Chapter 5: Problem 15
Solve the equation. $$ x y^{\prime}-y=x^{k} y^{n}, \text { where } n \neq 1 \text { and } k+n \neq 1. $$
Short Answer
Expert verified
The solution involves substitution to linearize and solve using the integrating factor.
Step by step solution
01
Recognize the Type of Equation
The given equation \( x y^{\prime} - y = x^{k} y^{n} \) is a first-order differential equation. It can be rearranged into separable form or solved using integrating factors, but here it is a Bernoulli differential equation because of its non-linear terms.
02
Transform into Standard Bernoulli Form
Initially, recognize the Bernoulli form from the equation: \( y' + P(x)y = Q(x)y^n \). Here, \( P(x) = -\frac{1}{x} \) and \( Q(x) = x^{k-1} \).
03
Use Substitution
To linearize it, use the substitution \( v = y^{1-n} \), which implies \( y = v^{\frac{1}{1-n}} \) and \( y' = \frac{1}{1-n} v^{\frac{n}{1-n}} v' \). Substitute into the original equation.
04
Substitute Derivatives
Substituting \( y' = \frac{1}{1-n} v^{\frac{n}{1-n}} v' \) into the transformed equation gives:\[ xv \left( \frac{1}{1-n} v^{\frac{n}{1-n}} v' \right) - v^{\frac{1}{1-n}} = x^{k} \left(v^{\frac{1}{1-n}}
ight)^{n} \]Simplify to make it linear after using the substitution.
05
Simplify the Equation
After substitution and simplification, arrive at a new linear differential equation in terms of \( v \):\[ xv' - (1-n)v = x^{k}(1-n) \]This is a first-order linear differential equation in terms of \( v \).
06
Solve the Linear Equation
This linear equation can be solved using an integrating factor. The integrating factor \( \mu(x) \) is given by \( e^{\int P(x) \, dx} \), where \( P(x) = -\frac{1-n}{x} \). Calculate the integrating factor and solve the linear equation for \( v \).
07
Back-substitute to Solve for \( y \)
Once \( v(x) \) is found, substitute back to find \( y \) using the relation \( v = y^{1-n} \). Solve for \( y \) to get the general solution of the original differential equation.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
First-Order Differential Equation
A first-order differential equation is a relation that involves the first derivative of a function. In our original exercise, the equation is presented as \( x y^{\prime} - y = x^{k} y^{n} \). This is first-order because the highest derivative present is the first derivative \( y' \).
Typically, such equations can be categorized as either linear or non-linear, depending on the nature of the terms involved. Linear first-order differential equations take the form \( y' + P(x)y = Q(x) \). However, when non-linear terms like \( y^n \) appear, the equation is not straightforwardly linear but still belongs to the first-order category due to the degree of the derivative.
Understanding the distinction between linear and non-linear is essential because it guides the choice of methods for solving the equation. For linear equations, integrating factors are often used, while non-linear equations might require different approaches, such as substitution or transformation techniques like with Bernoulli equations.
Typically, such equations can be categorized as either linear or non-linear, depending on the nature of the terms involved. Linear first-order differential equations take the form \( y' + P(x)y = Q(x) \). However, when non-linear terms like \( y^n \) appear, the equation is not straightforwardly linear but still belongs to the first-order category due to the degree of the derivative.
Understanding the distinction between linear and non-linear is essential because it guides the choice of methods for solving the equation. For linear equations, integrating factors are often used, while non-linear equations might require different approaches, such as substitution or transformation techniques like with Bernoulli equations.
Integrating Factor
The integrating factor is a crucial tool for solving linear differential equations. It is particularly useful when dealing with first-order linear equations which can be tricky to solve otherwise. The general form of such equations is \( y' + P(x)y = Q(x) \).
In our step-by-step solution, the integrating factor is utilized to solve the linear form of the transformed non-linear equation \( xv' - (1-n)v = x^{k}(1-n) \). To find the integrating factor \( \mu(x) \), we calculate \( e^{\int P(x) \, dx} \), where \( P(x) \) is the coefficient of \( y \) in the rearranged equation.
This mathematical tool allows us to transform a difficult-to-solve differential equation into one where separation of variables is possible, enabling integration with ease.
In our step-by-step solution, the integrating factor is utilized to solve the linear form of the transformed non-linear equation \( xv' - (1-n)v = x^{k}(1-n) \). To find the integrating factor \( \mu(x) \), we calculate \( e^{\int P(x) \, dx} \), where \( P(x) \) is the coefficient of \( y \) in the rearranged equation.
This mathematical tool allows us to transform a difficult-to-solve differential equation into one where separation of variables is possible, enabling integration with ease.
- Calculate the integrating factor using the exponent of the integral of the coefficient of \( y \).
- Use this factor to rewrite the equation in a form that is straightforward to integrate.
- Integrate to find the solution for the transformed variable, in this case, \( v \).
Non-linear Differential Equation
A non-linear differential equation involves a mathematical relation where either the function or its derivatives appear raised to a power or multiplied together, among other non-linear operations. In our example, the equation \( x y^{\prime} - y = x^{k} y^{n} \) is non-linear due to the presence of \( y^n \) as it represents an instance where the function is raised to a power.
Non-linear differential equations cannot be solved using straightforward linear methods such as separation or direct integration. Instead, they often require clever transformations or substitutions. In particular, Bernoulli's differential equation provides a way to handle certain non-linear equations by transforming them into linear equations.
By using a substitution like \( v = y^{1-n} \), the non-linear nature is moderated, turning the equation into a form that can then be tackled with linear solution techniques such as integrating factors. Understanding when and how to employ these transformations is key to successfully navigating and solving non-linear differential equations.
Non-linear differential equations cannot be solved using straightforward linear methods such as separation or direct integration. Instead, they often require clever transformations or substitutions. In particular, Bernoulli's differential equation provides a way to handle certain non-linear equations by transforming them into linear equations.
By using a substitution like \( v = y^{1-n} \), the non-linear nature is moderated, turning the equation into a form that can then be tackled with linear solution techniques such as integrating factors. Understanding when and how to employ these transformations is key to successfully navigating and solving non-linear differential equations.
Differential Equation Solution Steps
Solving a differential equation often involves a series of methodical steps aimed at transforming and simplifying the original problem. For the given equation \( x y^{\prime} - y = x^{k} y^{n} \), the steps consist of recognizing the type of equation, transforming it, simplifying, solving, and interpreting solutions.
**Key Steps Include:**
**Key Steps Include:**
- Recognize the Type: Identify whether the differential equation is linear or non-linear and categorize it accordingly. This forms the basis for selecting the appropriate solving method.
- Transform: In the case of non-linear equations like the Bernoulli type in our example, use substitutions to transform into linear form.
- Simplify: Substitute and simplify the expression to work with a linear equation. This often involves derivatives and algebraic manipulation.
- Solve the Linear Equation: Apply the integrating factor or other linear solution techniques to solve the simplified equation.
- Back-substitute: After solving for the transformed variable, revert back to the original function using substitution relationships.