( A \) gives:\( 1 + A = \frac{8500}{500} = 17 \) which implies \( A = 16. \)
04
Write the Specific Solution
Substituting the value of \( A = 16 \) into the general solution, the specific solution is:\[P(t) = \frac{8500}{1 + 16e^{-0.8t}}.\]
05
Verify the Solution
To ensure the solution satisfies both the differential equation and the initial condition, differentiate \( P(t) = \frac{8500}{1 + 16e^{-0.8t}} \) and substitute back into the original differential equation. Check that \( P(0) = 500 \) holds true, confirming the solution is valid.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Initial Condition
The initial condition in a differential equation is like the starting point of a journey. It's the value that the solution of the equation will pass through at a specific time, usually at time zero. For our problem, the initial condition is given as \( P_0 = 500 \). This means that at time \( t = 0 \), the population \( P \) is 500.
In solving logistic differential equations, initial conditions are crucial. They allow us to find specific solutions that fit real-world scenarios. Here's how it works:
In solving logistic differential equations, initial conditions are crucial. They allow us to find specific solutions that fit real-world scenarios. Here's how it works:
- Start with the general solution of the differential equation.
- Plug in the initial condition to solve for any constants in the equation.
- Get a specific solution that tailors the general formula to the given scenario.
Carrying Capacity
The carrying capacity in biological terms is like a ceiling, the maximum limit the population can reach in an environment. This is represented by \( K \) in logistic equations. In this problem, the carrying capacity is specified as 8500.
Carrying capacity is vital because:
Carrying capacity is vital because:
- It reflects the environmental limitations that restrict population growth.
- It helps predict the behavior of populations over time. As population approaches \( K \), the growth slows down.
- A logistic curve flattens as it nears the carrying capacity, demonstrating sustainability of the population.
Growth Rate
The growth rate in a logistic differential equation quantifies how quickly a population can increase under ideal conditions without constraints. Represented by \( r \), the growth rate influences the steepness of the logistic curve.
Here it's given as 0.8, which means:
Here it's given as 0.8, which means:
- It's how much the population is growing when limited by resources.
- The constant impacts how quickly the population reaches the carrying capacity.
General Solution
The general solution of the logistic differential equation is a versatile formula that can describe various growth scenarios based on initial conditions and other parameters. For a logistic differential equation like \( \frac{dP}{dt} = rP\left(1 - \frac{P}{K}\right) \), the general solution is expressed as:\[P(t) = \frac{K}{1 + Ae^{-rt}}\]This expression:
- Helps describe how population changes over time.
- Includes constants like \( K \) for carrying capacity and \( r \) for growth rate.
- Is adaptable, ready to be adjusted using specific conditions like the initial population.
