Problem 4 - Calculus Early Transcendentals

Question

Solve the differential equation by the method of integrating factors. $$ 2 \frac{d y}{d x}+4 y=1 $$

Step-by-Step Solution

Answer

The solution is \( y(x) = \frac{1}{4} + Ce^{-2x} \).

Step 1: Identify Standard Form

First, recognize that the given differential equation is a first-order linear ordinary differential equation (ODE). The standard form for such an equation is \( \frac{dy}{dx} + P(x)y = Q(x) \). Rewriting the given equation: \[ 2 \frac{dy}{dx} + 4y = 1 \] Divide the entire equation by 2 to match the standard form:\[ \frac{dy}{dx} + 2y = \frac{1}{2} \] Here, \( P(x) = 2 \) and \( Q(x) = \frac{1}{2} \).

Step 2: Find the Integrating Factor

To solve the differential equation, we need to find the integrating factor, \( \mu(x) \). The integrating factor for a linear ODE is given by:\[ \mu(x) = e^{\int P(x) \, dx} = e^{\int 2 \, dx} \]Calculate the integral:\[ \int 2 \, dx = 2x \]Thus, the integrating factor:\[ \mu(x) = e^{2x} \]

Step 3: Multiply Through by the Integrating Factor

Multiply the entire differential equation by the integrating factor \( e^{2x} \):\[ e^{2x} \frac{dy}{dx} + 2e^{2x} y = \frac{1}{2} e^{2x} \]

Step 4: Recognize and Simplify Left Side

The left side of the equation now becomes the derivative of \( y \cdot e^{2x} \). Therefore, \[ \frac{d}{dx}(y \cdot e^{2x}) = \frac{1}{2} e^{2x} \]

Step 5: Integrate Both Sides

Now, integrate both sides of the equation with respect to \( x \):\[ \int \frac{d}{dx}(y \cdot e^{2x}) \, dx = \int \frac{1}{2} e^{2x} \, dx \]The left side simplifies to:\[ y \cdot e^{2x} \]The right side becomes:\[ \frac{1}{2} \int e^{2x} \, dx = \frac{1}{2} \cdot \frac{1}{2} e^{2x} = \frac{1}{4} e^{2x} \]So, we have:\[ y \cdot e^{2x} = \frac{1}{4} e^{2x} + C \]

Step 6: Solve for y(x)

Solve for \( y \) by dividing through by \( e^{2x} \):\[ y = \frac{1}{4} + Ce^{-2x} \] Thus, the general solution to the differential equation is:\[ y(x) = \frac{1}{4} + Ce^{-2x} \]

Key Concepts

First-Order Linear Differential Equation

A first-order linear differential equation is a type of ordinary differential equation (ODE) that involves the first derivative of a function. It is called "linear" because both the function and its derivative appear linearly in the equation. In mathematical terms, a first-order linear ODE has a standard form that looks like this: \( \frac{dy}{dx} + P(x)y = Q(x) \). Here, \( P(x) \) and \( Q(x) \) are functions of \( x \).
To solve such an equation, you often use a technique called the method of integrating factors. This method cleverly transforms the original equation into one that is easier to solve.
By identifying the first-order linear ODE's form, you can determine \( P(x) \) and \( Q(x) \), simplifying the approach to finding the solution.

Ordinary Differential Equations (ODE)

Ordinary differential equations (ODEs) are a type of mathematical equation that describes the relationship between a function and its derivatives. They play a crucial role in modeling dynamic systems in fields such as physics, engineering, and biology.
An ODE involves ordinary derivatives, as opposed to partial derivatives, which occur in partial differential equations (PDEs). The degree of an ODE is determined by the highest order derivative present.
In our specific case, we're dealing with a first-order linear ODE. This indicates the presence of only the first derivative \( \frac{dy}{dx} \). Solving these equations can provide significant insights into how a particular system changes over time, capturing the rates at which different quantities evolve.

Solution to Differential Equations

The solution to a differential equation is a function or a family of functions that satisfy the equation. In the realm of first-order linear ODEs, this solution process typically involves finding particular and general solutions.
To achieve this, the method of integrating factors is commonly used. By multiplying the entire equation by a specially derived function, known as the integrating factor, it becomes possible to rewrite the differential equation in a simplified form that permits easy integration.

• The integrating factor \( \mu(x) \) is usually \( e^{\int P(x) \, dx} \).
• After applying the integrating factor, the equation can often be expressed as the derivative of a product of terms.
• The resulting expression is then integrated to find the solution.

In the example exercise, the general solution is found to be \( y(x) = \frac{1}{4} + Ce^{-2x} \), where \( C \) is a constant determined by initial or boundary conditions.