Problem 31 - Advanced Engineering Mathematics

Question

$$ x y^{\prime \prime}+2 y^{\prime}=0 $$

Step-by-Step Solution

Answer

The general solution is \( y(x) = -\frac{C_1}{x} + C_2 \).

Step 1: Identify the Type of Differential Equation

The given equation is a second-order linear homogeneous ordinary differential equation of the form \( x y'' + 2 y' = 0 \). Notice that it is not with constant coefficients as there is a variable \( x \) in front of \( y'' \).

Step 2: Simplify the Differential Equation

First, let's divide through by \( x \) to simplify the equation: \[ y'' + \frac{2}{x} y' = 0 \]. This form is easier to recognize in terms of a standard, simpler form.

Step 3: Use a Substitution Method

Let's set \( v = y' \); then, \( v' = y'' \). Substituting these into the equation gives us: \[ v' + \frac{2}{x} v = 0 \]. This is now a first-order linear differential equation in \( v \).

Step 4: Recognize and Solve the First-Order Linear Equation

The equation \( v' + \frac{2}{x} v = 0 \) can be solved using the standard integrating factor method. The integrating factor \( \mu(x) \) is given by \( e^{\int \frac{2}{x} \, dx} = x^2 \).

Step 5: Solve for \( v(x) \)

Multiply both sides of the differential equation for \( v \) by the integrating factor \( x^2 \): \[ x^2 v' + 2x v = 0 \]. This simplifies to \( (x^2 v)' = 0 \). Integrate to find \( x^2 v = C_1 \). Hence, \( v = \frac{C_1}{x^2} \).

Step 6: Integrate to Find \( y(x) \)

Since \( v = y' \), we have \( y' = \frac{C_1}{x^2} \). Integrating with respect to \( x \): \( y = \int \frac{C_1}{x^2} \, dx = -\frac{C_1}{x} + C_2 \). This gives the general solution.

Step 7: Conclusion: Write the General Solution

The general solution to the differential equation \( xy'' + 2y' = 0 \) is \( y(x) = -\frac{C_1}{x} + C_2 \), where \( C_1 \) and \( C_2 \) are constants determined by boundary conditions if given.

Key Concepts

Second-Order Differential Equation

A second-order differential equation involves the second derivative of a function. In the equation \( x y'' + 2 y' = 0 \), the term \( y'' \) signifies that it is a second-order equation. Such equations can describe various phenomena such as motion dynamics and electrical circuits. Typically, they come in the form \( a(x) y'' + b(x) y' + c(x) y = g(x) \). In this exercise, it's homogeneous because there is no separate function of \( x \) (i.e., \( g(x) = 0 \)).To solve such an equation, we often reduce it to a simpler form, making it easier to handle. This specific type is nonconstant because of the variable coefficient \( x \), differing from the typical constant coefficient equations.

First-Order Linear Differential Equation

After simplifying the second-order equation, we derive a first-order linear differential equation for the function \( v \), where \( v = y' \). A first-order linear differential equation generally appears as \( v' + p(x) v = q(x) \). In this exercise, our simplified version is \( v' + \frac{2}{x} v = 0 \). First-order linear differential equations are easier to solve because they require fewer steps and tend to have fewer complexities compared to second-order equations. Key to their solution is the method of the integrating factor.

Integrating Factor

An integrating factor is used to simplify the process of solving first-order linear differential equations. It's found using the function \( \mu(x) = e^{\int p(x) \, dx} \). Its purpose is to transform a differential equation into an easily integrable form. For our exercise, the integrating factor \( \mu(x) \) was calculated as \( x^2 \) from \( p(x) = \frac{2}{x} \), transforming \( v' + \frac{2}{x}v = 0 \) into \( (x^2 v)' = 0 \). This new equation allows a straightforward integration, ultimately leading to \( v = \frac{C_1}{x^2} \). The integrating factor method is a powerful tool in tackling linear equations, particularly when coefficients depend on \( x \).

General Solution

The general solution provides a comprehensive solution encompassing all particular solutions of a differential equation. Once we determined \( v = \frac{C_1}{x^2} \), we integrated \( y' = v \) to find \( y(x) = -\frac{C_1}{x} + C_2 \). This expression encompasses the general solution to the original differential equation.The constants \( C_1 \) and \( C_2 \) indicate that without further conditions, solutions could vary infinitely. To find a particular solution, additional information such as initial conditions or boundary conditions would be necessary.