Question

Consider the given differential equation on the interval \(-\infty

Short answer

Answer: Yes, the solutions of the given differential equation form a fundamental set.

Step-by-step solution

Write the differential equation and initial conditions

We are given the differential equation: \(y^{\prime\prime\prime}+(\sin t)y=0\) and the initial conditions: \(y_1(0)=1\), \(y_1^{\prime}(0)=-1\), \(y_1^{\prime\prime}(0)=0\) \(y_2(0)=0\), \(y_2^{\prime}(0)=0\), \(y_2^{\prime\prime}(0)=2\) \(y_3(0)=2\), \(y_3^{\prime}(0)=-2\), \(y_3^{\prime\prime}(0)=1\)

Construct the Wronskian matrix

To check for linear independence we can use the Wronskian of the three functions - \(y_1(t)\), \(y_2(t)\), and \(y_3(t)\). The Wronskian matrix is given by: $ \begin{vmatrix} y_1 & y_2 & y_3 \\ y_1^{\prime} & y_2^{\prime} & y_3^{\prime} \\ y_1^{\prime\prime} & y_2^{\prime\prime} & y_3^{\prime\prime} \end{vmatrix} $

Substitute the initial conditions into the Wronskian matrix

We now plug in the given initial conditions into the Wronskian matrix: $ \begin{vmatrix} 1 & 0 & 2 \\ -1 & 0 & -2 \\ 0 & 2 & 1 \end{vmatrix} $

Calculate the determinant of the Wronskian matrix

The determinant of the Wronskian matrix will indicate if the solutions are linearly independent (non-zero determinant) or not (zero determinant). We compute the determinant of the given matrix: $W(y_1, y_2, y_3) = \begin{vmatrix} 1 & 0 & 2 \\ -1 & 0 & -2 \\ 0 & 2 & 1 \end{vmatrix} = 1 \begin{vmatrix} 0 & -2 \\ 2 & 1 \end{vmatrix} - 0 \begin{vmatrix} -1 & -2 \\ 0 & 1 \end{vmatrix} + 2 \begin{vmatrix} -1 & 0 \\ 0 & 2 \end{vmatrix}$ \(W(y_1, y_2, y_3) = 1(0-(-4)) + 0 - 4 = 8\) Since the determinant of the Wronskian is non-zero, the solutions are linearly independent.

Conclusion

The solutions of the given differential equation form a fundamental set, as the Wronskian of the solutions is non-zero. This indicates that the solutions are linearly independent.

Key concepts

Differential Equations

A differential equation is a mathematical equation that involves functions and their derivatives. It describes the relationship between a function and its rates of change. For example, in the given exercise, we are dealing with the third-order differential equation \(y^{\prime\prime\prime}+(\sin t)y=0\), which means the rate of change of the rate of change of the rate of change of \(y\) is related to \(y\) itself by the sine of \(t\). Solving these equations often requires finding functions that satisfy the given relationship. Additionally, understanding the behavior of the solutions, such as whether they remain bounded or oscillate over time, is crucial in applications ranging from physics to engineering.

One common approach to solving such equations is through the method of undetermined coefficients or variation of parameters. The objective is always to find a solution or set of solutions that satisfies the equation under certain constraints or initial conditions, which leads us to our next core concept.

Initial Conditions

Initial conditions specify values of the solution and its derivatives at a certain point, usually at the beginning of the interval of interest. They are essential for finding a particular solution to a differential equation. In our exercise, the initial conditions given for three functions at \(t=0\) allow us to uniquely determine each of the functions. These conditions help us in determining the constants when we have a general solution for a differential equation. Without initial conditions, we could have infinitely many solutions to a differential equation, as we would have a family of curves that satisfy the equation, but with initial conditions, we can pinpoint the exact solution that fits the scenario.

Linear Independence

The concept of linear independence is key to understanding the uniqueness and span of solutions in differential equations. A set of functions is linearly independent if no function in the set can be written as a linear combination of the others. This implies that each function adds a new dimension to the solution space, and there's no redundancy among the solutions provided. The Wronskian determinant, as seen in the exercise, provides a test for linear independence. If the Wronskian of a set of functions is non-zero at any point in the interval, the functions are linearly independent on that interval. The Wronskian is especially useful when dealing with solutions to linear differential equations where we are often interested in finding a set of linearly independent solutions to describe all possible solutions.

Fundamental Set of Solutions

A fundamental set of solutions of a linear differential equation is a set of linearly independent solutions that spans the entire solution space. In other words, any solution to the differential equation can be expressed as a linear combination of functions from this set. The exercise shows that if the Wronskian is non-zero, we have such a fundamental set. Hence, knowing the fundamental set is crucial for describing all possible behaviors of solutions to the differential equation. It’s like having a complete toolkit, where each tool plays a unique role in constructing any solution to the equation. As we conclude that we have a fundamental set of solutions due to a non-zero Wronskian, we are assured that our toolkit is neither missing any tools nor does it have any redundant ones.