Problem 32 - Calculus

Question

Solve with \(y_{0}=0\) and graph the solution. $$ \frac{d y}{d t}=y-1 $$

Step-by-Step Solution

Answer

The solution is \( y = 1 - e^{t} \), starting at 0 and approaching 1 as \( t \to \infty \).

Step 1: 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.

Step 2: 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 \( Q(t) = -1 \).

Step 3: Determine the Integrating Factor

The integrating factor \( \mu(t) \) is given by \( e^{\int P(t) \, dt} = e^{-\int 1 \, dt} = e^{-t} \).

Step 4: Multiply the Equation by the Integrating Factor

Multiply both sides of the rearranged differential equation by the integrating factor: \( e^{-t} \frac{dy}{dt} - e^{-t} y = -e^{-t} \).

Step 5: 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 \( t \):\[ e^{-t}y = \int -e^{-t} dt = e^{-t} + C \]

Step 6: Solve for \(y\)

Multiply through by \( e^{t} \):\[ y = 1 + Ce^{t} \].

Step 7: Apply the Initial Condition

Use the initial condition \( y(0) = 0 \):\[ 0 = 1 + Ce^{0} \]Simplify to find \( C = -1 \).

Step 8: Write the Particular Solution

Substitute \( C = -1 \) into the general solution:\[ y = 1 - e^{t} \].

Step 9: 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 \( t \to \infty \).

Key Concepts

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:

• 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 \( e^{\int P(t) \, dt} = e^{-t} \).

This technique ensures the left side of the equation becomes a derivative of a single expression, paving the way for straightforward integration.

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.

• 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 \( C \).

Substituting these values into the general solution gives us the particular solution needed to plot and understand the behavior of the differential equation over time.

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:

• \( \frac{dy}{dt} + P(t)y = Q(t) \).
• Here, \( \frac{dy}{dt} = y - 1 \) is rewritten as \( \frac{dy}{dt} - y = -1 \).

Recognizing the structure of an ODE is crucial because it helps us determine the method for finding the solution, such as using an integrating factor.

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:

• Start from the initial condition, which is \( y(0) = 0 \).
• As \( t \) increases, \( y \) approaches 1. This demonstrates the steady-state behavior.

The graph illustrates how quickly the function approaches the horizontal asymptote, showing the dynamic behavior of the solution over time.