Problem 24 - Advanced Engineering Mathematics

Question

Find the general solution of the given differential equation. Give the largest interval over which the general solution is defined. Determine whether there are any transient terms in the general solution. $$ \left(x^{2}-1\right) \frac{d y}{d x}+2 y=(x+1)^{2} $$

Step-by-Step Solution

Answer

The general solution is \( y = \left(\frac{x^2}{2} + x + C \right)\frac{|x+1|}{|x-1|} \). Largest interval: \((-\infty, -1) \cup (-1, 1) \cup (1, \infty)\). Transient term: possible if \(C\) specific.

Step 1: Identify the Type of Differential Equation

The given differential equation is \((x^2 - 1)\frac{dy}{dx} + 2y = (x+1)^2\). It's a first-order linear differential equation of the form \(P(x)\frac{dy}{dx} + Q(x)y = R(x)\), where \(P(x) = x^2 - 1\), \(Q(x) = 2\), and \(R(x) = (x+1)^2\).

Step 2: Transform to Standard Form

We divide the entire equation by \(P(x) = x^2 - 1\) to convert it to the standard form. This gives:\[\frac{dy}{dx} + \frac{2}{x^2 - 1}y = \frac{(x+1)^2}{x^2 - 1}.\]

Step 3: Find Integrating Factor

To solve the standard linear differential equation, we need an integrating factor \(\mu(x)\), given by:\[\mu(x) = e^{\int \frac{2}{x^2 - 1} \, dx}.\]Perform partial fraction decomposition for \(\frac{2}{x^2 - 1}\) as \(\frac{1}{x-1} - \frac{1}{x+1}\) and integrate to get:\[\mu(x) = e^{\ln|x-1| - \ln|x+1|} = \frac{|x-1|}{|x+1|}.\]

Step 4: Solve the Differential Equation

Multiply the standard form by the integrating factor \(\mu(x)\), yielding:\[\frac{|x-1|}{|x+1|} \frac{dy}{dx} + \frac{2|x-1|}{(x^2 - 1)|x+1|}y = \frac{|x-1|}{(x^2 - 1)|x+1|}(x+1)^2.\]Recognizing the left side as a derivative, it is:\[\frac{d}{dx}\left(\frac{|x-1|}{|x+1|}y\right) = (x+1).\]Integrating both sides:\[\frac{|x-1|}{|x+1|}y = \int (x+1) \, dx = \frac{x^2}{2} + x + C,\]where \(C\) is the constant of integration.

Step 5: Solve for y

Solve the equation for \(y\):\[y = \left(\frac{x^2}{2} + x + C\right)\frac{|x+1|}{|x-1|}.\]This is the general solution of the differential equation.

Step 6: Determine Interval of the Solution

Since \(P(x) = x^2 - 1 = 0\) at \(x = \pm 1\), these points are not in the domain of the solution since it would make the denominator zero. The largest interval where the solution is defined and not zero at the denominator are \((-\infty, -1) \cup (-1, 1) \cup (1, \infty)\).

Step 7: Identify Transient Terms

In the solution \(y = \left(\frac{x^2}{2} + x + C\right)\frac{|x+1|}{|x-1|}\), transient terms may appear if the terms tend to zero as \(x\) approaches infinity. Here, \(\frac{x^2}{2} + x\) and \(\frac{|x+1|}{|x-1|}\) do not tend to zero or infinity. However, for specific \(C\), the term \(C\frac{|x+1|}{|x-1|}\) can contribute to a transient term.

Key Concepts

First-Order Linear Differential Equation

A first-order linear differential equation is one of the simplest types of differential equations and is expressed in the form: \[ A(x)\frac{dy}{dx} + B(x)y = C(x) \] where \( A(x) \), \( B(x) \), and \( C(x) \) are functions of \( x \). This equation involves only the first derivative of \( y \), making it 'first-order'.
The main goal with these equations is to find the function \( y \) that satisfies the given equation. The equation in our example: \[ (x^2 - 1)\frac{dy}{dx} + 2y = (x+1)^2 \] has a structure that matches the first-order linear differential equation form. By identifying \( P(x) = x^2 - 1 \), \( Q(x) = 2 \), and \( R(x) = (x+1)^2 \), we acknowledge it is a first-order linear differential equation, which sets the stage for solving it using specific techniques.
Understanding this form is crucial because it allows us to use methods like the Integrating Factor Method to find solutions.

Integrating Factor Method

The Integrating Factor Method is a powerful technique used to solve first-order linear differential equations. Once an equation is in its standard form: \[ \frac{dy}{dx} + p(x)y = q(x) \] where \( p(x) = \frac{B(x)}{A(x)} \) and \( q(x) = \frac{C(x)}{A(x)} \), the next step involves finding an integrating factor, \( \mu(x) \).
The integrating factor \( \mu(x) \) is calculated using: \[ \mu(x) = e^{\int p(x) \, dx} \] This factor helps transform the differential equation into an easily integrable form.
In our differential equation, after dividing by \( P(x) = x^2 - 1 \), we get: \[ \frac{dy}{dx} + \frac{2}{x^2 - 1}y = \frac{(x+1)^2}{x^2 - 1} \] Here, \( p(x) = \frac{2}{x^2 - 1} \). By performing partial fraction decomposition and integrating, we find: \[ \mu(x) = \frac{|x-1|}{|x+1|} \] Applying this integrating factor simplifies solving the equation. It helps combine the terms into a single derivative expression, making integration straightforward.

General Solution of Differential Equations

The general solution of a differential equation represents a family of solutions. It incorporates a constant of integration \( C \) which can vary.
For the given first-order linear differential equation, using the integrating factor method resulted in the general solution: \[ y = \left(\frac{x^2}{2} + x + C\right)\frac{|x+1|}{|x-1|} \] This expression includes \( C \), indicating infinitely many solutions depending on its value.
Understanding the general solution is crucial as it provides a complete set of possible functions \( y \) that satisfy the original differential equation for any constant \( C \). By assigning specific values to \( C \), you determine particular solutions, which can be applicable to initial conditions if provided.

Interval of Definition

In mathematics, the interval of definition describes the range of values of \( x \) over which the solution is valid.
For this differential equation, an integral step is identifying where the solution can exist without causing any undefined operations, such as division by zero.
In the process, \( P(x) = x^2 - 1 \) determines critical points where the denominator becomes zero—specifically at \( x = 1 \) and \( x = -1 \). Therefore, these points must be excluded from our interval of definition, as the solution becomes undefined there.
Thus, the largest interval over which the solution maintains validity is:

• \(( -\infty, -1)\)
• \((-1, 1)\)
• \((1, \infty)\)

Transient Terms in Differential Equations

Transient terms in a differential equation are temporary components of the solution that may diminish over time or as \( x \) approaches infinity.
In our example, the general solution is: \[ y = \left(\frac{x^2}{2} + x + C\right)\frac{|x+1|}{|x-1|} \] This involves constant \( C \), which allows variations in the solution’s behavior. However, since neither \( \frac{x^2}{2} + x \) nor \( \frac{|x+1|}{|x-1|} \) tends to zero or infinity as \( x \) approaches infinity, these terms themselves are not considered transient.
Transient effects appear if specific values of \( C \) cause parts of the solution to diminish over time or distance. Here, \( C\frac{|x+1|}{|x-1|} \) is particularly sensitive and may act as a transient term for certain \( C \). Therefore, if a transient term is present, it appears due to particular choices of \( C \) and can reflect transient phenomena in application-specific scenarios.