Problem 23 - A First Course in Differential Equations with Modeling Applications

Question

Verify that the given functions form a fundamental set of solutions of the differential equation on the indicated interval. Form the general solution. $$y^{\prime \prime}-y^{\prime}-12 y=0 ; \quad e^{-3 x}, e^{4 x},(-\infty, \infty)$$

Step-by-Step Solution

Answer

The functions \( e^{-3x} \) and \( e^{4x} \) form a fundamental set, and the general solution is \( y(x) = c_1 e^{-3x} + c_2 e^{4x} \).

Step 1: Verify the Solutions

To verify that \( e^{-3x} \) and \( e^{4x} \) form a fundamental set of solutions for the differential equation, we need to make sure they are solutions. Substitute \( y = e^{-3x} \) into the differential equation \( y'' - y' - 12y = 0 \).Calculate first and second derivatives:\(y' = -3e^{-3x}, \quad y'' = 9e^{-3x}\)Substitute into the differential equation:\[ 9e^{-3x} + 3e^{-3x} - 12e^{-3x} = 0 \]\[ 0 = 0 \]Since the equation holds true, \( e^{-3x} \) is a solution.Do the same for \( e^{4x} \):\(y' = 4e^{4x}, \quad y'' = 16e^{4x}\)Substitute into the differential equation:\[ 16e^{4x} - 4e^{4x} - 12e^{4x} = 0 \]\[ 0 = 0 \]Since the equation holds true, \( e^{4x} \) is also a solution.

Step 2: Check Linear Independence

To confirm that \( e^{-3x} \) and \( e^{4x} \) form a fundamental set, they must be linearly independent. We use the Wronskian determinant:\[ W(e^{-3x}, e^{4x}) = \begin{vmatrix} e^{-3x} & e^{4x} \ -3e^{-3x} & 4e^{4x} \end{vmatrix} \]Calculate the determinant:\[ = e^{-3x}(4e^{4x}) - e^{4x}(-3e^{-3x}) \]\[ = 4e^{x} + 3e^{x} \]\[ = 7e^{x} \]Since the Wronskian is nonzero for all \( x \, \in \, (-\infty, \infty) \), the functions are linearly independent and form a fundamental set.

Step 3: Form the General Solution

Since \( e^{-3x} \) and \( e^{4x} \) are linearly independent solutions, the general solution of the differential equation is a linear combination:\[ y(x) = c_1 e^{-3x} + c_2 e^{4x} \]where \( c_1 \) and \( c_2 \) are arbitrary constants.

Key Concepts

Fundamental Set of Solutions

When solving a second-order linear differential equation, finding the fundamental set of solutions is key. This set consists of two or more functions that are solutions to the differential equation and can be used to express any other solution as a linear combination. In our exercise, we have determined that the functions \( e^{-3x} \) and \( e^{4x} \) form such a set for the given differential equation.

Why are these specific functions considered a fundamental set? Because they satisfy the equation and, together, they capture the entire solution space of the differential equation. When you combine them linearly, you can reach every possible solution to the differential equation. Understanding this concept allows us to generalize solutions from specific instances.

Linear Independence

A crucial step in determining if a set of solutions is fundamental is checking for linear independence. Two functions are linearly independent if they cannot be expressed as a multiple of one another.

In mathematical terms, linear independence is verified using the Wronskian determinant. For the functions \( e^{-3x} \) and \( e^{4x} \), we found that their Wronskian is \( 7e^{x} \), which is non-zero for all real numbers.

This non-zero value confirms that these functions do not resemble each other as constant multiples and thus, they are indeed linearly independent. Linear independence ensures that our solutions are distinct and uniquely cover the solution space of the differential equation.

Wronskian

The Wronskian is a determinant used to see if a set of functions is linearly independent. It involves constructing a matrix from the functions and their derivatives. In our exercise, we calculated the Wronskian for \( e^{-3x} \) and \( e^{4x} \) using their derivatives.

The Wronskian appeared as:

• Calculate the first derivatives: \( -3e^{-3x} \) and \( 4e^{4x} \).
• Form the Wronskian matrix: \( \begin{vmatrix} e^{-3x} & e^{4x} \ -3e^{-3x} & 4e^{4x} \end{vmatrix} \)
• Compute the determinant: \( 7e^{x} \).

Since the determinant is never zero, \( e^{-3x} \) and \( e^{4x} \) are linearly independent across their domain.

General Solution

Once we have a fundamental set of linearly independent solutions, forming the general solution is straightforward. The general solution is essentially a linear combination of these independent solutions. For our exercise, the general solution is given as:

• \( y(x) = c_1 e^{-3x} + c_2 e^{4x} \)

Here, \( c_1 \) and \( c_2 \) are arbitrary constants. These constants allow the solution to fit any initial conditions or specific problem requirements. This flexibility is why finding the general solution is so valuable. It's a template that can be adjusted to solve a wide range of related problems, simply by selecting suitable values for \( c_1 \) and \( c_2 \).