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Magazine Chapter 6: Problem 32
Solve with \(y_{0}=0\) and graph the solution. $$ \frac{d y}{d t}=y-1 $$
Short Answer
Expert verified
The solution is \( y = 1 - e^{t} \), starting at 0 and approaching 1 as \( t \to \infty \).
Step by step solution
01
Recognize the Type of Differential Equation
The given differential equation is \( \frac{dy}{dt} = y - 1 \). This is a first-order linear ordinary differential equation.
02
Rewrite the Differential Equation
Rearrange the equation as \( \frac{dy}{dt} - y = -1 \). This allows us to recognize it in the standard form \( \frac{dy}{dt} + P(t)y = Q(t) \), with \( P(t) = -1 \) and \( Q(t) = -1 \).
03
Determine the Integrating Factor
The integrating factor \( \mu(t) \) is given by \( e^{\int P(t) \, dt} = e^{-\int 1 \, dt} = e^{-t} \).
04
Multiply the Equation by the Integrating Factor
Multiply both sides of the rearranged differential equation by the integrating factor: \( e^{-t} \frac{dy}{dt} - e^{-t} y = -e^{-t} \).
05
Simplify and Integrate
The left side becomes \( \frac{d}{dt}(e^{-t}y) \), so the equation is now \( \frac{d}{dt}(e^{-t}y) = -e^{-t} \). Integrate both sides with respect to \( t \):\[ e^{-t}y = \int -e^{-t} dt = e^{-t} + C \]
06
Solve for \(y\)
Multiply through by \( e^{t} \):\[ y = 1 + Ce^{t} \].
07
Apply the Initial Condition
Use the initial condition \( y(0) = 0 \):\[ 0 = 1 + Ce^{0} \]Simplify to find \( C = -1 \).
08
Write the Particular Solution
Substitute \( C = -1 \) into the general solution:\[ y = 1 - e^{t} \].
09
Graph the Solution
To graph the solution \( y = 1 - e^{t} \), plot \( y \) against \( t \). The curve will start from \( y(0) = 0 \) and approach \( y = 1 \) as \( t \to \infty \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Integrating Factor
When working with first-order linear ordinary differential equations, the integrating factor is a crucial concept. It helps to simplify the differential equation, allowing for easier integration. An integrating factor is a special function that, when multiplied by the original equation, transforms it into an exact differential equation.
For the differential equation presented:
For the differential equation presented:
- Firstly, rearrange it to the standard form: \( \frac{dy}{dt} + P(t)y = Q(t) \).
- Here, \( P(t) = -1 \) and \( Q(t) = -1 \).
- The integrating factor, \( \mu(t) \), is calculated as \( e^{\int P(t) \, dt} = e^{-t} \).
Initial Condition
In solving ordinary differential equations, initial conditions are essential to find a specific solution. Initial conditions refer to the values of a function and its derivatives at a particular point, which are used to solve for the constants in the solution.
For this exercise, the initial condition is \( y(0) = 0 \), meaning when \( t = 0 \), \( y \) must be equal to 0.
For this exercise, the initial condition is \( y(0) = 0 \), meaning when \( t = 0 \), \( y \) must be equal to 0.
- After finding the general solution, which is \( y = 1 + Ce^{t} \), substitute \( t = 0 \) and \( y = 0 \).
- This process allows us to solve for the constant \( C \).
Ordinary Differential Equations
An ordinary differential equation (ODE) involves functions of a single independent variable and their derivatives. In this scenario, the exercise provided involves a first-order linear ODE, which means it involves the first derivative of the function.
The general form of a first-order linear ODE is:
The general form of a first-order linear ODE is:
- \( \frac{dy}{dt} + P(t)y = Q(t) \).
- Here, \( \frac{dy}{dt} = y - 1 \) is rewritten as \( \frac{dy}{dt} - y = -1 \).
Solution Graphing
Once the particular solution of a differential equation is found, graphing it can provide a visual interpretation of the solution over a range of values. For the equation solved, the particular solution is \( y = 1 - e^{t} \).
Graphing this function helps you observe significant characteristics:
Graphing this function helps you observe significant characteristics:
- Start from the initial condition, which is \( y(0) = 0 \).
- As \( t \) increases, \( y \) approaches 1. This demonstrates the steady-state behavior.
