Question

Use variation of parameters to find a particular solution, given the solutions \(y_{1}, y_{2}\) of the complementary equation. $$ x^{2} y^{\prime \prime}-(2 a-1) x y^{\prime}+a^{2} y=x^{a+1} ; \quad y_{1}=x^{a}, \quad y_{2}=x^{a} \ln x $$

Short answer

Based on the step-by-step solution above, answer the following question: Question: Using the variation of parameters method, find a particular solution to the given second-order linear differential equation with two solutions to the complementary equation. Solution: The particular solution, \(y_p\), for the given second-order linear differential equation is: $$ y_p=-\int \frac{x^{-(a+1)}\ln x}{x^{a-1}(1-\ln x)} dx\cdot x^a + \int \frac{x^{-(a+1)}}{x^{a-1}(1-\ln x)} dx \cdot x^a \ln x $$

Step-by-step solution

Find the Wronskian of the given solutions

To find the Wronskian, \(W(y_1, y_2)\), we use the following formula: $$ W(y_1, y_2) = \begin{vmatrix} y_{1} & y_{2} \\ y_{1}^{\prime} & y_{2}^{\prime} \end{vmatrix} $$ For our solutions, we first need the derivatives: $$ y_1^{\prime} = ax^{a-1} \\ y_2^{\prime} = ax^{a-1} \ln x + x^{a-1} $$ Now we can compute the Wronskian: $$ W(y_1, y_2) = \begin{vmatrix} x^a & x^a \ln x \\ ax^{a-1} & ax^{a-1} \ln x + x^{a-1} \end{vmatrix} $$

Find the expressions for the parameters

To find the parameters \(u_1\) and \(u_2\), we will use the following expressions: $$ u_{1}^{\prime} \cdot y_{1} + u_{2}^{\prime} \cdot y_{2} = 0 \\ u_{1}^{\prime} \cdot y_{1}^{\prime} + u_{2}^{\prime} \cdot y_{2}^{\prime} = x^{-(a+1)} $$ Rewriting the expressions using the values of \(y_1\), \(y_2\), \(y_1^{\prime}\), and \(y_2^{\prime}\), we have: $$ u_{1}^{\prime} \cdot x^{a} + u_{2}^{\prime} \cdot x^{a} \ln x = 0 \\ u_{1}^{\prime} \cdot ax^{a-1} + u_{2}^{\prime} \cdot (ax^{a-1} \ln x + x^{a-1}) = x^{-(a+1)} $$ Now we solve for \(u_1^{\prime}\) and \(u_2^{\prime}\). From the first equation, we have \(u_{1}^{\prime}=-u_{2}^{\prime} \ln x\). Substituting this into the second equation, we get: $$ (-u_{2}^{\prime} \ln x) \cdot ax^{a-1} + u_{2}^{\prime} \cdot (ax^{a-1} \ln x + x^{a-1}) = x^{-(a+1)} $$ Now, solve for \(u_{2}^{\prime}\): $$ u_{2}^{\prime} = \frac{x^{-(a+1)}}{x^{a-1}(1-\ln x)} $$ And for \(u_{1}^{\prime}\): $$ u_{1}^{\prime} = -u_{2}^{\prime} \ln x = -\frac{x^{-(a+1)}\ln x}{x^{a-1}(1-\ln x)} $$

Integrate the parameters and find the particular solution

Now we will integrate \(u_1^{\prime}\) and \(u_2^{\prime}\) to find \(u_1\) and \(u_2\): $$ u_{1} = \int u_{1}^{\prime} dx = -\int \frac{x^{-(a+1)}\ln x}{x^{a-1}(1-\ln x)} dx \\ u_{2} = \int u_{2}^{\prime} dx = \int \frac{x^{-(a+1)}}{x^{a-1}(1-\ln x)} dx $$ To get the particular solution, we multiply the parameters by the corresponding solutions and sum them: $$ y_p = u_{1} \cdot y_{1} + u_{2} \cdot y_{2} $$ We get: $$ y_p=-\int \frac{x^{-(a+1)}\ln x}{x^{a-1}(1-\ln x)} dx\cdot x^a + \int \frac{x^{-(a+1)}}{x^{a-1}(1-\ln x)} dx \cdot x^a \ln x $$

Write down the final result

Our final result for the particular solution is: $$ y_p=-\int \frac{x^{-(a+1)}\ln x}{x^{a-1}(1-\ln x)} dx\cdot x^a + \int \frac{x^{-(a+1)}}{x^{a-1}(1-\ln x)} dx \cdot x^a \ln x $$

Key concepts

Wronskian

The Wronskian is a critical concept in the study of differential equations. It is a determinant that helps determine whether a set of functions are linearly independent. In our context, if the Wronskian of functions is non-zero at some point, the functions are linearly independent on the interval where it is non-zero. The Wronskian is named after the Polish mathematician Hoene Wronski and is a fundamental tool used in the method called variation of parameters.

For two functions, the Wronskian is calculated using their values and the values of their first derivatives, arranged in a 2x2 matrix. When solving differential equations, finding the Wronskian is typically one of the first steps in using the variation of parameters method to find a particular solution. In our given exercise, the Wronskian of the solutions to the complementary equation, denoted as \( y_1 \) and \( y_2 \), is computed to further derive the parameters needed for the particular solution.

Particular Solution

A particular solution to a differential equation is a specific solution that satisfies both the differential equation and non-homogeneous conditions, like initial or boundary values. It's important to distinguish this from the general solution, which includes the complementary (or homogeneous) solution plus the particular solution and accounts for all possible solutions.

In the context of our exercise, we seek a particular solution using the method of variation of parameters. This method is employed after obtaining the complementary solution, which solves the associated homogeneous equation. To find a particular solution, we first calculate expressions for the parameters—represented by functions of \( u_1 \) and \( u_2 \)—and then integrate these to get the functions themselves. These functions are then used to construct the particular solution by multiplying them with the corresponding solutions of the complementary equation and summing up the results.

Complementary Equation

In the realm of differential equations, a complementary equation refers to the homogeneous part of a differential equation. Homogeneous equations are those without a non-homogeneous term, in other words, set to zero on one side of the equation. Solutions to these homogeneous equations are known as complementary solutions and play a role in finding the general solution of a non-homogeneous differential equation.

The complementary equation is simpler to solve and its solutions, which form the complementary function or homogeneous solution, are typically used as a stepping stone in the variation of parameters technique. The homogeneous solution consists of a linear combination of functions that solve the complementary equation. In our exercise, the given solutions \( y_{1}=x^{a} \) and \( y_{2}=x^{a} \ln x \) represent the complementary solution for the corresponding homogeneous differential equation.

Differential Equations

Differential equations are mathematical equations that describe relationships involving functions and their derivatives. They play a significant role in various fields such as physics, engineering, biology, economics, and many others since they can model a wide range of phenomena. A differential equation can be either ordinary, involving derivatives with respect to one variable, or partial, involving derivatives with respect to more than one variable.

Differential equations are broadly categorized as homogeneous when the equation equals zero, and non-homogeneous when there's an additional non-zero term. Solutions to these equations often require finding a particular solution that addresses the non-homogeneous part and a general solution that encompasses all potential solutions, including the homogeneous part.

In solving differential equations, it is sometimes necessary to resort to methods such as variation of parameters, separation of variables, or integrating factors, depending on the nature of the equation. Our example exercise demonstrates the application of variation of parameters to solve a non-homogeneous ordinary differential equation by first identifying the complementary solution and then constructing the particular solution.