Problem 84 - Calculus Volume 3

Question

Two linearly independent solutions \(-y_{1}\) and \(y_{2}\) -are given that satisfy the corresponding homogeneous equation. Use the method of variation of parameters to find a particular solution to the given nonhomogeneous equation. Assume \(x>0\) in each exercise. \(x^{2} y^{\prime \prime}+2 x y^{\prime}-2 y=3 x\), \(y_{1}(x)=x, \quad y_{2}(x)=x^{-2}\)

Step-by-Step Solution

Answer

The particular solution is \( y_p = \frac{x^3}{2} - \frac{x^2}{4} \).

Step 1: Identify the Wronskian

To apply the method of variation of parameters, we first need to calculate the Wronskian of the given solutions. The Wronskian, \( W(y_1, y_2) \), of two functions \( y_1 \) and \( y_2 \) is given by the determinant \( W(y_1, y_2) = \begin{vmatrix} y_1 & y_2 \ y_1' & y_2' \end{vmatrix} \). For the functions \( y_1(x) = x \) and \( y_2(x) = x^{-2} \), we first find their derivatives: \( y_1'(x) = 1 \) and \( y_2'(x) = -2x^{-3} \). The Wronskian is \( W = \begin{vmatrix} x & x^{-2} \ 1 & -2x^{-3} \end{vmatrix} = x(-2x^{-3}) - x^{-2} \cdot 1 = -2x^{-2} - x^{-2} = -3x^{-2} \).

Step 2: Set up the Variation of Parameters

Next, we employ the method of variation of parameters. We look for solutions of the form \( y_p = u_1 y_1 + u_2 y_2 \) where \( u_1 \) and \( u_2 \) are functions to be determined. The formulas for \( u_1' \) and \( u_2' \) are given by: \( u_1' = -\frac{y_2(x)g(x)}{W(y_1, y_2)} \) and \( u_2' = \frac{y_1(x)g(x)}{W(y_1, y_2)} \), with \( g(x) = 3x \) being the non-homogeneous term. Substitute all known parts to find \( u_1' = -\frac{x^{-2} \cdot 3x}{-3x^{-2}} = x \) and \( u_2' = \frac{x \cdot 3x}{-3x^{-2}} = -x^3 \).

Step 3: Integrate to find u1 and u2

Integrate \( u_1' \) and \( u_2' \) to find \( u_1 \) and \( u_2 \). For \( u_1 \), integrate \( u_1' = x \) to get \( u_1 = \frac{x^2}{2} + C_1 \), where \( C_1 \) is a constant of integration (which can be set to zero since we only seek a particular solution). For \( u_2 \), integrate \( u_2' = -x^3 \) to obtain \( u_2 = -\frac{x^4}{4} + C_2 \), where \( C_2 \) can also be set to zero.

Step 4: Calculate the Particular Solution

Using the integrated functions \( u_1 \) and \( u_2 \), form the particular solution \( y_p = u_1 y_1 + u_2 y_2 \). Substitute \( u_1 = \frac{x^2}{2} \) and \( u_2 = -\frac{x^4}{4} \): \( y_p = \left(\frac{x^2}{2}\right)x + \left(-\frac{x^4}{4}\right)x^{-2} = \frac{x^3}{2} - \frac{x^2}{4} \). Simplify to find the particular solution: \( y_p = \frac{x^3}{2} - \frac{x^2}{4} \).

Key Concepts

Nonhomogeneous Differential Equation

Differential equations can be classified into two major categories: homogeneous and nonhomogeneous. A nonhomogeneous differential equation contains terms without derivatives of the unknown function, which means there is an "extra" term added to the equation that can be thought of as an external influence or force. In the exercise, the given nonhomogeneous differential equation is \[ x^2 y'' + 2x y' - 2y = 3x. \]Here, the term on the right, 3x, represents the nonhomogeneous, or external, term. The goal when solving such an equation is to find a solution that satisfies the entire equation, including the influence of this extra term. To achieve this, we often find a particular solution, in addition to the general solution of the corresponding homogeneous equation.

Wronskian

The Wronskian is a crucial tool in determining whether given solutions to a differential equation are linearly independent. For two solutions, \( y_1 \) and \( y_2 \), of a differential equation, the Wronskian, \( W(y_1, y_2) \), is calculated as the determinant:\[W(y_1, y_2) = \begin{vmatrix} y_1 & y_2 \ y_1' & y_2' \end{vmatrix} = y_1 y_2' - y_2 y_1'.\]In this particular exercise, the solutions given are \( y_1 = x \) and \( y_2 = x^{-2} \). By calculating their derivatives, \( y_1' = 1 \) and \( y_2' = -2x^{-3} \), we find the Wronskian as \(-3x^{-2}\). A non-zero Wronskian at a point indicates that the solutions are linearly independent in some interval around that point, which is fundamental for applying the method of variation of parameters.

Linearly Independent Solutions

In differential equations, solutions are said to be linearly independent if they cannot be expressed as a linear combination of each other over an interval. This concept is very important because it allows us to build a general solution of a differential equation using these solutions. For the given exercise, the solutions \( y_1 = x \) and \( y_2 = x^{-2} \) are proven to be linearly independent because their Wronskian is non-zero (\(-3x^{-2}\)). When the solutions are linearly independent, they form a fundamental set of solutions for the corresponding homogeneous equation, meaning any solution to the homogeneous equation can be written as a linear combination of these solutions. This fact is crucial when finding the particular solution using variation of parameters, as it ensures the validity of the solutions used.

Particular Solution

A particular solution to a nonhomogeneous differential equation is a solution that specifically satisfies the nonhomogeneous part of the equation. It accounts for the external force or non-zero term causing the equation to be nonhomogeneous. In our exercise, we use the method of variation of parameters to find this particular solution. We express a general form for the particular solution \( y_p = u_1 y_1 + u_2 y_2 \), where \( u_1 \) and \( u_2 \) are functions to determine. This involves using the known linearly independent solutions of the related homogeneous equation along with integrating resulting expressions based on the nonhomogeneous term \( g(x) = 3x \). After working through the integration and substituting back, we arrive at the particular solution \( y_p = \frac{x^3}{2} - \frac{x^2}{4} \). This solution directly satisfies the nonhomogeneous equation, providing that specific solution that balances the equation with the external term included.