Problem 26 - Advanced Engineering Mathematics

Question

In Problems 23-26, verify that the indicated function is an explicit solution of the given differential equation. Give an interval of definition \(I\) for each solution. $$ \begin{aligned} &x^{2} y^{\prime \prime}+x y^{\prime}+y=\sec (\ln x) \\ &y=\cos (\ln x) \ln (\cos (\ln x))+(\ln x) \sin (\ln x) \end{aligned} $$

Step-by-Step Solution

Answer

The interval of definition is \( I = (0, e^{\pi/2}) \).

Step 1: Differentiate the Function y

Given the function \( y = \cos(\ln x) \ln(\cos(\ln x)) + (\ln x)\sin(\ln x) \), we first find the first and second derivatives. Let \( u = \cos(\ln x) \) and \( v = (\ln x) \sin(\ln x) \). Differentiate each term separately, then sum the results for the full derivative.

Step 2: Compute y' using the Product Rule

For \( u = \cos(\ln x) \), \( u' = -\sin(\ln x)/x \). For \( v = (\ln x)\sin(\ln x) \), apply the product rule: \( v' = \sin(\ln x) + \ln x \cos(\ln x) \). The first derivative \( y' = u' \ln u + u/v \times u' + v' \). Simplify this expression.

Step 3: Compute y'' for the Function

Differentiate \( y' \) to find \( y'' \). Again, use the product rule, as \( y' \) consists of products of functions. Differentiate each term in \( y' \) carefully, then sum their contributions to find \( y'' \).

Step 4: Substitute y, y', y'' into the Differential Equation

Plug the expressions for \( y \), \( y' \), and \( y'' \) into the given differential equation \( x^2 y'' + x y' + y = \sec(\ln x) \). Verify that each term satisfies the equation. Simplify and combine like terms to verify that both sides of the equation are equal.

Step 5: Identify the Interval of Definition

The function \( y = \cos(\ln x) \ln(\cos(\ln x)) + (\ln x)\sin(\ln x) \) is defined where all its components are defined. Consider where \( \ln x \), \( \cos(\ln x) \), and \( \sin(\ln x) \) are valid. Since \( \ln x \) is defined for \( x > 0 \) and \( \ln(\cos(\ln x)) \) requires the argument of \( \cos \) to be between \(-\pi/2\) and \( \pi/2 \), the interval \( I = (0, e^{\pi/2}) \).

Key Concepts

Explicit Solution

An explicit solution of a differential equation is a function that is expressed directly in terms of the independent variable and which satisfies the equation for all values within a specific interval. Here, we have a differential equation with a given explicit solution:

• Equation: \( x^2 y'' + x y' + y = \sec(\ln x) \)
• Explicit Solution: \( y = \cos(\ln x) \ln(\cos(\ln x)) + (\ln x) \sin(\ln x) \)

To verify that the given function is an explicit solution, it must satisfy the original differential equation when substituted into it. This involves calculating the first and second derivatives of the function and substituting them into the equation alongside the function itself.

Interval of Definition

The interval of definition for a solution is the range of the independent variable over which the solution is valid. This can be crucial for understanding where the solution applies:

• The independent variable here is \( x \).
• For the function \( y = \cos(\ln x) \ln(\cos(\ln x)) + (\ln x) \sin(\ln x) \), certain conditions must be met.
• \( \ln x \) is defined only for \( x > 0 \).
• \( \cos(\ln x) \) needs \( \ln x \) to remain within the domain of the cosine function where it is not undefined.

By analyzing these conditions, we determine that the function is defined in the interval \( I = (0, e^{\pi/2}) \). This means the solution is valid for all \( x \) in this range where both the logarithmic and trigonometric parts are defined.

Second Derivative

The second derivative of a function, denoted as \( y'' \), tells you how the rate of change of a function is changing. It provides insight into the curvature and concavity of the original function. To find the second derivative, you take the derivative of the first derivative:

• Start with the function \( y \): \( y = \cos(\ln x) \ln(\cos(\ln x)) + (\ln x) \sin(\ln x) \).
• Find the first derivative, \( y' \): This involves using the product rule as the function is composed of products of functions.
• Finally, differentiate \( y' \) again to get \( y'' \).
• This step is often more complex, requiring careful application of differentiation rules.
The second derivative \( y'' \) is used in the original differential equation to verify that the proposed function satisfies it.

Product Rule

The product rule is an essential rule in calculus used to differentiate functions that are the product of two or more functions. When differentiating such a product, the product rule is applied as follows:

• If we have two functions, \( u(x) \) and \( v(x) \), their derivative is given by: \( (uv)' = u'v + uv' \).

In our exercise, the use of the product rule is crucial due to the complicated nature of the function \( y \). For instance:

• The function contains \( \cos(\ln x) \) and \( (\ln x) \sin(\ln x) \) as components that are multiplied together.
• To differentiate \( y \) to find \( y' \) and ultimately \( y'' \), you apply the product rule multiple times.
• First to get the derivatives of the individual components.
• Then combine them into the complete derivative expressions.
This allows us to construct the necessary expressions for verifying the solution of the differential equation.