Chapter 4: Problem 24
Solve the given differential equation by variation of parameters. $$x^{2} y^{\prime \prime}+x y^{\prime}-y=\frac{1}{x+1}$$
Short Answer
Expert verified
The general solution is \( y = C_1 x + \frac{C_2}{x} + x - \frac{1}{x+1} + \ln|x+1| \frac{1}{x} - \frac{1}{x} \).
Step by step solution
01
Determine the Homogeneous Solution
First, solve the homogeneous equation, which is obtained by setting the right-hand side to zero: \[ x^2 y'' + x y' - y = 0. \]This is an Euler-Cauchy (or equidimensional) equation. We assume a solution of the form \( y = x^m \). Substituting this into the homogeneous equation gives:\[ m(m-1) x^m + mx^m - x^m = 0. \]Simplify to find the characteristic equation:\[ m^2 - 1 = 0, \]which factors to \((m - 1)(m + 1) = 0\). Therefore, the roots are \( m_1 = 1 \) and \( m_2 = -1 \). Thus, the general solution for the homogeneous equation is:\[ y_h = C_1 x + \frac{C_2}{x}. \]
02
Set Up for Variation of Parameters
We use the method of variation of parameters to find the particular solution, starting from the homogeneous solution:\[ y_h = C_1 x + \frac{C_2}{x}. \]Assume a particular solution of the form:\[ y_p = u_1(x) x + u_2(x) \frac{1}{x}. \]Where \( u_1(x) \) and \( u_2(x) \) are functions to be determined. The derivatives are:\[ y_p' = u_1' x + u_1 + u_2' \frac{1}{x} - u_2 \frac{1}{x^2}, \]\[ y_p'' = u_1' + u_1' x \frac{1}{x^2} - 2 u_2' \frac{1}{x^2} + 2 u_2 \frac{1}{x^3}. \]
03
Apply Variation of Parameters Formula
Substitute expressions for \( y_p \), \( y_p' \), and \( y_p'' \) into the original equation, and simplify using the conditions for variation of parameters:\[ x^2 (u_1' x + u_1 + u_2' \frac{1}{x} - u_2 \frac{1}{x^2})' + x (u_1' x + u_1 + u_2' \frac{1}{x} - u_2 \frac{1}{x^2}) - (u_1 x + u_2 \frac{1}{x}) = \frac{1}{x+1}. \]Using known conditions:\[ u_1'(x) \cdot x + u_2'(x) \cdot \frac{1}{x} = 0, \]\[ u_1'(x) x^2 - u_2'(x) = \frac{1}{x+1}. \]Solve these equations for \( u_1' \) and \( u_2' \).
04
Solve for Functions \( u_1(x) \) and \( u_2(x) \)
From the equation:\[ u_1' x + u_2' \frac{1}{x} = 0, \]\[ u_1'(x) x^2 - u_2'(x) = \frac{1}{x+1}, \]solve for \( u_1' \) and \( u_2' \):\[ u_1' = \frac{1}{x(x+1)}, \]\[ u_2' = -\frac{x^2}{x+1}. \]Integrate to find \( u_1(x) \) and \( u_2(x) \):\[ u_1(x) = \int \frac{1}{x(x+1)} \, dx, \]\[ u_2(x) = \int -\frac{x^2}{x+1} \, dx. \]
05
Perform Integration
For \( u_1(x) \) and \( u_2(x) \), perform the integrations:1. For \( u_1(x) \): The integral: \[ \int \frac{1}{x(x+1)} \, dx = \frac{1}{x} - \frac{1}{x+1}, \] after partial fraction decomposition.2. For \( u_2(x) \): The integral: \[ \int -\frac{x^2}{x+1} \, dx = -x - x^2 + \ln|x+1|. \]Substitute back to find the particular solution \( y_p \).
06
Find the Particular Solution and General Solution
Substitute \( u_1(x) \) and \( u_2(x) \) back to find the particular solution:\[ y_p = \left( \frac{1}{x} - \frac{1}{x+1} \right) x + \left( -x - x^2 + \ln|x+1| \right) \frac{1}{x}. \]Simplify, and then add to the general solution of the homogeneous equation to get the full solution:\[ y = C_1 x + \frac{C_2}{x} + \left( x - \frac{1}{x+1} \right) + \left( \ln|x+1| - 1 \right) \frac{1}{x}. \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Variation of Parameters
Variation of parameters is a powerful method used to find particular solutions to non-homogeneous differential equations. It builds upon the general solution of the corresponding homogeneous equation to construct a specific solution that satisfies the entire equation.
To apply this method, follow these steps:
To apply this method, follow these steps:
- First, solve the homogeneous equation to find the general solution. This will usually involve finding two linearly independent solutions.
- Next, replace the arbitrary constants in the homogeneous solution with functions, say \( u_1(x) \) and \( u_2(x) \), to form a particular solution \( y_p \).
- Determine these unknown functions \( u_1(x) \) and \( u_2(x) \) by substituting the particular solution into the original differential equation.
- Use conditions from variation of parameters to simplify and solve for \( u_1'(x) \) and \( u_2'(x) \).
Euler-Cauchy Equation
The Euler-Cauchy equation, also known as the equidimensional equation, is a special type of second-order linear differential equation. It usually takes the form:\[ ax^2 y'' + bxy' + cy = 0. \]
The distinctive feature of Euler-Cauchy equations is how each term's degree matches the degree of the derivatives. It's particularly common in problems where the independent variable has a logarithmic relationship with the dependent variable or in physical applications.
To solve this equation, assume a solution of the form \( y = x^m \). Substituting into the Euler-Cauchy equation reveals the characteristic equation. This polynomial in \( m \) will dictate the nature of the solution:
The distinctive feature of Euler-Cauchy equations is how each term's degree matches the degree of the derivatives. It's particularly common in problems where the independent variable has a logarithmic relationship with the dependent variable or in physical applications.
To solve this equation, assume a solution of the form \( y = x^m \). Substituting into the Euler-Cauchy equation reveals the characteristic equation. This polynomial in \( m \) will dictate the nature of the solution:
- If the roots are real and distinct, the general solution is a linear combination of powers of \( x \).
- If the roots are repeated, logarithmic terms appear in the solution.
- If the roots are complex, sinusoidal functions might be part of the solution.
Homogeneous Equation
Homogeneous equations are a critical part of differential equation analysis. They represent a scenario where the differential equation equals zero:\[ x^2 y'' + x y' - y = 0. \]
The role of homogeneous equations is two-fold when dealing with non-homogeneous equations:
For instance, in the given problem, the roots \( m_1 = 1 \) and \( m_2 = -1 \) lead to a homogeneous solution of:\[ y_h = C_1 x + C_2 \frac{1}{x}. \] These solutions, combined with the particular solution, give the comprehensive solution to the original differential equation.
The role of homogeneous equations is two-fold when dealing with non-homogeneous equations:
- Understanding the nature of the complete solution to the entire differential equation.
- Providing solutions that form the basis for constructing a particular solution using methods like variation of parameters.
For instance, in the given problem, the roots \( m_1 = 1 \) and \( m_2 = -1 \) lead to a homogeneous solution of:\[ y_h = C_1 x + C_2 \frac{1}{x}. \] These solutions, combined with the particular solution, give the comprehensive solution to the original differential equation.
Particular Solution
A particular solution is a specific solution to a non-homogeneous differential equation. Unlike the homogeneous solution which applies to any case of the equation, the particular solution satisfies the specific non-homogeneous conditions.
To find a particular solution using variation of parameters, start with the homogeneous solution and adjust it by replacing constants with functions that can make the entire equation correct. These functions, \( u_1(x) \) and \( u_2(x) \), are found by solving the system derived from substituting the assumed form of the particular solution back into the original equation.
To find a particular solution using variation of parameters, start with the homogeneous solution and adjust it by replacing constants with functions that can make the entire equation correct. These functions, \( u_1(x) \) and \( u_2(x) \), are found by solving the system derived from substituting the assumed form of the particular solution back into the original equation.
- The particular solution generally complements the homogeneous solution by accounting for the specific additional term (like \( \frac{1}{x+1} \) in this equation).
- Once \( u_1(x) \) and \( u_2(x) \) are integrated and obtained, they replace the constants in the solution of the homogeneous equation.