Problem 24 - A First Course in Differential Equations with Modeling Applications

Question

Solve the given differential equation by variation of parameters. $$x^{2} y^{\prime \prime}+x y^{\prime}-y=\frac{1}{x+1}$$

Step-by-Step Solution

Answer

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 1: 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:\[ y_h = C_1 x + \frac{C_2}{x}. \]

Step 2: 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:\[ y_p' = u_1' x + u_1 + u_2' \frac{1}{x} - u_2 \frac{1}{x^2}, \]\[ y_p'' = u_1' + u_1' x \frac{1}{x^2} - 2 u_2' \frac{1}{x^2} + 2 u_2 \frac{1}{x^3}. \]

Step 3: 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 \( u_2' \).

Step 4: 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 \( u_2(x) \):\[ u_1(x) = \int \frac{1}{x(x+1)} \, dx, \]\[ u_2(x) = \int -\frac{x^2}{x+1} \, dx. \]

Step 5: 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 \( y_p \).

Step 6: 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:\[ y = C_1 x + \frac{C_2}{x} + \left( x - \frac{1}{x+1} \right) + \left( \ln|x+1| - 1 \right) \frac{1}{x}. \]

Key Concepts

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:

• 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 \( u_2'(x) \).

By integrating \( u_1'(x) \) and \( u_2'(x) \), you'll find \( u_1(x) \) and \( u_2(x) \). Together with the homogeneous solution, this provides the complete solution to the differential equation.

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:

• 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.

Recognizing and solving the characteristic equation helps establish the fundamental framework needed for solving the differential equation.

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:

• 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 the Euler-Cauchy equation, the homogeneous solution involves finding linearly independent solutions by assuming a form like \( y = x^m \). Solving for \( m \) generates the characteristic equation, whose roots define the general solution.

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.

• 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.

Combining both solutions provides a complete solution to the differential equation as displayed in the given solution steps.