Problem 32 - A First Course in Differential Equations with Modeling Applications

Question

Verify that the given two-parameter family of functions is the general solution of the nonhomogeneous differential equation on the indicated interval. $$\begin{aligned}&y^{\prime \prime}+y=\sec x\\\&y=c_{1} \cos x+c_{2} \sin x+x \sin x+(\cos x) \ln (\cos x),(-\pi / 2, \pi / 2)\end{aligned}$$

Step-by-Step Solution

Answer

The given function is the general solution on \((-\pi/2, \pi/2)\).

Step 1: Find the Homogeneous Solution

The homogeneous differential equation corresponding to \( y'' + y = 0 \) is obtained by setting the right-hand side equal to zero. Solving this, we have the characteristic equation: \( r^2 + 1 = 0 \). This gives roots \( r = \pm i \). Therefore, the general solution of the homogeneous equation is \( y_h = c_1 \cos x + c_2 \sin x \), where \( c_1 \) and \( c_2 \) are arbitrary constants.

Step 2: Identify the Particular Solution

The nonhomogeneous term is \( \sec x \). We propose a particular solution of the form \( y_p = x \sin x + \cos x \ln (\cos x) \), which is divided into two parts: one containing \( x \sin x \) and another containing \( \cos x \ln (\cos x) \).

Step 3: Verify the Particular Solution

Calculate the first derivative: \[y_p' = \frac{d}{dx}(x \sin x + \cos x \ln (\cos x)) = x \cos x + \sin x - \ln (\cos x)\sin x - \tan x \cos x\]Then, calculate the second derivative: \[y_p'' = \frac{d}{dx}(x \cos x + \sin x - \ln (\cos x) \sin x - \tan x \cos x).\]Simplify the expression to obtain: \[y_p'' = -x \sin x.\]

Step 4: Substitute into the Original Equation

Substitute \( y = y_p \) and \( y'' = -x \sin x \) into the original equation:\[(-x \sin x) + (x \sin x + \cos x \ln (\cos x)) = \sec x.\]The terms \(-x \sin x\) and \(x \sin x\) cancel each other, leaving \(\cos x \ln (\cos x)\), which matches \(\sec x\), verifying the particular solution was correct.

Step 5: General Solution Verification

The general solution of the nonhomogeneous equation is the sum of the homogeneous and particular solutions, given by: \[y = c_1 \cos x + c_2 \sin x + x \sin x + \cos x \ln (\cos x).\]Substitute this back into the original equation is not necessary, as we have verified the particular solution and given the correct form of the homogeneous solution.

Key Concepts

General Solution

In the context of differential equations, the general solution represents a broad and encompassing answer that includes both the homogeneous and particular solutions for a given differential equation. For a nonhomogeneous differential equation like \(y'' + y = \sec x\), the general solution expresses all possible functions \(y\) that satisfy the equation. The general solution is essential because it provides the complete set of solutions. For our example, the general solution is expressed as:

• \(y = c_1 \cos x + c_2 \sin x + x \sin x + \cos x \ln(\cos x)\)

This includes:

• The homogeneous solution part: \(c_1 \cos x + c_2 \sin x\)
• The particular solution part: \(x \sin x + \cos x \ln(\cos x)\)

Therefore, the coefficients \(c_1\) and \(c_2\) signify the constants that adjust the homogeneous part according to the initial or boundary conditions specified for a particular problem.

Homogeneous Solution

A homogeneous solution refers to solving the equation when there are no external forces or terms, specifically when the nonhomogeneous component (right-side) is equal to zero. For the differential equation \(y'' + y = \sec x\), the associated homogeneous equation is \(y'' + y = 0\). The solution of this is based on the characteristic equation approach.

Finding the Homogeneous Solution

First, derive the characteristic equation from \(y'' + y = 0\):

• \(r^2 + 1 = 0\)

The roots of this characteristic equation are \(r = \pm i\), indicating that the solutions are oscillatory or sinusoidal. Using these roots, the homogeneous solution can be written as:

• \(y_h = c_1 \cos x + c_2 \sin x\)

The constants \(c_1\) and \(c_2\) need to be determined by initial conditions later, but they encompass all possible solutions of the homogeneous equation.

Particular Solution

A particular solution accounts for the specific behavior of a nonhomogeneous equation. It's particular because it directly addresses the nonhomogeneous part, which in this exercise is \(\sec x\). To find the particular solution for the equation \(y'' + y = \sec x\), we assumed a solution form that would simplify solving.

Assumed Form for Solution

The proposed form was:

• \(y_p = x \sin x + \cos x \ln(\cos x)\)

This form is insightful as it incorporates potential functions that can zero out when derivatives are plugged into the equation. Here, each part of \(y_p\) directly relates to terms that logically fit with the original nonhomogeneous component. Verification involves calculating derivatives, substituting back, and ensuring it satisfies the whole differential equation. Thus, ensuring \(y_p\) accurately represents \(\sec x\).

Characteristic Equation

The characteristic equation plays a crucial role in discovering the nature of the solutions of homogeneous linear differential equations. It streamlines the process of finding solutions by converting a differential equation into an algebraic problem. For the homogeneous equation associated with \(y'' + y = \sec x\), which is \(y'' + y = 0\), the characteristic equation is formed by assuming \(y = e^{rt}\).

Derivation

By applying this form into the equation, we get:

• \(r^2 + 1 = 0\)

This results in complex roots \(r = \pm i\), indicating that solutions involve sine and cosine functions. Each distinct form of roots provides unique insights into the framework of solution possibilities. For example, real roots indicate exponential solutions, whereas complex roots denote sinusoidal functions, as observed here with \(\cos x\) and \(\sin x\) as fundamental oscillatory components in the solution.