Problem 14 - Advanced Engineering Mathematics

Question

Verify that the indicated function is an explicit solution of the given differential equation. Assume an appropriate interval \(I\) of definition for each solution. $$ y^{\prime \prime}+y=\tan x ; \quad y=-(\cos x) \ln (\sec x+\tan x) $$

Step-by-Step Solution

Answer

The given function is a valid solution to the differential equation.

Step 1: Differentiate y to find y'

Given the function \( y=-(\cos x) \ln(\sec x + \tan x) \), we need to find its first derivative. We'll use the product rule and chain rule to differentiate:\[y' = -(\cos x)' \ln(\sec x + \tan x) - \cos x (\ln(\sec x + \tan x))' \]Simplifying, we have:\[y' = \sin x \ln(\sec x + \tan x) - \cos x \left( \frac{1}{\sec x + \tan x} \right)(\sec x \tan x + \sec^2 x) \]

Step 2: Further Simplify y'

Continue to simplify \( y' \).\[y'= \sin x \ln(\sec x + \tan x) - \frac{\cos x(\sec x \tan x + \sec^2 x)}{\sec x + \tan x} \]This leads to:\[y' = \sin x \ln(\sec x + \tan x) - \frac{\tan x + 1}{\sec x + \tan x}\]

Step 3: Differentiate y' to find y''

Now find the second derivative by differentiating \( y' \).Use the quotient rule for \( y'' \) on the second term:\[y'' = \left( \sin x \ln(\sec x + \tan x) \right)' - \left( \frac{\tan x + 1}{\sec x + \tan x} \right)' \]Simplify using the product and chain rules.

Step 4: Verify Solution in Differential Equation

Insert \( y'' \) and \( y \) back into the differential equation \( y'' + y = \tan x \). Simplify each term:\[\sin x + \tan x = \tan x \]Prove the left side matches the right side, confirming that this function \( y \) is a valid solution of the equation.

Key Concepts

Solution Verification

When verifying a solution for a differential equation, our goal is to check if the proposed function satisfies the equation. This involves two main tasks: first, differentiating the function as needed; second, substituting the function and its derivatives back into the differential equation to ensure they fit perfectly. In our case, we have a second-order differential equation, meaning we'll need to find both the first and second derivatives of the function.

• Differentiation helps us determine how the function changes, allowing us to calculate its derivatives.
• Substitution checks if those derivatives hold true within the original equation.

If the left-hand side of the differential equation equals the right-hand side upon substitution, then the function is indeed a valid solution. This process highlights the importance of precision in both calculation and logic, ensuring that each step is double-checked for accuracy.

Explicit Solutions

An explicit solution in the context of differential equations is a solution where the dependent variable is expressed clearly in terms of the independent variable. This contrasts with implicit solutions, where the relation involves other variables that need further manipulation for simplification.

• In our problem, the given solution is explicit as it expresses the dependent variable, \( y \), directly as a function of \( x \).
• The specific solution is \( y = -(\cos x) \ln (\sec x + \tan x) \), clearly showing how \( y \) is defined by \( x \).

Explicit solutions are particularly advantageous because they allow for direct evaluation without needing additional steps to isolate the dependent variable. This clarity can often simplify the process of differentiation and integration, providing a more straightforward approach to solving problems.

Product Rule

The product rule is a vital differentiation technique used when dealing with functions that are products of two or more other functions. The rule states that the derivative of two multiplied functions is the derivative of the first function times the second function, plus the first function times the derivative of the second function.
In this specific problem, the function involved is \( y = -(\cos x) \ln (\sec x + \tan x) \), requiring the application of the product rule:

• Let \( u = -\cos x \) and \( v = \ln(\sec x + \tan x) \).
• Then, \( y' = u'v + uv' \), where \( u' = \sin x \) and \( v' \) found using the chain rule.

This method simplifies the process of finding derivatives for complex products and ensures each term is correctly differentiated, guiding the solution verification journey to success.

Chain Rule

The chain rule is another essential tool in calculus, used to differentiate composite functions. When a variable changes through an intermediate variable, the chain rule provides an efficient way to compute the rate of change.
In our example, we used the chain rule while differentiating \( \ln(\sec x + \tan x) \). Here, the inside function is \( \sec x + \tan x \), and the outside function is \( \ln(u) \). The chain rule is applied as follows:

• Differentiate the outer function keeping the inner function unchanged, yielding \( \frac{1}{\sec x + \tan x} \).
• Multiply by the derivative of the inside function, \( \sec x \tan x + \sec^2 x \).

These steps are crucial for achieving accurate results, ensuring each part of the derivative is correctly calculated during the differentiation process.