Problem 6 - Calculus Early Transcendentals

Question

Solve the differential equation by the method of integrating factors. $$ \frac{d y}{d x}+y+\frac{1}{1-e^{x}}=0 $$

Step-by-Step Solution

Answer

The solution is \( y = -\frac{\ln|1-e^x|}{e^x} + \frac{C}{e^x} \).

Step 1: Identify the Standard Form

The given differential equation is \( \frac{dy}{dx} + y + \frac{1}{1-e^x} = 0 \). A first-order linear differential equation should be in the form \( \frac{dy}{dx} + P(x)y = Q(x) \). Thus, rewrite it to identify \( P(x) \) and \( Q(x) \).\( \frac{dy}{dx} + y = -\frac{1}{1-e^x} \).Here, \( P(x) = 1 \) and \( Q(x) = -\frac{1}{1-e^x} \).

Step 2: Calculate the Integrating Factor

The integrating factor \( \mu(x) \) is calculated using the formula\[ \mu(x) = e^{\int P(x) \, dx} \].Given \( P(x) = 1 \), we have:\[ \int 1 \, dx = x \], so \( \mu(x) = e^x \).

Step 3: Multiply the Equation by the Integrating Factor

Multiply the entire differential equation by the integrating factor \( e^x \):\[ e^x \frac{dy}{dx} + e^x y = -\frac{e^x}{1-e^x} \].

Step 4: Recognize the Left-Hand Side as a Derivative

Notice that the left-hand side of the equation \( e^x \frac{dy}{dx} + e^x y \) is the derivative of \( e^x y \) with respect to \( x \):\[ \frac{d}{dx}(e^x y) = -\frac{e^x}{1-e^x} \].

Step 5: Integrate Both Sides

Integrate both sides with respect to \( x \):\[ \int \frac{d}{dx}(e^x y) \, dx = \int -\frac{e^x}{1-e^x} \, dx \].The left side simplifies to \( e^x y \), and you integrate the right-hand side separately.

Step 6: Integration of the Right-Hand Side

Integrate the right side \( \int -\frac{e^x}{1-e^x} \, dx \). This integral might need substitution methods, let's consider the substitution \( u = 1-e^x \), giving \( du = -e^x dx \). The integral becomes:\[ \int \frac{1}{u} \, du = \ln|u| = \ln|1-e^x| \].So,\[ e^x y = -\ln|1-e^x| + C \]where \( C \) is the constant of integration.

Step 7: Solve for \( y \)

Divide both sides by \( e^x \) to solve for \( y \):\[ y = -\frac{\ln|1-e^x|}{e^x} + \frac{C}{e^x} \].And that's the general solution to the differential equation.

Key Concepts

Integrating Factors

An integrating factor is a vital concept in solving first-order linear differential equations. It helps transform a non-exact equation into an exact one, making it more manageable to solve.
Understanding how to find and apply integrating factors is crucial in this context.
To compute the integrating factor, denoted as \( \mu(x) \), you use the expression \( \mu(x) = e^{\int P(x) \, dx} \). Here's a breakdown of the steps:

• First, identify \( P(x) \) from the differential equation.
• Carry out the integration of \( P(x) \) to obtain an expression to exponentiate.
• The result, \( e^x \) in the given exercise, becomes the integrating factor.

This factor is then used to multiply through the entire differential equation. The advantage is that once you apply the integrating factor, the left-hand side of the equation often transforms into the derivative of a product, which is simpler to handle.

First-Order Linear Differential Equations

First-order linear differential equations are a fundamental type of differential equation. It's characterized by the format \( \frac{dy}{dx} + P(x)y = Q(x) \).
These equations involve derivatives of first-order and typically contain a variable, its derivative, and a function.
Identifying your equation in this form helps in utilizing techniques like integrating factors. Follow these essential steps:

• Rewrite your equation in the standard format if needed.
• pinpoint \( P(x) \) and \( Q(x) \).
• In the given problem, \( P(x) = 1 \) and \( Q(x) = -\frac{1}{1-e^x} \).

Once the structure is apparent, applying techniques for solving becomes a straightforward process. These equations model numerous real-world phenomena where rates of change are proportional to current values.

Constant of Integration

The constant of integration is an integral part of solving differential equations. Whenever you solve an indefinite integral, you include this constant to account for all possible solutions. This is because multiple functions can share the same derivative, differing only by a constant.
The constant of integration is typically represented by \( C \).
Here's how it works:

• After integrating the equation, don't forget to append \( + C \) to your solution.
• For instance, after integrating \( -\frac{e^x}{1-e^x} \), you obtain \( -\ln|1-e^x| + C \).
• The constant is critical, especially for determining initial conditions or specific solutions.

This constant ensures your solution remains as general as possible, adapting to a variety of potential initial value problems or applications.