Problem 54 - Calculus Early Transcendentals

Question

Plot a solution to the initial-value problem $$ \frac{d y}{d t}=0.98\left(1-\frac{y}{5}\right) y, \quad y_{0}=1 $$

Step-by-Step Solution

Answer

The solution is \( y(t) = \frac{5}{1+4e^{-0.98t}} \), starting at \( y=1 \) and \( t=0 \).

Step 1: Understand the differential equation

The differential equation given is \( \frac{dy}{dt} = 0.98 \left(1 - \frac{y}{5} \right) y \), which is a separable first-order differential equation. The initial condition is \( y_0 = 1 \).

Step 2: Rewrite the equation in separable form

Express the equation such that all \( y \) terms are on one side and \( t \) terms are on the other: \( \frac{dy}{y(1 - \frac{y}{5})} = 0.98 \, dt \).

Step 3: Integrate both sides

Integrate the left side with respect to \( y \) and the right side with respect to \( t \): \( \int \frac{dy}{y(1 - \frac{y}{5})} = \int 0.98 \, dt \).

Step 4: Solve the integrals

The left integral can be solved using partial fractions, and the right is straightforward: \( \ln|y| - \ln|5-y| = 0.98t + C \).

Step 5: Solve for the constant C

Use the initial condition \( y(0) = 1 \) to find \( C \): Substitute \( y = 1 \) and \( t = 0 \) into \( \ln|1| - \ln|5-1| = C \), resulting in \( C = -\ln 4 \).

Step 6: Find the explicit solution

Solve the equation for \( y \): \( \frac{|y|}{|5-y|} = e^{0.98t - \ln 4} \), which simplifies to \( y = \frac{5}{1 + 4e^{-0.98t}} \).

Step 7: Plot the solution

Using the function \( y(t) = \frac{5}{1 + 4e^{-0.98t}} \), plot \( y \) versus \( t \). This function describes an increasing sigmoid curve starting from \( y = 1 \) at \( t = 0 \).

Key Concepts

Initial Value Problem

In the field of differential equations, an initial value problem is a common scenario where you need to find a function that satisfies a given differential equation and meets a specified initial condition. The initial condition provides a starting point for solving the differential equation, ensuring we find a particular solution instead of the general one.

For example, in the exercise provided, the initial value problem is stated as \( \frac{dy}{dt} = 0.98(1-\frac{y}{5})y \) with the initial condition \( y_0 = 1 \). This means that when \( t = 0 \), \( y(t) = 1 \). The aim is to find the specific function \( y(t) \) that not only satisfies the differential equation but also starts at the initial point \( y_0 = 1 \).

This setup is crucial because it dictates how the function will behave over time, often requiring specific techniques to ensure the solution is both accurate and meaningful within the context of the problem.

Partial Fraction Decomposition

Partial fraction decomposition is a technique used to simplify the process of integrating complex rational expressions. When dealing with separable differential equations, like the one in our original exercise, this method comes in handy to tackle the integration.In our example, the separation gave us \( \frac{dy}{y(1-\frac{y}{5})} = 0.98\, dt \). To integrate the left-hand side, we need to decompose \( \frac{1}{y(1-\frac{y}{5})} \) into simpler fractions. This expression can be broken down into two parts:

• \( \frac{A}{y} \)
• \( \frac{B}{1-\frac{y}{5}} \)

By solving for \( A \) and \( B \), you rewrite the fraction in terms of these simpler components, making the integration process straightforward. This decomposition enables the use of basic integration techniques, converting a daunting integral into an easily solvable one.

Integrating Differential Equations

The integration of differential equations is a fundamental skill needed to solve initial value problems. In our example, after splitting the variables, we arrive at two integrals: one in terms of \( y \) and the other in terms of \( t \).The integration of \( \int \frac{dy}{y} \) results in \( \ln|y| \) and the integration of \( \int \frac{dy}{5-y} \) results in \( -\ln|5-y| \). This is achieved by using partial fraction decomposition, which simplifies the integrals.On the right side, \( \int 0.98 \, dt \) is straightforward and results in \( 0.98t + C \), where \( C \) is the constant of integration. Using the initial condition given \( y_0 = 1 \), we can solve for this constant. Substituting back into the equation, we link the solution to the initial condition, ensuring the particular solution behaves as required from the initial point \( t = 0 \).This combination of techniques facilitates solving differential equations, transforming them from complex puzzles into manageable problems.