Problem 6 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

Solve the given differential equation. \(2 y^{\prime}-y=1\)

Step-by-Step Solution

Answer

The solution to the given differential equation is \(y = e^{-0.5x}+\frac{C}{e^{0.5x}}\), where \(C\) is a constant.

Step 1: Writing The Equation In Standard Form

First, divide through by the coefficient of \(y^\prime\) to write the equation in standard form. For the given equation \(2y^\prime - y = 1\), divide every term by 2 to give: \(y^\prime - 0.5y = 0.5\).

Step 2: Identify p(x) and g(x)

In the standard form \(y^\prime + p(x)y = g(x)\), identify \(p(x)\) and \(g(x)\) in our equation. We have \(p(x) = -0.5\) and \(g(x) = 0.5\).

Step 3: Calculation of Integrating Factor

An integrating factor is an expression we multiply through the linear differential equation to allow it to be solved using direct integration. It is calculated as \(e^{\int p(x) \, dx}\). For our equation, that would be \(e^{\int -0.5 \, dx}\) which simplifies to \(e^{-0.5x}\). To make calculations easier, we can use the equivalent form \(e^{0.5x}\).

Step 4: Multiply Through by the Integrating Factor

Multiply every term in our equation \(y^\prime - 0.5y = 0.5\) by the integrating factor, \(e^{0.5x}\). Our equation becomes \(e^{0.5x}y^\prime - 0.5e^{0.5x}y = 0.5e^{0.5x}\).

Step 5: Integration

Now we integrate the equation. The left side of the equation integrates to \(e^{0.5x}y\) and integrating the right side with respect to \(x\) gives \(e^{0.5x}\).

Step 6: Solving for y

Finally, we solve for \(y\) by dividing through by \(e^{0.5x}\) to give \(y = e^{-0.5x}+\frac{C}{e^{0.5x}}\), where \(C\) is the constant of integration.

Key Concepts

Integrating Factor

One of the most powerful tools for solving linear first order differential equations is the integrating factor. It simplifies the process by transforming the differential equation into a simpler form that can be easily integrated. Essentially, the integrating factor is a specific function—often written as \( \mu(x) = e^{\int p(x) \, dx} \)—which, when multiplied through the entire differential equation, turns it into a form that is easily solvable. In our example, the integrating factor was derived as \( e^{0.5x} \). This factor is vital for allowing us to simplify and integrate the differential equation efficiently.

• Purpose: Turns the differential equation into a product of a derivative, which is easier to integrate.
• Calculation: Determined by the function \( p(x) \) from the standard form of the equation.
• Simplification: Multiplying through the differential equation aligns terms and allows direct integration.

By using the integrating factor, we can move forward with solving the differential equation more effectively and efficiently.

Linear First Order Differential Equations

Linear first order differential equations are equations that exhibit a specific structure, making them amenable to particular solving techniques like the integrating factor method. These equations can be written in the form \( y' + p(x)y = g(x) \), where \( y' \) is the derivative of the function \( y \) with respect to \( x \), and \( p(x) \) and \( g(x) \) are continuous functions.The key characteristics of these equations include:

• Linearity: The variable \( y \) and its first derivative \( y' \) appear to power of 1.
• Standard Form: Easily structured into a format that highlights \( p(x) \) and \( g(x) \).
• Solving Techniques: Solutions are often attainable using the integrating factor method.

In our example, the equation \( 2y' - y = 1 \) was transformed into standard form by dividing by the coefficient of \( y' \), resulting in \( y' - 0.5y = 0.5 \). This transformation is crucial as it allows for straightforward application of the integrating factor method.

Solving Differential Equations

Solving differential equations, particularly linear first order ones, involves several important steps to systematically reach a solution. Next, we'll breakdown these steps.### Process Overview

1. Standard Form: Begin by writing the equation in standard form \( y' + p(x)y = g(x) \).
2. Integrating Factor: Calculate the integrating factor \( \mu(x) = e^{\int p(x) \, dx} \).
3. Multiply and Integrate: Multiply the equation by the integrating factor and integrate both sides.
4. Solve for \( y \): Finally, solve for \( y \) to find the particular solution \( y = e^{-0.5x} + \frac{C}{e^{0.5x}} \), where \( C \) is the constant of integration.

The integration step is key as it yields a straightforward solution great for handling initial value problems. Solving differential equations efficiently involves combining these techniques into a cohesive problem-solving strategy.