Problem 24 - Advanced Engineering Mathematics

Question

In Problems 23 and 24, the indicated functions are known linearly independent solutions of the associated homogeneous differential equation on the interval \((0, \infty)\). Find the general solution of the given nonhomogeneous equation. $$ x^{2} y^{\prime \prime}+x y^{\prime}+y=\sec (\ln x) ; \quad y_{1}=\cos (\ln x), y_{2}=\sin (\ln x) $$

Step-by-Step Solution

Answer

The general solution is \( y(x) = C_1 \cos(\ln x) + C_2 \sin(\ln x) - \ln x \cos(\ln x) + \int \frac{\tan(\ln x)}{x} \sin(\ln x) \, dx \).

Step 1: Identify the Homogeneous Solution

First, recognize that the given homogeneous differential equation is:\[ x^2 y'' + x y' + y = 0 \] The linearly independent solutions provided are \( y_1 = \cos(\ln x) \) and \( y_2 = \sin(\ln x) \). These solutions will form the homogeneous part of the general solution.

Step 2: Write the Homogeneous Solution

The general solution to the homogeneous equation is a combination of the independent solutions. Thus, the homogeneous solution is:\[ y_{h} = C_1 \cos(\ln x) + C_2 \sin(\ln x) \]where \( C_1 \) and \( C_2 \) are arbitrary constants.

Step 3: Set Up a Particular Solution for the Nonhomogeneous Equation

To find the general solution to the nonhomogeneous equation, we use the method of variation of parameters. Assume a particular solution of the form:\[ y_p = u_1(x) \cos(\ln x) + u_2(x) \sin(\ln x) \] where \( u_1(x) \) and \( u_2(x) \) are functions to be determined.

Step 4: Variation of Parameters - Finding Derivatives

First, find the derivatives needed for the method of variation of parameters:\[ y_p' = u_1'(x) \cos(\ln x) - u_1 \sin(\ln x)/x + u_2'(x) \sin(\ln x) + u_2 \cos(\ln x)/x \]because derivatives of \( \cos(\ln x) \) and \( \sin(\ln x) \) with respect to \( x \) involve \( x^{-1} \).

Step 5: Set Up System of Equations for Variation of Parameters

To solve for \( u_1'(x) \) and \( u_2'(x) \), setup the following system:1. \( u_1'(x) \cos(\ln x) + u_2'(x) \sin(\ln x) = 0 \)2. \( -u_1'(x) \sin(\ln x) + u_2'(x) \cos(\ln x) = \sec(\ln x)/x^2 \)This system comes from eliminating \( y_p(x) \) using the properties of \( y_1 \) and \( y_2 \) and matching \( y'' + py' + qy \) to the nonhomogeneous part.

Step 6: Solve for Functions u1 and u2

The system of equations can be solved using methods of solving linear systems, resulting in:\[ u_1'(x) = -\cos(\ln x) \sec(\ln x)/x = -1/x \]\[ u_2'(x) = \sin(\ln x) \sec(\ln x)/x = \tan(\ln x)/x \]Integrate \( u_1'(x) \) and \( u_2'(x) \) to find \( u_1(x) \) and \( u_2(x) \):\[ u_1(x) = -\ln(x) \]\[ u_2(x) = \int \tan(\ln x)/x \, dx \] (Note: This requires knowledge of an antiderivative for \( \tan(\ln x) \).)

Step 7: Construct the Particular Solution

After finding \( u_1(x) \) and \( u_2(x) \), substitute back to find:\[ y_p = (-\ln x) \cos(\ln x) + \left(\int \tan(\ln x)/x \, dx\right) \sin(\ln x) \]The specific integral for \( \int \tan(\ln x)/x \, dx \) may be addressed by involving partial fraction decomposition or numeric approximation, depending on problem constraints.

Step 8: General Solution to the Nonhomogeneous Equation

Combine the homogeneous solution and particular solution:\[ y(x) = C_1 \cos(\ln x) + C_2 \sin(\ln x) - \ln x \cos(\ln x) + \left(\int \tan(\ln x)/x \, dx\right) \sin(\ln x) \]This is the full general solution of the nonhomogeneous differential equation.

Key Concepts

Linearly Independent Solutions

In the context of differential equations, especially those pertaining to homogeneous solutions, the idea of linearly independent solutions plays a crucial role. Consider two functions, say \( y_1 \) and \( y_2 \). These functions are deemed linearly independent if their Wronskian is non-zero over a certain interval. The Wronskian is derived from a determinant involving the functions and their derivatives. For example, let's say we have \( y_1 = \cos(\ln x) \) and \( y_2 = \sin(\ln x) \). To verify their independence, compute the Wronskian: \[ W(y_1, y_2) = \begin{vmatrix} \cos(\ln x) & \sin(\ln x) \ -\sin(\ln x)/x & \cos(\ln x)/x \end{vmatrix} \] This results in a non-zero value, confirming that \( y_1 \) and \( y_2 \) are indeed linearly independent on the given interval \((0, \infty) \). This independent nature ensures that linear combinations such as \( y_h = C_1 y_1 + C_2 y_2 \) form a complete solution basis for the homogeneous differential equation. This characteristic is vital for solving the differential equations effectively and becomes a stepping stone in finding general solutions.

Homogeneous Solution

A homogeneous differential equation is one without any additional forcing term, meaning all terms involve the unknown function and its derivatives only. These are situations where the right-hand side of the equation is zero. For example, in the given exercise, the homogeneous equation is:\[ x^2 y'' + x y' + y = 0 \] To find a homogeneous solution, we use linearly independent solutions of the corresponding homogeneous equation, which, in this case, are \( y_1 = \cos(\ln x) \) and \( y_2 = \sin(\ln x) \). The general solution to this homogeneous equation is then a linear combination of these solutions:\[ y_h = C_1 \cos(\ln x) + C_2 \sin(\ln x) \] where \( C_1 \) and \( C_2 \) are arbitrary constants. This general solution represents all possible solutions that satisfy the homogeneous part of the differential equation, which we can use in solving the nonhomogeneous equation by extending it with a particular solution.

Nonhomogeneous Differential Equation

When a differential equation includes a non-zero function on its right-hand side, it transitions from homogeneous to nonhomogeneous. These equations are often encountered as:\[ x^2 y'' + x y' + y = \sec(\ln x) \] The strategy to solve such equations often involves finding a **particular solution** that satisfies the nonhomogeneous equation. One effective method for finding this solution is the variation of parameters technique. In this method, we assume a particular solution of the form:\[ y_p = u_1(x) \cos(\ln x) + u_2(x) \sin(\ln x) \] where \( u_1(x) \) and \( u_2(x) \) are functions determined through solving a system of equations derived from the differential equation.The general solution to the nonhomogeneous equation combines this particular solution with the homogeneous solution:\[ y(x) = C_1 \cos(\ln x) + C_2 \sin(\ln x) - \ln x \cos(\ln x) + \left(\int \tan(\ln x)/x \, dx\right) \sin(\ln x) \] Thus, solving nonhomogeneous equations typically involves blending unique tools with the foundational solution obtained from the homogeneous equation.