Question

Solve each differential equation. $$ \frac{d y}{d x}+y=e^{-x} $$

Short answer

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

Step-by-step solution

Identify the type of differential equation

The given differential equation is \( \frac{dy}{dx} + y = e^{-x} \). This is a first-order linear differential equation.

Write the standard form

First, identify the standard form of a first-order linear differential equation, which is \( \frac{dy}{dx} + P(x)y = Q(x) \). For the given equation, \( P(x) = 1 \) and \( Q(x) = e^{-x} \).

Calculate the integrating factor

The integrating factor \( \mu(x) \) is calculated as \( e^{\int P(x) \, dx} \). Since \( P(x) = 1 \), we get \( \mu(x) = e^{\int 1 \, dx} = e^{x} \).

Multiply through by the integrating factor

Multiply every term of the differential equation by the integrating factor \( e^{x} \). The equation becomes \( e^{x} \frac{dy}{dx} + e^{x} y = e^{x} e^{-x} = 1 \).

Simplify the equation

Notice that the left-hand side of the equation is the derivative of \( e^{x} y \), or \( \frac{d}{dx}(e^{x} y) = 1 \).

Integrate both sides

Integrate both sides with respect to \( x \):\[ \int \frac{d}{dx}(e^{x} y) \, dx = \int 1 \, dx \]This gives \( e^{x} y = x + C \), where \( C \) is the constant of integration.

Solve for \( y \)

To find \( y \), divide both sides by \( e^{x} \):\[ y = e^{-x}(x + C) \].

Key concepts

Integrating Factor

An integrating factor is a mathematical tool used to solve first-order linear differential equations. It simplifies the process by converting the original equation into a form that is easier to integrate. To determine the integrating factor, identify the function \( P(x) \) from the standard form \( \frac{dy}{dx} + P(x)y = Q(x) \). Once identified, calculate the integrating factor \( \mu(x) \) using the formula:

• \( \mu(x) = e^{\int P(x) \, dx} \)

In our example, \( P(x) = 1 \), so the integrating factor is \( e^{x} \). With this factor, multiplying through the given differential equation aligns the left-hand side into the derivative of a product, simplifying further integration.

Standard Form of Differential Equations

The standard form of a first-order linear differential equation is crucial for applying the method of solving these equations. It provides a clear structure, which is:

• \( \frac{dy}{dx} + P(x)y = Q(x) \)

In the given exercise, we can directly compare our equation \( \frac{dy}{dx} + y = e^{-x} \) to the standard form. Here,

• \( P(x) = 1 \)
• \( Q(x) = e^{-x} \)

Recognizing these components is the first step in determining how to approach solving the equation. It helps identify the integrating factor and set up the equation for further manipulation.

Solution of Differential Equations

After identifying the standard form and integrating factor, solving the differential equation involves integrating both sides of the equation. By multiplying the entire equation by the integrating factor, we transform it into a more manageable form:

• \( (e^{x} \cdot \frac{dy}{dx} + e^{x}y) = e^{x} \cdot e^{-x} = 1 \)

This simplifies to the derivative \( \frac{d}{dx}(e^{x}y) = 1 \). The next step involves integrating this result:

• \( \int \frac{d}{dx}(e^{x} y) \, dx = \int 1 \, dx \)

The integration yields \( e^{x} y = x + C \), where \( C \) is the constant of integration.

Constant of Integration

The constant of integration, represented by \( C \), arises whenever solving differential equations through integration. It represents an arbitrary constant that accounts for the family of solutions possible in indefinite integration. In our solution, the integral gives

• \( e^{x} y = x + C \)

To express \( y \), we divide by \( e^{x} \), resulting in:

• \( y = e^{-x}(x + C) \)

Here, \( C \) allows the generalization of the solution, accommodating any initial conditions that might be applied later. It captures the idea that multiple solutions can satisfy the differential equation, depending on the initial value or boundary conditions.