Problem 20 - Advanced Engineering Mathematics

Question

In Problems 1-22, solve the given differential equation by separation of variables. $$ \frac{d y}{d x}=\frac{x y+2 y-x-2}{x y-3 y+x-3} $$

Step-by-Step Solution

Answer

The implicit solution is \( y-1 + 2 \ln|y-1| = x-3 + 5 \ln|x-3| + C \).

Step 1: Simplify the Equation

First, we simplify the initial differential equation \( \frac{d y}{d x} = \frac{x y + 2 y - x - 2}{x y - 3 y + x - 3} \). Notice that each term in the numerator and denominator could be grouped and factored. Let's factor both: the numerator can be written as \( y(x + 2) - (x + 2) = (y-1)(x+2) \) and the denominator as \( y(x - 3) + (x - 3) = (y+1)(x-3) \). Thus, the equation becomes \( \frac{d y}{d x} = \frac{(y-1)(x+2)}{(y+1)(x-3)} \).

Step 2: Separate Variables

Now, we rearrange the equation to separate variables. This involves setting the equation so that all \( y \) terms are on one side (with \( dy \) and \( dx \) on opposite sides). From \( \frac{d y}{d x} = \frac{(y-1)(x+2)}{(y+1)(x-3)} \), rearrange to get \( \frac{y+1}{y-1} dy = \frac{x+2}{x-3} dx \).

Step 3: Integrate Both Sides

Integrate both sides separately. The left side is \( \int \frac{y+1}{y-1} \, dy \), and the right side is \( \int \frac{x+2}{x-3} \, dx \). Use the substitution method to integrate these. For the left side, let \( u=y-1 \), then \( du=dy \), making it \( \int\left(1 + \frac{2}{u}\right) du \). For the right side, let \( v=x-3 \), then \( dv=dx \), making it \( \int\left(1 + \frac{5}{v}\right) dv \).

Step 4: Evaluate the Integrals

Evaluate both integrals. The left integral becomes \( u+2 \ln|u| = y-1 + 2 \ln|y-1| \). The right integral becomes \( v+5 \ln|v| = x-3 + 5 \ln|x-3| \). Thus, after integration, we have \( y-1 + 2 \ln|y-1| = x-3 + 5 \ln|x-3| + C \).

Step 5: Solve for \( y \)

To solve for \( y \), we leave it implicit due to the complexity arising from the logarithmic terms. The relationship is \( y-1 + 2 \ln|y-1| = x-3 + 5 \ln|x-3| + C \). This is the solution; \( y \) can be expressed explicitly with further simplification or numeric approximation if required.

Key Concepts

Separation of Variables

Separation of variables is a handy technique used to solve first-order differential equations. It involves rearranging the equation so that all terms involving one variable end up on one side of the equation, and the terms involving the other variable end up on the opposite side. This allows us to integrate each side separately.
Here's a simple breakdown of how it works:

• Identify the equation in the form where one side involves only the variables you want to separate, typically expressed like \( \frac{dy}{dx} \).
• Rearrange the terms so that all \( y \) terms (along with \( dy \)) are on one side of the equation and all \( x \) terms (with \( dx \)) are on the other side. This results in an expression like \( f(y) \, dy = g(x) \, dx \).
• Once separated, you can integrate both sides independently.

The primary goal is to transform a differential equation that may look complex into two simpler integration problems. It's important to perform each step carefully to ensure correct separation.
This technique, though straightforward, forms the backbone of many more complex solutions in calculus.

Integration Techniques

Integration is the fundamental skill needed to solve separated differential equations. Sometimes, direct integration works, but in many cases, we need special techniques to solve different integrals.
For variable separable equations, some useful integration methods include:

• Basic Integration: If the integral directly matches a basic formula like \( \int x^n \, dx \), it can be solved directly without extra steps.
• Substitution: When the integral is composite, use a substitution to simplify it. For example, if the integral involves \( (y - 1) \), substitute \( y - 1 \) as \( u \), making it easier to solve.
• Partial Fractions: Break down complex rational expressions into simpler fractions that can be integrated easily.
• Integration by Parts: Useful when an integral is a product of functions that need decomposition, but less common in variable separation problems.

Understanding which technique to use and how to apply it effectively is crucial when dealing with integration in calculus, especially when handling non-linear differential equations. Each method has its rules, so practicing with various integrals and problems is key to mastering them.

Implicit Solutions

In many cases, after integrating a separated differential equation, the solution comes out in an implicit form. Implicit solutions mean the relationship between the dependent and independent variable is expressed without explicitly solving for one variable in terms of the other.
Here's why an implicit solution might be used:

• Complex Equations: Sometimes, after integration, the resulting equation might be too complex to solve for a variable explicitly, especially if it involves logarithmic or exponentional terms.
• Logarithms and Exponentials: These often make it difficult or impossible to isolate one variable on one side of the equation purely symbolically.
• Easier Representation: Implicit forms often provide a more concise way to understand the relationship between variables without additional transformations.

While it is preferred to find explicit solutions, implicit solutions still fully describe the behavior of variables in the equation. In practice, numerical methods can be used to analyze these solutions further, or in some cases, simplifications can be made to reach an explicit form. Implicit solutions remain highly valid and useful if they complete the goal of explaining a system's behavior.