91Ó°ÊÓ

When dealing with differential equations, finding the right substitution is key to simplifying complex expressions. For this particular problem, we begin by identifying a substitution that can help us make the equation more manageable. In our exercise, the substitution used is \( y = vx \), where \( v \) is a function of \( x \). This choice is inspired by the presence of \( y \) and \( yx \) in the original equation, suggesting that expressing \( y \) in terms of \( x \) could streamline our equation.

By substituting \( y = vx \), we automatically introduce a new variable that relates \( y \) to \( x \). To replace \( dy \), we differentiate \( y = vx \) to get \( dy = v \, dx + x \, dv \). These cleverly chosen substitutions transform the original differential equation into a more familiar form, allowing for further manipulation and simplification.

Choosing an appropriate substitution is all about recognizing patterns and being able to deduce which relationship might simplify the terms involved. It's a critical step that can determine how easily you can solve a differential equation.

Separation of variables is a powerful technique used in solving differential equations that allows us to separate the variables on different sides of the equation. In the given problem, after substituting and simplifying, we reached a point where the equation becomes \( x \, dv = -(v^2 + 2v) \, dx \). The next step is to isolate terms involving \( v \) on one side and terms involving \( x \) on the other.

This separation results in \( \frac{dv}{v^2 + 2v} = -\frac{dx}{x} \). Notice how we've managed to isolate the differentials such that each variable is on opposite sides, facilitating straightforward integration.

Why is this useful? It transforms the problem into a task of integrating two simpler equations. The integration process for each side further untangles the equation into a form where its solution is more apparent. Remember, separation of variables is applicable only when it's possible to algebraically separate the different terms like this, allowing for independent integration.

Partial fraction decomposition is a technique used to simplify complex rational expressions, primarily useful when integrating, which is exactly what we encounter in this problem. When we separated the variables, we got an integral \( \int \frac{dv}{v(v+2)} \). To resolve this integral, partial fraction decomposition helps us break \( \frac{1}{v(v+2)} \) into simpler fractions that can be integrated individually.

By expressing \( \frac{1}{v(v+2)} \) as \( \frac{A}{v} + \frac{B}{v+2} \) and solving for constants \( A \) and \( B \), we found that \( A = \frac{1}{2}, B = -\frac{1}{2} \). This allows us to write the integral as two separate integrals: \( \int \frac{1/2}{v} \, dv - \int \frac{1/2}{v+2} \, dv \). Each of these integrals is a basic logarithmic function, which is much simpler to compute.

Partial fraction decomposition is essential in the toolkit for solving differential equations, especially when dealing with expressions that can be broken down into simpler components. It makes the process of integration straightforward and more approachable.