Question

Solving a Logistic Differential Equation In Exercises \(55-58\) , find the logistic equation that passes through the given point. $$ \frac{d y}{d t}=\frac{3 y}{20}-\frac{y^{2}}{1600}, \quad(0,15) $$

Short answer

The final solution is the logistic equation passing through the given point, achieved by solving the differential equation and substitifying the given point into the equation to determine the constant of integration.

Step-by-step solution

Understand the given differential Equation

The given differential equation is \(\frac{d y}{d t} = \frac{3 y}{20}-\frac{y^{2}}{1600}\). It can be rewritten as \(\frac{d y}{d t} = ry - \frac{y^{2}}{K}\), where r is called the specific growth rate, and K is called the carrying capacity of the environment. The task is to find a solution to this equation that passes through the point (0,15).

Solve the differential Equation

By rearranging, we can write equation as: \[\frac{dy}{y(\frac{r}{K}-\frac{y}{K})} = dt\] Integrate both sides to get the solution, which will generate a natural logarithm on the left-hand side and t plus the constant of integration C on the right-hand side.

Use the given point to find the constant of integration

With the solution from Step 2 having a constant of integration, use the given point (t=0, y=15) to find the exact value of the constant. Substitute the coordinates of the point into the equation, solve for C.

Write down the final solution

Substitute the constant C back into the equation derived in Step 2. This gives the logistic equation that passes through the given point.

Key concepts

Specific Growth Rate

The specific growth rate ( ) is a vital concept in the logistic differential equation. It represents how quickly a population can grow in an ideal environment without any limiting factors. In mathematical terms, it is denoted by and appears as a coefficient next to the population variable. The specific growth rate affects how fast the population reaches its maximum potential size when no restrictions are in place.

In our logistic differential equation, this rate is measured by the term rac{3y}{20}. It describes the potential rate of growth for a given population at an initial stage, providing insight into how fast the population would expand under ideal circumstances. Understanding this concept helps in modeling and predicting population dynamics over time.

Carrying Capacity

Carrying capacity, symbolized by K, is another crucial component of the logistic differential equation. It refers to the maximum population size that an environment can sustain indefinitely given food, habitat, water, and other resources. Essentially, it is the limit of population growth in realistic conditions.

In the logistic model, carrying capacity ( K) is found in the denominator of the quadratic term, as seen in rac{y^2}{1600}. This term modifies the growth rate, slowing it down as the population approaches the carrying capacity. Understanding carrying capacity allows us to predict how populations interact with their environments and what thresholds could lead to drastic shifts in population stability.

Constant of Integration

The constant of integration plays an important role when solving differential equations. After integrating, solutions include a constant of integration ( C) that accounts for initial conditions. This constant can be understood as an arbitrary value added during integration that makes the solution fit specific circumstances.

In our problem, after integrating both sides of the equation, we introduce C on the right-hand side. To determine its value, we use the given initial condition or specific point (for instance, (0,15)). By substituting this point into the equation, C can be solved, allowing us to tailor the generic solution to this particular instance. The constant of integration is essential for finding a specific solution that fits the initial or boundary conditions of a problem.

Integration of Differential Equations

Integration of differential equations is a fundamental process in solving them. It involves finding a function whose derivative matches the given differential form. For logistic differential equations, this usually contains integrating both sides to relate the population to time.

In the given exercise, integrating rac{d y}{y(rac{r}{K}-rac{y}{K})} with respect to dt allows us to solve for yt over t. This integral results in a natural logarithm term and a constant of integration to be determined later. The logistics of integrating correctly provide us with a solvable equation that denotes population size at any given time. It helps in understanding how different factors like specific growth rate and carrying capacity play into the broader equation and model, allowing better predictive capabilities for population studies.