Problem 23 - 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. $$ y^{\prime \prime}-y^{\prime}-12 y=0 ; e^{-3 x}, e^{4 x},(-\infty, \infty) $$

Step-by-Step Solution

Answer

The general solution is \(y(x) = c_1 e^{-3x} + c_2 e^{4x}\).

Step 1: Verify that functions are solutions

To verify that \(e^{-3x}\) and \(e^{4x}\) are solutions, substitute them into the differential equation \(y'' - y' - 12y = 0\). First, for \(y = e^{-3x}\), find \(y' = -3e^{-3x}\) and \(y'' = 9e^{-3x}\). Substitute into the equation: \(9e^{-3x} + 3e^{-3x} - 12e^{-3x} = 0\). This simplifies to 0, verifying it is a solution. Repeat for \(y = e^{4x}\): \(y' = 4e^{4x}\), \(y'' = 16e^{4x}\). Substitute to get \(16e^{4x} - 4e^{4x} - 12e^{4x} = 0\), which also simplifies to 0, confirming it's a solution.

Step 2: Check linear independence

To check that \(e^{-3x}\) and \(e^{4x}\) are linearly independent, compute the Wronskian. The Wronskian for the functions \(e^{-3x}\) and \(e^{4x}\) is given by: \( W(e^{-3x}, e^{4x}) = \begin{vmatrix} e^{-3x} & e^{4x} \ -3e^{-3x} & 4e^{4x} \end{vmatrix} \). This evaluates to \(e^{-3x} \cdot 4e^{4x} - e^{4x} \cdot (-3e^{-3x}) = 4e^{x} + 3e^{x} = 7e^{x}\). Since the Wronskian is not zero anywhere on \((-\infty, \infty)\), the functions are linearly independent.

Step 3: Form the general solution

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

Key Concepts

Fundamental Set of Solutions

A fundamental set of solutions for a homogeneous linear differential equation like \(y'' - y' - 12y = 0\) consists of solutions that can be combined to express every possible solution of the differential equation.
In our example, the functions \(e^{-3x}\) and \(e^{4x}\) form such a set because they solve the equation independently.
When checking whether a given set of functions is a fundamental one, they must solve the differential equation when substituted into it.
This step was achieved through verifying substitution, and fortunately, both functions satisfied the differential equation, reinforcing their role as a fundamental set of solutions.
This means any solution can be expressed as a combination of \(e^{-3x}\) and \(e^{4x}\).

Linear Independence

Linear independence is a core property needed for the fundamental set of solutions.
It ensures that no function in the set can be written as a linear combination of others.
For our functions \(e^{-3x}\) and \(e^{4x}\), this is confirmed by showing that there is no non-trivial way to mix these functions to get zero.
To check this, you need specifically to calculate the Wronskian.
In simpler terms, two functions are linearly independent if their Wronskian is not zero.

• Linearly independent functions make sure you can construct various solutions.
• If they were dependent, it would imply no diversity in solutions, which is not useful.

Wronskian

The Wronskian is a determinant used in the process of confirming linear independence. For two functions, \(f\) and \(g\), it is formulated as:\[W(f, g) = \begin{vmatrix} f & g \ f' & g' \end{vmatrix}\]
In our case with \(e^{-3x}\) and \(e^{4x}\), we found\[W(e^{-3x}, e^{4x}) = \begin{vmatrix} e^{-3x} & e^{4x} \ -3e^{-3x} & 4e^{4x} \end{vmatrix} = 7e^x \]The nonzero result of \(7e^x\) confirms that the two functions are indeed linearly independent.
This is crucial for guaranteeing their ability to form all possible solutions to the differential equation, as they are not redundant or overlapping.

General Solution

The general solution of a homogeneous linear differential equation with a fundamental set of solutions is a combination of those solutions.
In our example, since \(e^{-3x}\) and \(e^{4x}\) are the solutions, the general solution will take the form:\[y(x) = c_1 e^{-3x} + c_2 e^{4x}\]
Here, \(c_1\) and \(c_2\) are constants that can be adjusted to fit specific conditions or requirements of the problem.
The beauty of this approach is its flexibility.

• Different choices for these constants will allow us to cover every possible solution.
• It provides a comprehensive view of all solutions in one simplified expression, which is key in mathematics and engineering applications.

Understanding this makes it clear why finding linearly independent functions is vital—each unique combination gives a valid solution!