Question

Determine whether the given functions form a fundamental set of solutions for the linear system. $$ \mathbf{y}^{\prime}=\left[\begin{array}{ll} 1 & -1 \\ 5 & -1 \end{array}\right] \mathbf{y}, \quad \mathbf{y}_{1}(t)=\left[\begin{array}{c} \cos 2 t \\ \cos 2 t+\sin 2 t \end{array}\right], \quad \mathbf{y}_{2}(t)=\left[\begin{array}{c} \sin 2 t \\ \sin 2 t-2 \cos 2 t \end{array}\right] $$

Short answer

Answer: No, the given functions do not form a fundamental set of solutions for the linear system, as they are linearly independent but do not satisfy the given linear system.

Step-by-step solution

Compute the Wronskian

To check if \(\mathbf{y}_1(t)\) and \(\mathbf{y}_2(t)\) are linearly independent, calculate their Wronskian: $$ W(t) = \det\begin{bmatrix} \cos 2t & \sin 2t \\ \cos 2t + \sin 2t & \sin 2t - 2\cos 2t \end{bmatrix} $$ Compute the determinant: $$ W(t) = (\cos 2t)(\sin 2t - 2\cos 2t) - (\sin 2t)(\cos 2t + \sin 2t) $$ Simplify the expression: $$ W(t) = -3\cos 2t \sin 2t $$

Check for linear independence

The functions \(\mathbf{y}_1(t)\) and \(\mathbf{y}_2(t)\) are linearly independent if the Wronskian \(W(t)\) is non-zero for any value of \(t\). Since the Wronskian is \(-3\cos 2t \sin 2t\), it is non-zero for any value of \(t\), where \(\cos 2t\) and \(\sin 2t\) are both non-zero. Hence, the functions are linearly independent.

Verify if given functions are solutions

Now, we need to check if the given functions are solutions to the linear system. To do that, we will differentiate each function and substitute it into the system. First, differentiate \(\mathbf{y}_1(t)\) and \(\mathbf{y}_2(t)\): $$ \mathbf{y}_1^{\prime}(t) = \begin{bmatrix} -2\sin 2t \\ -2\sin 2t + 2\cos 2t \end{bmatrix}, \quad \mathbf{y}_2^{\prime}(t) = \begin{bmatrix} 2\cos 2t \\ 2\cos 2t + 4\sin 2t \end{bmatrix} $$ Now, substitute these into the linear system: $$ \left[\begin{array}{ll} 1 & -1 \\\ 5 & -1 \end{array}\right] \mathbf{y}_1(t) = \begin{bmatrix} -2\sin 2t \\ -2\sin 2t + 2\cos 2t \end{bmatrix}, \quad \left[\begin{array}{ll} 1 & -1 \\\ 5 & -1 \end{array}\right] \mathbf{y}_2(t) = \begin{bmatrix} 2\cos 2t \\ 2\cos 2t + 4\sin 2t \end{bmatrix} $$ Check if the matrix-vector multiplication gives the required result: $$ \left[\begin{array}{ll} 1 & -1 \\\ 5 & -1 \end{array}\right] \mathbf{y}_1(t) = \begin{bmatrix} -\sin 2t \\ -\sin 2t + 6\cos 2t \end{bmatrix} \ne \begin{bmatrix} -2\sin 2t \\ -2\sin 2t + 2\cos 2t \end{bmatrix} $$ $$ \left[\begin{array}{ll} 1 & -1 \\\ 5 & -1 \end{array}\right] \mathbf{y}_2(t) = \begin{bmatrix} 2\cos 2t \\ 2\cos 2t + 6\sin 2t \end{bmatrix} \ne \begin{bmatrix} 2\cos 2t \\ 2\cos 2t + 4\sin 2t \end{bmatrix} $$ As we can see, the given functions do not satisfy the given linear system.

Conclusion

The functions \(\mathbf{y}_1(t)\) and \(\mathbf{y}_2(t)\) are linearly independent, but they do not satisfy the given linear system. Therefore, these functions do not form a fundamental set of solutions for the given linear system.

Key concepts

Linear Differential Equations

Linear differential equations are equations involving derivatives of an unknown function and the function itself, where the unknown function and its derivatives appear linearly. In the context of the classroom exercise, we are dealing with a system of linear differential equations which can be expressed in matrix form as \[ \mathbf{y}' = A\mathbf{y} \.\] Here, \(A\) is a matrix of coefficients, \(\mathbf{y}\) is a vector of the unknown functions, and \(\mathbf{y}'\) represents their derivatives. Solving such systems involves finding functions that satisfy the equation for all values of the independent variable, often time \(t\). These solutions, especially when they form a fundamental set, provide valuable insight into the system's behavior and are crucial for understanding complex dynamic systems.

Wronskian

The Wronskian is a crucial tool in the study of differential equations, named after the Polish mathematician Józef Hoene-Wroński. It is used to determine if a set of functions is linearly independent over an interval. For two functions, the Wronskian \(W\) is defined as the determinant of a matrix composed of the functions and their derivatives:

\[ W(t) = \det\begin{bmatrix} y_1(t) & y_2(t) \ y_1'(t) & y_2'(t) \end{bmatrix} \]
If the Wronskian is non-zero on an interval, then the functions \(y_1\) and \(y_2\) are linearly independent on that interval. In the exercise, by showing that the Wronskian \(W(t) = -3\cos 2t \sin 2t\) is not identically zero, we confirmed the linear independence of the functions \(\mathbf{y}_1(t)\) and \(\mathbf{y}_2(t)\) at the points where neither \(\sin 2t\) nor \(\cos 2t\) are zero.

Linear Independence

Linear independence is a fundamental concept in linear algebra and it applies to solutions of differential equations as well. A set of functions is said to be linearly independent if no function in the set can be written as a linear combination of the others. In simpler terms, none of the functions in the set can be created by adding or subtracting multiples of the other functions.

For the exercise, verifying linear independence is critical since only linearly independent solutions can form a fundamental set. The Wronskian helped us conclude that the functions \(\mathbf{y}_1(t)\) and \(\mathbf{y}_2(t)\) are linearly independent, which is a step towards forming a fundamental solution set, although it does not guarantee that they are valid solutions to the system.

Matrix Differential System

A matrix differential system is a way to represent a set of simultaneous first-order linear differential equations using matrices. The form \(\mathbf{y}' = A\mathbf{y}\), which we saw in the given exercise, is an example of such a system. The matrix \(A\) contains the coefficients that define how each function in vector \(\mathbf{y}\) interacts with every other function.

To solve a matrix differential system, one typically looks for a set of solutions that satisfies the system of equations. These solutions should ideally form a fundamental set, which in the two-dimensional case, would consist of two linearly independent vectors. In our exercise, even though the two provided functions turned out to be linearly independent, they failed to satisfy the matrix differential system, hence they did not make up a fundamental set of solutions.