Problem 5 - Calculus for Biology and Medicine

Question

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

Step-by-Step Solution

Answer

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

Step 1: Identify the Standard Form

The differential equation given is \( \frac{dy}{dx} + \frac{y}{x+2} = x - 1 \). We need to rewrite this equation in the standard linear form \( \frac{dy}{dx} + P(x)y = Q(x) \). It is already in this form with \( P(x) = \frac{1}{x+2} \) and \( Q(x) = x-1 \).

Step 2: Calculate the Integrating Factor

The integrating factor \( \mu(x) \) is calculated using the formula \( \mu(x) = e^{\int P(x) \, dx} \). Here, \( P(x) = \frac{1}{x+2} \), so we compute the integral: \( \int \frac{1}{x+2} \, dx = \ln|x+2| \). Thus, the integrating factor is \( \mu(x) = e^{\ln|x+2|} = |x+2| \).

Step 3: Rewrite the Differential Equation Using the Integrating Factor

Multiply the entire differential equation by the integrating factor \( |x+2| \):\[|x+2| \left( \frac{dy}{dx} + \frac{y}{x+2} \right) = |x+2|(x-1)\]Simplifying, the left-hand side becomes \( \frac{d}{dx}(|x+2|y) \).

Step 4: Integrate Both Sides

Now integrate both sides:\[\int \frac{d}{dx}(|x+2|y) \, dx = \int |x+2|(x-1) \, dx\]The left side yields \(|x+2|y\), and we need to solve the integral on the right. The integral expands to:\[\int (x^2 + x - 2) \, dx = \frac{x^3}{3} + \frac{x^2}{2} - 2x + C\]

Step 5: Solve for y

Equate both the integrated sides:\[|x+2|y = \frac{x^3}{3} + \frac{x^2}{2} - 2x + C\]To solve for \( y \), divide both sides by \(|x+2|\):\[y = \frac{1}{|x+2|} \left( \frac{x^3}{3} + \frac{x^2}{2} - 2x + C \right)\]This gives us the general solution for \( y \).

Key Concepts

Integrating Factor

The integrating factor is a powerful technique used to solve certain types of linear differential equations. It helps in making the equation simpler to solve via integration. An integrating factor is a function, typically denoted as \( \mu(x) \), constructed to simplify the integration process when handling linear first-order differential equations.

To find the integrating factor, we use the formula:

• \( \mu(x) = e^{\int P(x) \, dx} \)

where \( P(x) \) is the coefficient of \( y \) in the standard form of the differential equation. This formula helps in transforming the left-hand side of the equation into the derivative of a product, making it easier to integrate. In our example, the integrating factor came out to be \(|x+2|\), simplifying the process significantly.

Linear Differential Equation

A linear differential equation is an equation involving derivatives of a function that only has terms proportional to the function and its derivatives. These equations are essential in mathematics and are used to describe various physical systems.

Such equations can often be expressed in a standard form:

• \( \frac{dy}{dx} + P(x)y = Q(x) \)

where \( P(x) \) and \( Q(x) \) are functions of \( x \). The primary goal when solving these equations is to make them more amenable to integration, which often involves finding an integrating factor.

General Solution

Finding the general solution to a differential equation involves determining a function that encompasses all possible solutions. For linear differential equations, after applying an integrating factor, the solution generally exists in the form of:

• \( y = \frac{1}{\mu(x)} \, \left( \int \mu(x)Q(x) \, dx + C \right) \)

Here, \( C \) is an arbitrary constant that represents an infinite family of particular solutions due to the indefinite nature of integration. In the provided problem, once we figured out the integrating factor and integrated successfully, dividing by \( |x+2| \) allowed us to express \( y \) as the general solution.

Integration

Integration is a fundamental operation in calculus, being the inverse process of differentiation. It enables the finding of functions when their derivatives are known, among other applications.

In solving differential equations, once the equation is in a suitable form using an integrating factor, we perform integration on both sides:

• The left-hand side simplifies to a straightforward integral of a product of functions.
• The right-hand side requires careful integration, often involving polynomial terms or other expressions.

This step often yields an expression containing an arbitrary constant \( C \), indicative of the indefinite integral. The process of integrating in these problems is key as it unfolds the potential solutions needed to satisfy the differential equation.