Question

Solve the following differential equations by using integrating factors. $$ y^{\prime}=y+e^{x} $$

Short answer

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

Step-by-step solution

Identify the Standard Form

The given differential equation is \( y^{\prime} = y + e^x \). The standard form for using integrating factors is \( y' + P(x)y = Q(x) \). Rewrite the equation as \( y' - y = e^x \), where \( P(x) = -1 \) and \( Q(x) = e^x \).

Determine the Integrating Factor

The integrating factor, \( \mu(x) \), is given by \( e^{\int P(x) \, dx} \). Here, \( P(x) = -1 \), so the integrating factor is \( \mu(x) = e^{\int -1 \, dx} = e^{-x} \).

Multiply Equation by Integrating Factor

Multiply the entire differential equation by the integrating factor: \( e^{-x}(y' - y) = e^{-x}e^x \), which simplifies to \( e^{-x}y' - e^{-x}y = 1 \).

Simplify the Left-Hand Side

Notice that the left-hand side is the derivative of the product of the integrating factor and \( y \): \( \frac{d}{dx}(e^{-x}y) = e^{-x}y' - e^{-x}y \). Thus, we can rewrite the equation as \( \frac{d}{dx}(e^{-x}y) = 1 \).

Integrate Both Sides

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

Solve for \( y \)

Isolate \( y \) by multiplying both sides by \( e^x \): \( y = e^{x}(x + C) \). This is the general solution of the differential equation.

Key concepts

Integrating Factors

Integrating factors are a powerful technique used to solve linear first-order differential equations. These equations often take the form \( y' + P(x)y = Q(x) \). To make the equation solvable, we use a specific function, known as the integrating factor. This function transforms the left-hand side of the equation into a product rule derivative.

• Step 1: Identify the standard form of the equation.
• Step 2: Calculate the integrating factor \( \mu(x) = e^{\int P(x) \, dx} \).

After calculating the integrating factor, multiply every term in the differential equation by this factor. This makes the equation easier to solve, as the left side will simplify into the derivative of a product of two functions. By doing so, you can directly integrate both sides to find the solution.

Ordinary Differential Equations

Ordinary Differential Equations (ODEs) involve relations between a function and its derivatives. They're termed "ordinary" because they deal with functions of a single variable, unlike partial differential equations, which involve multiple variables.ODEs can appear in various forms and complexities:

• First-order ODEs: involve the first derivative \( y' \).
• Second-order ODEs: involve up to the second derivative \( y'' \).
• Linear ODEs: can be written in the form \( y' + P(x)y = Q(x) \).
• Non-linear ODEs: cannot be reduced to a linear form.

The process of solving ODEs often involves finding a function that satisfies the equation across its entire range. This can sometimes be straightforward, while other times require more advanced methods, like integrating factors.

General Solution

The general solution of a differential equation represents all possible solutions, typically including a constant of integration, denoted as \( C \). For linear first-order ODEs, the general solution can be obtained by integrating the transformed equation and solving for the dependent variable.For the differential equation \( y' = y + e^x \), after finding the integrating factor and simplifying, the equation \( e^{-x}y = x + C \) arises. To solve for \( y \):- Multiply both sides by \( e^x \) to isolate \( y \).- Here, \( y = e^x(x + C) \) forms the general solution.General solutions are crucial because they encapsulate every possible specific solution. Specifying \( C \) will yield a particular solution that satisfies specified initial conditions, making it increasingly important in practical applications.