Problem 24 - Advanced Engineering Mathematics

Question

From \(\frac{1}{y^{2}-1} d y=\frac{1}{x^{2}-1} d x\) or \(\frac{1}{2}\left(\frac{1}{y-1}-\frac{1}{y+1}\right) d y=\frac{1}{2}\left(\frac{1}{x-1}-\frac{1}{x+1}\right) d x\) we obtain \(\ln |y-1|-\ln |y+1|=\ln |x-1|-\ln |x+1|+\ln c\) or \(\frac{y-1}{y+1}=\frac{c(x-1)}{x+1} . \quad\) Using \(y(2)=2\) we find \(c=1 .\) A solution of the initial-value problem is \(\frac{y-1}{y+1}=\frac{x-1}{x+1}\) or \(y=x\).

Step-by-Step Solution

Answer

The solution is \( y = x \).

Step 1: Rewriting the Differential Equation

We begin by considering the differential equation \( \frac{1}{y^2-1} \, dy = \frac{1}{x^2-1} \, dx \), and its equivalent transformed expression \( \frac{1}{2}\left(\frac{1}{y-1}-\frac{1}{y+1}\right) dy = \frac{1}{2}\left(\frac{1}{x-1}-\frac{1}{x+1}\right) dx \). Both expressions are ways to represent the differential equation in separable form, suitable for integration.

Step 2: Integrating Each Side

Integrate both sides of the equation. For the left side, integrate \( \frac{1}{2}\left(\frac{1}{y-1}-\frac{1}{y+1}\right) \, dy \) to obtain \( \frac{1}{2}(\ln|y-1| - \ln|y+1|) \). Similarly, for the right side, integrate \( \frac{1}{2}\left(\frac{1}{x-1}-\frac{1}{x+1}\right) \, dx \) to obtain \( \frac{1}{2}(\ln|x-1| - \ln|x+1|) \).

Step 3: Equating the Integrated Sides

Set the integrated expressions equal to each other: \( \ln|y-1| - \ln|y+1| = \ln|x-1| - \ln|x+1| + \ln c \), where \( c \) is the constant of integration.

Step 4: Simplifying the Logarithmic Expression

Utilize logarithmic properties to simplify \( \ln|y-1| - \ln|y+1| \) and \( \ln|x-1| - \ln|x+1| \). This simplifies to \( \ln \left( \frac{y-1}{y+1} \right) = \ln \left( \frac{c(x-1)}{x+1} \right) \).

Step 5: Exponential Form

Remove the logarithms by taking exponentials of both sides: \( \frac{y-1}{y+1} = \frac{c(x-1)}{x+1} \).

Step 6: Finding the Constant of Integration \(c\)

Use the initial condition \( y(2) = 2 \) to find \( c \). Substitute \( x = 2 \) and \( y = 2 \) into the equation to get \( \frac{2-1}{2+1} = \frac{c(2-1)}{2+1} \), which simplifies to \( \frac{1}{3} = \frac{c}{3} \). Solving for \( c \), we find \( c = 1 \).

Step 7: Final Solution

With \( c = 1 \), our differential equation simplifies to \( \frac{y-1}{y+1} = \frac{x-1}{x+1} \). By cross-multiplying and simplifying, it becomes \( y = x \). Thus, the solution to the initial-value problem is \( y = x \).

Key Concepts

Differential Equations

Differential equations are mathematical equations that relate a function with its derivatives. These equations can describe various phenomena such as motion, growth, and decay. Essentially, they show how a particular quantity changes when related variables are altered. In the context of separation of variables, a type of differential equation, the goal is to rearrange the equation so that all terms involving one variable are on one side, and all terms involving another variable are on the opposite side. This method simplifies complex relationships into more manageable parts.

• Allows the separation of differential equations into simpler, solvable parts.
• Useful in practical problems including physics, engineering, and biology.
• Often require techniques such as integration to resolve.

Separation of variables is foundational because it provides a straightforward path to finding solutions by isolating variables and making direct integrations.

Integration

Integration is a fundamental concept in calculus, often seen as the reverse process of differentiation. While differentiation focuses on rates of change, integration focuses on the accumulation of quantities. In solving differential equations, integration helps to find the antiderivative or the original function before it was differentiated.

• Essential for solving separable differential equations.
• Involves finding functions whose derivative is the integrand.
• Integration constant, often denoted as 'c', accounts for all possible antiderivatives.

In this exercise, the integration of each side of the differential equation involves finding functions of logarithmic form that can later be equated to reach a common solution.

Initial Value Problem

An initial value problem is a type of differential equation that provides specific conditions for the solution to satisfy at a particular point. This is crucial because it ensures a unique solution rather than a family of solutions.

• Specifies conditions at the outset to isolate a single solution.
• Initial conditions such as 'y(2) = 2' are used to determine constants.
• Used widely in scenarios requiring precise results like climate models, population predictions, etc.

By using an initial condition, the solution becomes tailored to meet the requirements of the situation described. In our exercise, knowing that when "x = 2, y = 2" enabled finding the constant of integration 'c' accurately.

Logarithmic Properties

Logarithmic properties are important in simplifying complex equations. They transform multiplicative relationships into additive ones, which can often be easier to solve. This is particularly helpful in solving differential equations after integration.

• Key properties include: \( \ln(ab) = \ln(a) + \ln(b) \) and \( \ln\left(\frac{a}{b}\right) = \ln(a) - \ln(b) \).
• Simplify expressions by using properties of logarithms.
• Allow exponentiation as a means to remove logarithms and solve equations.

In the exercise, logarithmic properties combined with separation of variables assisted in reaching a form that could be more directly manipulated to find a solution like converting \( \ln \left( \frac{y-1}{y+1} \right) = \ln \left( \frac{c(x-1)}{x+1} \right) \) into an equation without logarithms.