Question

Consider the \(n\)th order homogeneous linear differential equation $$ y^{(n)}+a_{n-1} y^{(n-1)}+\cdots+a_{3} y^{\prime \prime \prime}+a_{2} y^{\prime \prime}+a_{1} y^{\prime}+a_{0} y=0, $$ where \(a_{0}, a_{1}, a_{2}, \ldots, a_{n-1}\) are real constants. In each exercise, several functions belonging to a fundamental set of solutions for this equation are given. (a) What is the smallest value \(n\) for which the given functions can belong to such a fundamental set? (b) What is the fundamental set? $$ y_{1}(t)=t, \quad y_{2}(t)=e^{t}, \quad y_{3}(t)=\cos t $$

Short answer

Answer: The smallest value of \(n\) is 3. What is the fundamental set of solutions for the given homogeneous linear differential equation of order 3? Answer: The fundamental set of solutions is \(\{t, e^t, \cos t\}\).

Step-by-step solution

Reviewing the Fundamental Set of Solutions

A fundamental set of solutions for a homogeneous linear differential equation consists of \(n\) linearly independent functions that, when combined to form a linear combination, provide the general solution to the given differential equation. For the given functions \(y_1(t) = t\), \(y_2(t) = e^t\), and \(y_3(t) = \cos t\), we want to find the smallest possible \(n\) that would allow these functions to belong to a fundamental set of solutions for the equation. To do this, we'll check the linear independence of the given functions and examine their derivatives.

Checking Linear Independence

In order for the given functions to be part of a fundamental set of solutions, they must be linearly independent. We can check this by computing the Wronskian \(W(y_1, y_2, y_3)\), which is defined for three functions as follows: $$ W(y_1, y_2, y_3) = \begin{vmatrix} y_1 & y_2 & y_3 \\ y_1' & y_2' & y_3' \\ y_1'' & y_2'' & y_3'' \\ \end{vmatrix}. $$ If \(W(y_1, y_2, y_3)\) is non-zero for at least one value of \(t\), the functions \(y_1(t)\), \(y_2(t)\), and \(y_3(t)\) are linearly independent, and we can proceed to find the value of \(n\) required.

Computing the Wronskian

To compute the Wronskian, we first need to find the first and second derivatives of the given functions: $$ y_1'(t) = 1, \quad y_1''(t) = 0 \\ y_2'(t) = e^t, \quad y_2''(t) = e^t \\ y_3'(t) = -\sin t, \quad y_3''(t) = -\cos t $$ Now we can compute the Wronskian: $$ W(y_1, y_2, y_3) = \begin{vmatrix} t & e^t & \cos t \\ 1 & e^t & -\sin t \\ 0 & e^t & -\cos t \\ \end{vmatrix} = -e^{2t} \cos t + e^t \cos^2 t + e^t \sin^2 t = e^t(-e^t\cos t +1). $$ The Wronskian is non-zero for many values of \(t\), so the functions are linearly independent.

Determine the smallest value of \(n\)

Now that we have established the linear independence of the functions, we can determine the smallest value of \(n\) for which these functions could be part of a fundamental set of solutions. Each linearly independent function represents one of the \(n\) orders of the homogeneous linear differential equation. Since there are three linearly independent functions, the smallest value of \(n\) is 3.

Finding the Fundamental Set of Solutions

Since we have determined that the smallest value for \(n\) is 3, we can conclude that the fundamental set of solutions for the given homogeneous linear differential equation of order 3 is: $$ \{y_1(t), y_2(t), y_3(t)\} = \{t, e^t, \cos t\}. $$ So, the answer to (a) is 3, and the answer to (b) is the set \(\{t, e^t, \cos t\}\).

Key concepts

Wronskian

The Wronskian is a crucial tool used in the study of differential equations to establish whether a set of functions is linearly independent. This concept is named after the Polish mathematician Hoene Wronski, and it takes the form of a determinant constructed from the functions and their derivatives.

When dealing with the question of linear independence for solutions to homogeneous linear differential equations, the Wronskian becomes vitally important. Its calculated value provides key insight. If the Wronskian is nonzero at any point within an interval, it implies that the functions are linearly independent within that interval. Calculating the Wronskian involves creating a square matrix, where each row is populated with the derivatives of the functions up to the \(n-1\)th derivative, for \(n\) functions, and then taking the determinant of this matrix. As demonstrated in the step-by-step solution:

\[W(y_1, y_2, y_3) = \begin{vmatrix} t & e^t & \cos t \ 1 & e^t & -\sin t \ 0 & e^t & -\cos t \end{vmatrix}\]
This computation yielded a non-zero result, reaffirming the linear independence of the functions \(t\), \(e^t\), and \(\cos t\).

Linear Independence

Linear independence is a foundational concept in differential equations, which indicates that no function in a set can be represented as a linear combination of the others. In simpler terms, if you cannot express one function as a mixture of any others within the group by multiplying them with constants and adding or subtracting them, the functions are considered linearly independent.

In the context of solving differential equations, particularly the homogeneous linear types, identifying whether the solution functions are linearly independent is essential. It ensures that the fundamental set of solutions is complete, meaning that any solution to the differential equation can be expressed as a linear combination of this set. The step-by-step solution provides a clear method to check for linear independence by using the Wronskian:

\[W(y_1, y_2, y_3) = e^t(-e^t\cos t +1)\].
The non-zero Wronskian concludes that the functions \(y_1(t)\), \(y_2(t)\), and \(y_3(t)\) do not overlap in their contributions to the solution of the differential equation, thus they are linearly independent.

Fundamental Set of Solutions

A fundamental set of solutions in the realm of differential equations is a suite of solutions that span the solution space for a homogeneous linear differential equation. This essentially means that any solution to the equation can be written as a combination of functions in the fundamental set. In the context of an \(n\)th order differential equation, a fundamental set would typically consist of \(n\) linearly independent functions.

The 'smallest value \(n\)' referred to in the exercise forms the basis for establishing how many functions you need for a complete fundamental set for a given differential order. As shown previously, for three linearly independent functions such as \(t\), \(e^t\), and \(\cos t\), the minimum value for \(n\) is 3, indicating a third-order homogeneous linear differential equation. This verified the functions given do indeed form a fundamental set of solutions for the equation:

\[\{y_1(t), y_2(t), y_3(t)\} = \{t, e^t, \cos t\}\].
Having a well-defined fundamental set is essential for dealing comprehensively with differential equations, as it allows for the construction of general solutions that cater to varying initial conditions or boundary values.