Problem 11 - Advanced Engineering Mathematics

Question

For \(y^{\prime}+\frac{4}{x} y=x^{2}-1\) an integrating factor is \(e^{\int(4 / x) d x}=x^{4}\) so that \(\frac{d}{d x}\left[x^{4} y\right]=x^{6}-x^{4}\) and \(y=\frac{1}{7} x^{3}-\frac{1}{5} x+c x^{-4}\) for \(0< x <\infty\).

Step-by-Step Solution

Answer

The solution is \(y = \frac{1}{7} x^3 - \frac{1}{5} x + C x^{-4}\), valid for \(0 < x < \infty\).

Step 1: Identify the type of differential equation

The given differential equation \(y^{\prime}+\frac{4}{x} y=x^{2}-1\) is a first-order linear ordinary differential equation. The standard form of such an equation is \(y' + P(x)y = Q(x)\), where \(P(x) = \frac{4}{x}\) and \(Q(x) = x^2 - 1\).

Step 2: Determine the integrating factor

The integrating factor \(\mu(x)\) is found by computing \(e^{\int P(x) \, dx}\). In this case, \(P(x) = \frac{4}{x}\), so the integrating factor is \(e^{\int \frac{4}{x} \, dx} = e^{4 \ln x} = x^4\).

Step 3: Multiply by the integrating factor

Multiply the entire differential equation by the integrating factor \(x^4\) to simplify it. The equation becomes \(x^4 y' + 4x^3 y = x^6 - x^4\).

Step 4: Recognize simplification through the integrating factor

Notice the left-hand side of the equation is the derivative of \(x^4 y\) with respect to \(x\), i.e., \(\frac{d}{dx}(x^4 y)\). Thus, the equation can be rewritten as: \[\frac{d}{dx}(x^4 y) = x^6 - x^4.\]

Step 5: Integrate both sides

To find \(x^4 y\), integrate both sides with respect to \(x\). The integration gives: \[x^4 y = \int (x^6 - x^4) \, dx = \frac{1}{7} x^7 - \frac{1}{5} x^5 + C,\] where \(C\) is the constant of integration.

Step 6: Solve for \(y\)

Divide both sides of the integrated equation by \(x^4\) to solve for \(y\). This results in: \[y = \frac{1}{7} x^3 - \frac{1}{5} x + C x^{-4}.\]

Step 7: State the solution condition

The solution \(y = \frac{1}{7} x^3 - \frac{1}{5} x + C x^{-4}\) is valid for the domain \(0 < x < \infty\).

Key Concepts

First-order Linear Differential Equations

First-order linear differential equations are a fundamental part of differential equations, often appearing in various scientific and engineering problems. They are called 'first-order' because they involve the first derivative of the unknown function. In general, these equations have the standard form:

• \( y' + P(x)y = Q(x) \)

Here, \( y' \) is the derivative of \( y \) with respect to \( x \), \( P(x) \) and \( Q(x) \) are functions of \( x \). The goal is usually to solve for \( y(x) \), which represents the dependent variable.
When we talk about making equations easier to solve, first-order linear differential equations offer a systematic approach. The structure of these equations allows us to use specific methods, such as the integrating factor method, to find solutions effectively. This can convert a seemingly complex equation into a more manageable form through multiplication with a chosen integrating factor.

Integrating Factor Method

The integrating factor method is a technique designed to solve first-order linear differential equations. It works by transforming the equation into a form that is easier to integrate directly. This method involves a special function called the 'integrating factor', typically denoted as \( \mu(x) \).

• The integrating factor is calculated as \( e^{\int P(x) \, dx} \).
• For the equation \( y' + \frac{4}{x}y = x^2 - 1 \), the integrating factor becomes \( x^4 \), as computed by integrating \( \frac{4}{x} \).

Once the integrating factor is found, it is multiplied through the entire differential equation. This simplifies the equation into a perfect derivative on one side, which can then be integrated easily to find the solution.
Using the integrating factor method is beneficial as it converts non-exact equations into exact ones, making the solution process much simpler. It's a straightforward method that provides practitioners with a clear path to solving these types of differential equations.

Ordinary Differential Equations

Ordinary differential equations (ODEs) are equations that involve functions of only one variable and their derivatives. They stand in contrast to partial differential equations, which involve multiple variables. In ODEs, each derivative represents the rate of change of the dependent variable with respect to the independent variable.

• The focus in ODEs is on the relationship between variables and how one changes with respect to another.
• They can be of various orders, with first-order being the simplest form.

In real-world applications, ODEs are used to model phenomena where the rate of change is a function of the current state. Examples include population dynamics, spring-mass systems, and electrical circuits.
Ordinary differential equations are a central tool in mathematical modeling, contributing to a deeper understanding of dynamic systems and processes. By solving these equations, we can predict the behavior of systems over time, providing vital insights across various fields of study.