Chapter 6: Problem 25
Find the solution \(y(t)\) by recognizing each differential equation as determining unlimited, limited, or logistic growth, and then finding the constants. \(y^{\prime}=2-0.01 y\) \(y(0)=0\)
Short Answer
Expert verified
The solution is \( y(t) = 200 - 200e^{-0.01t} \).
Step by step solution
01
Identify the Type of Growth Model
The differential equation given is \( y' = 2 - 0.01y \). By inspecting the structure, we can recognize this as a limited growth model, also known as the exponential decay towards a limit or linear differential equation. This type is characterized by the presence of a term proportional to \( y \).
02
Rearrange and Analyze the Equation
Rewrite the equation as \( y' + 0.01y = 2 \). This is a first-order linear differential equation, where the rate of change of \( y \) depends on \( y \) itself and a constant.
03
Solve Using Integrating Factor
To solve, identify the integrating factor \( \, \mu(t) = e^{ extstyle 0.01t} \, \).Multiply through by this factor:\[ e^{0.01t} y' + 0.01 e^{0.01t} y = 2 e^{0.01t} \]The left side is the derivative of \( e^{0.01t} y \):\[\frac{d}{dt} [ e^{0.01t} y ] = 2 e^{0.01t} \]
04
Integrate Both Sides
Integrate both sides with respect to \( t \):\[\int \! \frac{d}{dt} [ e^{0.01t} y ] \; dt = \int \! 2 e^{0.01t} \; dt\]This gives:\[e^{0.01t} y = 2 \cdot \frac{1}{0.01}e^{0.01t} + C\]Simplifying the right side:\[e^{0.01t} y = 200e^{0.01t} + C\]
05
Solve for y(t)
Divide both sides by \( e^{0.01t} \):\[ y = 200 + Ce^{-0.01t} \]Now use the initial condition \( y(0) = 0 \) to solve for \( C \).
06
Apply Initial Condition
Substitute \( t=0 \) and \( y=0 \):\[ 0 = 200 + C \cdot e^{0} \]This implies:\[ C = -200 \]Therefore, the solution is:\[ y(t) = 200 - 200e^{-0.01t} \]
07
Final Step: Interpret the Solution
This solution represents a limited growth model where \( y(t) \) approaches 200 as \( t \) approaches infinity, starting from an initial value of 0.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Limited Growth Model
The limited growth model is a concept in differential equations where the growth rate of a quantity is restricted by certain factors. In our given problem, we have the equation \(y' = 2 - 0.01y\). The limited growth model is characterized by terms proportional to \(y\) that exponentially decay toward a specific limit. In this case, as \(y\) increases, the term \(-0.01y\) starts to dominate, slowing down the growth until it approaches a limit, or ceiling, which is derived from balancing the terms on both sides of the equation. Some key properties of the limited growth model include:
- Exponential decay towards a ceiling value, where growth slows as it approaches this value.
- Often modeled by first-order linear differential equations with a constant term and a proportional term involving the dependent variable.
- Useful in simulating situations in biology, economics, and ecology where resources are limited.
Integrating Factor
An integrating factor is a technique used to solve first-order linear differential equations, bringing structure and symmetry to otherwise challenging equations. For the equation \(y' + 0.01y = 2\), we identify the integrating factor \(\mu(t) = e^{0.01t}\). The integrating factor is calculated based on the coefficient of \(y\) from the linear differential equation.The steps in applying the integrating factor:
- Find \(\mu(t) = e^{\int 0.01 \, dt} = e^{0.01t}\).
- Multiply every term in the equation by \(\mu(t)\) to facilitate integration.
- The left side will then become the derivative of a product, \(\frac{d}{dt} [ e^{0.01t} y ]\), simplifying integration.
Initial Condition
An initial condition is a foundational part of solving differential equations, allowing us to find the specific solution that fits a given problem. In this case, the initial condition provided is \(y(0) = 0\), meaning that when \(t = 0\), \(y\) starts at zero.With the general solution to the differential equation being \(y(t) = 200 + Ce^{-0.01t}\), we use the initial condition to determine the constant \(C\). Here's how this process works:
- Substitute the initial values into the general solution: \(0 = 200 + C \cdot e^0\)
- Solve for \(C\): \(C = -200\)