Problem 6 - Calculus for Biology and Medicine

Question

Find the general solution of the differential equations in Problems 1-12 using the method of integrating factors: $$ \frac{d y}{d x}+\frac{y}{x+2}=x $$

Step-by-Step Solution

Answer

The general solution is \( y = \frac{1}{|x+2|}(\frac{x^3}{3} + x^2 + C) \).

Step 1: Rewrite the Equation in Standard Form

The given differential equation is \( \frac{d y}{d x} + \frac{y}{x+2} = x \). We need to write it in standard form of a linear first-order differential equation: \( \frac{d y}{d x} + P(x)y = Q(x) \). Here, \( P(x) = \frac{1}{x+2} \) and \( Q(x) = x \). The equation is already in standard form.

Step 2: Identify the Integrating Factor

The integrating factor \( \mu(x) \) is defined as \( e^{\int P(x) \, dx} \). Calculate \( \int P(x) \, dx \), where \( P(x) = \frac{1}{x+2} \). Thus, \( \int \frac{1}{x+2} \, dx = \ln|x+2| \). Therefore, the integrating factor is \( \mu(x) = e^{\ln|x+2|} = |x+2| \).

Step 3: Multiply Through by the Integrating Factor

To solve the differential equation, multiply each term by the integrating factor \( |x+2| \). The equation becomes: \( |x+2| \frac{d y}{d x} + \frac{|x+2|}{x+2}y = x|x+2| \). Simplifying, we get \( |x+2| \frac{d y}{d x} + y = x|x+2| \).

Step 4: Recognize the Left Side as a Derivative

The left side of the equation is the derivative of the product \( |x+2|y \). Thus, we have: \( \frac{d}{dx}(|x+2|y) = x|x+2| \).

Step 5: Integrate Both Sides

To solve for \( y \), integrate both sides of the equation with respect to \( x \): \( \int \frac{d}{dx}(|x+2|y) \, dx = \int x|x+2| \, dx \). The left side integrates to \( |x+2|y \). The right side requires you to expand \( x|x+2| \) before integrating, resulting in: \( x|x+2| = x^2 + 2x \). Thus, \( \int (x^2 + 2x) \, dx = \frac{x^3}{3} + x^2 + C \), where \( C \) is the constant of integration.

Step 6: Solve for y

After integrating, we have \( |x+2|y = \frac{x^3}{3} + x^2 + C \). Finally, divide by \(|x+2|\) to solve for \( y \): \( y = \frac{1}{|x+2|}(\frac{x^3}{3} + x^2 + C) \).

Key Concepts

First-order Differential Equation

A first-order differential equation is an equation that relates a function with its first derivative. These kinds of equations are crucial in modeling various real-world phenomena such as population growth or radioactive decay. Typically, they are in the form

• \( \frac{dy}{dx} = f(x, y) \)

where \( y \) is the dependent variable and \( x \) is the independent variable. However, for easier analysis, they can be converted to a standard linear form, specifically for linear equations given as

• \( \frac{dy}{dx} + P(x)y = Q(x) \)

where \( P(x) \) and \( Q(x) \) are functions of \( x \). For example, in our exercise:

• \( \frac{dy}{dx} + \frac{y}{x+2} = x \)

The term \( \frac{1}{x+2} \) acts as our function \( P(x) \), and \( x \) is our function \( Q(x) \). Recognizing this form is the first step in applying the integrating factor method.

General Solution of Differential Equations

The general solution of a differential equation involves finding an expression for the dependent variable (e.g., \( y \)) in terms of the independent variable (e.g., \( x \)), along with an arbitrary constant that represents the 'family' of all possible solutions. In the context of a first-order linear differential equation of the form

• \( \frac{dy}{dx} + P(x)y = Q(x) \)

we use the integrating factor method to derive the general solution. An integrating factor transforms a non-exact differential equation into an equivalent exact equation.
To find this integrating factor \( \mu(x) \), we calculate

• \( \mu(x) = e^{\int P(x) \, dx} \)

For instance, for our exercise where \( P(x) = \frac{1}{x+2} \), we derive the integrating factor \( \mu(x) = e^{\ln|x+2|} = |x+2| \). This application allows you to rewrite and solve the equation in a manner that reveals the general solution:

• \( y = \frac{1}{|x+2|}(\frac{x^3}{3} + x^2 + C) \)

where \( C \) is an arbitrary constant representing an infinite set of solutions.

Linear Differential Equations

Linear differential equations are characterized by solutions that can be expressed as a linear combination of functions. They are immensely significant due to their wide applicability in physics, engineering, and other sciences.
A linear differential equation can be recognized by its standard form,

• \( \frac{dy}{dx} + P(x)y = Q(x) \)

where the function \( y \) and its derivative appear linearly. A crucial aspect of solving these equations is leveraging integrating factors, which convert them into integrable forms. This is what makes these equations simpler and more straightforward to solve.
Integrating factors, as seen in our exercise, are not only pivotal for solution strategies but also for understanding the behavior of physical systems modeled by these equations. They help in determining the complete and general solution of an equation. In our scenario, the application of the integrating factor \( |x+2| \) allows breaking down the differential equation into an easily solvable form, bringing us to the general solution which describes the behavior of \( y \) over all possible values of \( x \).