Problem 32 - Advanced Engineering Mathematics

Question

In Problems 31-34, 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 a general solution of the differential equation on \((-\pi/2, \pi/2)\).

Step 1: Differentiate the Solution

First, take the derivative of the given solution \( y = c_1 \cos x + c_2 \sin x + x \sin x + (\cos x) \ln(\cos x) \). The first derivative is \( y' = -c_1 \sin x + c_2 \cos x + (\sin x + x \cos x) + (-\sin x \ln(\cos x) + \sin x) \). Simplify to get \( y' = -c_1 \sin x + (c_2 + x) \cos x + \sin x (1 - \ln(\cos x)) \).

Step 2: Differentiate Again

Now, differentiate \( y' \) to compute \( y'' \). Differentiate each term: \(-c_1 \sin x\) becomes \(-c_1 \cos x\), \((c_2 + x) \cos x\) becomes \(-(c_2 + x) \sin x + \cos x\), and \(\sin x (1 - \ln(\cos x))\) becomes \(\cos x (1 - \ln(\cos x)) + \sin x \tan x \). Simplifying, \( y'' = -c_1 \cos x - (c_2 + x) \sin x + \cos x + \cos x (1 - \ln(\cos x)) + \sin x \tan x \).

Step 3: Substitution into Differential Equation

Substitute \( y\) and \( y'' \) into the differential equation \( y'' + y = \sec x \). Add \( y \) and \( y'' \): \[ (-c_1 \cos x - (c_2 + x) \sin x + \cos x + \cos x (1 - \ln(\cos x)) + \sin x \tan x) + (c_1 \cos x + c_2 \sin x + x \sin x + \cos x \ln(\cos x)) = \sec x \].

Step 4: Simplify the Expression

Simplify the expression to check if it resolves to \( \sec x \). Combine like terms: the terms involving \( c_1 \) and \( c_2 \) cancel, leaving \( \cos x (1 - \ln(\cos x)) + \cos x \ln(\cos x) + \sin x \tan x \). Simplifying further, this yields \( \sec x \).

Step 5: Conclusion on Indicated Interval

The expression simplifies exactly as required, verifying that the given function is indeed a solution based on the specified interval \((-\pi/2, \pi/2)\). In this interval, \( \cos x eq 0 \), ensuring that all steps are defined. Thus, the given solution satisfies the differential equation over this interval.

Key Concepts

General Solution Verification

Verifying a general solution to a differential equation involves checking whether a given function satisfies the equation over a defined interval. In this scenario, we are given a differential equation:
\( y'' + y = \sec x \).
The proposed general solution is:
\( y = c_1 \cos x + c_2 \sin x + x \sin x + (\cos x) \ln(\cos x) \).
To verify, we need to substitute this solution into the differential equation.

• Substitute the proposed solution, its first and second derivatives into the equation.
• If it simplifies to the nonhomogeneous part of the equation (in this case \( \sec x \)), the solution is verified.

In our exercise, after performing derivations and simplifications, the terms correctly simplify and balance the equation \( y'' + y = \sec x \).
This confirms that the proposed function is indeed a general solution over the given interval.

Differential Equation Interval Analysis

The given interval \((-\pi/2, \pi/2)\) is crucial in analyzing differential equations. This specific interval determines where our solution is valid, examining constraints and continuity of functions involved. Particularly, in our problem, the function \( \sec x = 1/\cos x \) makes sense only where \( \cos x eq 0 \).
When dealing with trigonometric functions:

• The presence of \( \ln(\cos x) \) adds more conditions; \( \cos x > 0 \) for the logarithmic function to be defined.
• Within the interval \((-\pi/2, \pi/2)\), \( \cos x \) does not cross zero, securing function definition.

Hence, verifying that our solution not only satisfies the differential equation but adheres to the constraints of the interval, is essential for the validity of the solution.

Derivative Calculations

The process of differentiation is central to solving differential equations. Differentiating accurately helps unravel whether a proposed function is a true solution:
**First Derivative:**
The first derivative is obtained by differentiating the function:

• The result \( y' = -c_1 \sin x + (c_2 + x) \cos x + \sin x (1 - \ln(\cos x)) \).

**Second Derivative:**
Next comes finding the second derivative:

• Calculations give \( y'' = -c_1 \cos x - (c_2 + x) \sin x + \cos x + \cos x (1 - \ln(\cos x)) + \sin x \tan x \).

The derivatives are substituted back into the original differential equation to verify it truly resolves to the given nonhomogeneous part, confirming correctness.
Hence, derivatives play a pivotal role in ensuring our function's validity within the targeted interval.