Problem 24 - Advanced Engineering Mathematics

Question

In Problems \(23-30\), verify that the given functions form a fundamental set of solutions of the differential equation on the indicated interval. Form the general solution. $$ y^{\prime \prime}-4 y=0 ; \cosh 2 x, \sinh 2 x,(-\infty, \infty) $$

Step-by-Step Solution

Answer

The functions \( \cosh 2x \) and \( \sinh 2x \) are linearly independent and form the general solution: \( y(x) = C_1 \cosh 2x + C_2 \sinh 2x \).

Step 1: Verify Linear Independence

To confirm that \( \cosh 2x \) and \( \sinh 2x \) form a fundamental set of solutions, we need to show that they are linearly independent. Calculate the Wronskian \( W(\cosh 2x, \sinh 2x) \): \[ W = \begin{vmatrix} \cosh 2x & \sinh 2x \ 2\sinh 2x & 2\cosh 2x \end{vmatrix} \]The Wronskian becomes \( W = (2\cosh 2x)(\cosh 2x) - (2\sinh 2x)(\sinh 2x) = 2(\cosh^2 2x - \sinh^2 2x) = 2\cdot 1 = 2 \). Since \( W eq 0 \) for all \( x \), these functions are linearly independent.

Step 2: Form General Solution

Since \( \cosh 2x \) and \( \sinh 2x \) are linearly independent solutions, the general solution of the differential equation is\[ y(x) = C_1 \cosh 2x + C_2 \sinh 2x \]where \( C_1 \) and \( C_2 \) are arbitrary constants.

Step 3: Verify Solutions

Substitute \( y = \cosh 2x \) and \( y = \sinh 2x \) into the differential equation \( y'' - 4y = 0 \):1. For \( y = \cosh 2x \), calculate derivatives: \( y' = 2\sinh 2x \) and \( y'' = 4\cosh 2x \). Substitute: \ \( 4\cosh 2x - 4\cosh 2x = 0 \).2. For \( y = \sinh 2x \), calculate derivatives: \( y' = 2\cosh 2x \) and \( y'' = 4\sinh 2x \). Substitute: \ \( 4\sinh 2x - 4\sinh 2x = 0 \).Both functions satisfy the differential equation, confirming they are indeed solutions.

Key Concepts

Understanding Linear Independence

Linear independence is a crucial concept in differential equations. It helps us understand whether a set of functions, like solutions to a differential equation, can independently span a solution space without being expressible as a linear combination of each other. In simpler terms, if two functions are linearly independent, they cannot be made from each other using constant multipliers.

For example, with functions like \( \cosh 2x \) and \( \sinh 2x \), we need to establish whether they are linearly independent on the interval \((-\infty, \infty)\). We do this by calculating their Wronskian. If the Wronskian is nonzero at every point in the interval, then these functions are truly linearly independent.

This concept is vital when forming general solutions because linearly independent functions ensure a unique set of solutions that covers all possibilities allowed by the differential equation.

The Role of the Wronskian

The Wronskian is named after the Polish mathematician Józef Hoene-Wroński and is used to test the linear independence of a set of functions. For two functions \( f(x) \) and \( g(x) \), the Wronskian is computed by:

• Differentiating each function: find \( f'(x) \) and \( g'(x) \).
• Creating a 2x2 matrix from these functions and their derivatives.
• Calculating the determinant of the matrix, which gives us the Wronskian.

For instance, with \( \cosh 2x \) and \( \sinh 2x \), the determinant of their Wronskian matrix is \( 2 \). Since this value isn't zero, the functions are linearly independent, one of the main tenets confirming a fundamental set of solutions.

Grasping how the Wronskian works aids students in verifying the independence of solutions, ensuring the completeness of the solution set for the given differential equation.

Finding the General Solution

After confirming that the functions are linearly independent, we can form the general solution to the differential equation. Linear independence ensures we have multiple valid solutions that, when combined, describe all possible behaviors described by the differential equation.

In our example, the linear independent solutions \( \cosh 2x \) and \( \sinh 2x \) allow us to write the general solution as:
\[y(x) = C_1 \cosh 2x + C_2 \sinh 2x\]
where \( C_1 \) and \( C_2 \) are constants. These constants can take on any real number depending on specific initial or boundary conditions given in a particular problem setting.

This general solution formulation covers all conceivable cases of the differential equation, granting students a comprehensive method to address more complex problem scenarios by simply tweaking the constants to fit specific situations. Developing an intuitive grasp of how to form and manipulate general solutions is an essential skill in understanding differential equations.