Problem 20 - Advanced Engineering Mathematics

Question

From \(\frac{y+1}{y-1} d y=\frac{x+2}{x-3} d x\) or \(\left(1+\frac{2}{y-1}\right) d y=\left(1+\frac{5}{x-3}\right) d x\) we obtain \(y+2 \ln |y-1|=x+5 \ln |x-3|+c\) or \(\frac{(y-1)^{2}}{(x-3)^{5}}=c_{1} e^{x-y}\).

Step-by-Step Solution

Answer

The solution is obtained as either \( y + 2 \ln|y-1| = x + 5 \ln|x-3| + C \) or \( \frac{(y-1)^2}{(x-3)^5} = C_1 e^{x-y} \).

Step 1: Analyze the Given Equation

We begin with the differential equation \( \frac{y+1}{y-1} dy = \frac{x+2}{x-3} dx \). This implies we have separate terms for \( y \) and \( x \) on each side.

Step 2: Integrate Both Sides

We integrate each side separately. The left side, \( \frac{y+1}{y-1} \), can be simplified and integrated as \( \int \left(1 + \frac{2}{y-1}\right) dy\). The right side, \( \frac{x+2}{x-3} \), can be integrated as \( \int \left(1 + \frac{5}{x-3}\right) dx\).

Step 3: Simplify the Integrals

For the left side, \( \int \left(1 + \frac{2}{y-1}\right) dy = y + 2\ln|y-1| + C_1 \). For the right side, \( \int \left(1 + \frac{5}{x-3}\right) dx = x + 5\ln|x-3| + C_2 \).

Step 4: Compare the Solutions

Set the results from both sides equal to each other: \( y + 2 \ln|y-1| = x + 5 \ln|x-3| + C \) where \( C = C_2 - C_1 \).

Step 5: Convert to Exponential Form

Using the exponential function, rearrange and manipulate terms to achieve the form \( \frac{(y-1)^2}{(x-3)^5} = C_1 e^{x-y} \), keeping consistency with the initial transformations and integrations.

Key Concepts

Integration Techniques

Integration is a crucial mathematical tool used to solve differential equations like the one given in the exercise. When you are tasked with integrating both sides of an equation, as seen in the solution, it means performing the integral operation on each side separately.

To work with the given equation, \[\int \left(1 + \frac{2}{y-1}\right) dy = \int \left(1 + \frac{5}{x-3}\right) dx,\]we break it into smaller parts that are easier to integrate. Let's focus on these parts:

• The term \(1\) integrates to simply \(y\) or \(x\), depending on the side of the equation.
• The term \(\frac{2}{y-1}\) integrates to \(2\ln|y-1|\).
• Similarly, the term \(\frac{5}{x-3}\) integrates to \(5\ln|x-3|\).

By adding these individual results, you understand how to simplify the integrals and derive the combined form like in the equation \(y + 2\ln|y-1| = x + 5\ln|x-3| + C\). Regular practice with integration techniques can greatly improve your skills in solving similar problems.

Separable Equations

A separable differential equation is one in which the variables can be separated onto different sides of the equation, allowing you to integrate each variable with respect to its own axis. In the given exercise, this principle is clearly shown in\(\frac{y+1}{y-1} dy = \frac{x+2}{x-3} dx.\)

The main goal with separable equations is to rewrite the equation so that all terms involving \(y\) are on one side and all terms involving \(x\) are on the other. By doing this, you transform the equation into a form that is ready for direct integration:

• Rearrange terms to isolate one variable on each side of the equation.
• Carefully apply integration to both sides separately, integrating the terms with respect to their own variable.

Understanding separable equations is crucial because it simplifies many complex differential equations to forms that are more manageable. It is a foundational technique that aids in solving a wide array of problems effectively.

Initial Value Problems

An initial value problem (IVP) involves a differential equation along with specified values for the variables at a particular point. This helps in finding unique solutions to differential equations.

While the exercise provided did not explicitly present an initial value problem, the concepts used, like adding a constant of integration \(C\), align with solving such problems. Here's how it connects:

• After integrating both sides of a differential equation, you obtain a general solution involving an unknown constant, \(C\).
• An additional condition, such as a specific value of \(y\) at a certain \(x\), can be used to solve for this constant, thereby yielding a unique solution.

By learning initial value problems, you gain the ability to solve differential equations completely. This skill is essential when dealing with real-world scenarios where conditions at specific points dictate the solutions to mathematical models.