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Determine whether the given functions form a fundamental set of solutions for the linear system. $$ \mathbf{y}^{\prime}=\left[\begin{array}{rr} 0 & 2 \\ -2 & 0 \end{array}\right] \mathbf{y}, \quad \mathbf{y}_{1}(t)=\left[\begin{array}{c} \cos 2 t \\ -\sin 2 t \end{array}\right], \quad \mathbf{y}_{2}(t)=\left[\begin{array}{l} -2 \sin 2 t \\ -2 \cos 2 t \end{array}\right] $$

Short Answer

Expert verified
Answer: Yes, the functions \(\mathbf{y}_1(t)\) and \(\mathbf{y}_2(t)\) form a fundamental set of solutions for the given linear system.

Step by step solution

01

Check linear independence

We are given two functions \(\mathbf{y}_1(t)\) and \(\mathbf{y}_2(t)\). To check if they are linearly independent, we will calculate their Wronskian and see if it is nonzero. The Wronskian of two functions is given by: $$W(\mathbf{y}_1, \mathbf{y}_2)(t) = \begin{vmatrix} \cos(2t) & -2\sin(2t) \\ -\sin(2t) & -2\cos(2t) \end{vmatrix}$$
02

Calculate the Wronskian

Now, calculate the determinant of the matrix obtained in Step 1: \begin{align*} W(\mathbf{y}_1, \mathbf{y}_2)(t) &= \cos(2t) \cdot (-2\cos(2t)) - (-2\sin(2t)) \cdot (-\sin(2t)) \\ &= -2\cos^2(2t) + 2\sin^2(2t) \end{align*} Now let's rewrite the expression, using the identity \(\sin^2(x) + \cos^2(x) = 1\) $$W(\mathbf{y}_1, \mathbf{y}_2)(t) = 2(\sin^2(2t) - \cos^2(2t))$$ As we can see, the Wronskian is not identically zero. Hence, the functions \(\mathbf{y}_1(t)\) and \(\mathbf{y}_2(t)\) are linearly independent.
03

Verify if they are solutions of the linear system

Let's differentiate both \(\mathbf{y}_1(t)\) and \(\mathbf{y}_2(t)\): $$\mathbf{y}_1'(t)=\left[\begin{array}{c} -2 \sin 2 t \\ -2 \cos 2 t \end{array}\right], \quad \mathbf{y}_2'(t)=\left[\begin{array}{c} -4 \cos 2 t \\ 2 \sin 2 t \end{array}\right]$$ Now, let's check if both functions satisfy the given linear system: $$\mathbf{A}\mathbf{y}_1(t) = \left[\begin{array}{rr} 0 & 2 \\ -2 & 0 \end{array}\right]\left[\begin{array}{c} \cos 2 t \\ -\sin 2 t \end{array}\right] = \left[\begin{array}{c} -2 \sin 2 t \\ -2 \cos 2 t \end{array}\right] = \mathbf{y}_1'(t)$$ $$\mathbf{A}\mathbf{y}_2(t) = \left[\begin{array}{rr} 0 & 2 \\ -2 & 0 \end{array}\right]\left[\begin{array}{c} -2 \sin 2 t \\ -2 \cos 2 t \end{array}\right] = \left[\begin{array}{c} -4 \cos 2 t \\ 2 \sin 2 t \end{array}\right] = \mathbf{y}_2'(t)$$ Both functions satisfy the given linear system, meaning they are solutions.
04

Conclusion

Since the functions \(\mathbf{y}_1(t)\) and \(\mathbf{y}_2(t)\) are linearly independent and they are solutions of the given linear system, they form a fundamental set of solutions.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Linear System of Differential Equations
Understanding a linear system of differential equations is paramount when dealing with various physical and engineering problems. In essence, such a system consists of multiple differential equations that involve the derivatives of several functions, often represented as vector components, related to one another through linear combinations.

For any given linear system, the solution set forms a vector space, and a fundamental set of solutions provides a basis for this space. This means that any solution to the linear system can be expressed as a linear combination of the fundamental solutions. Typically, for a system of n first-order differential equations, a fundamental set of solutions will consist of n linearly independent solutions. Ensuring that these are indeed solutions, as well as checking their independence, forms the core of solving such systems analytically.
Wronskian
The Wronskian is a critical tool in determining whether a set of functions forms a fundamental set of solutions for a linear system of differential equations. To envision the Wronskian, one must imagine a square matrix whose columns are the given functions and their derivatives up to order n-1, with n being the number of functions in the set.

The determinant of this matrix is known as the Wronskian, and it provides a straightforward test for linear independence. If the Wronskian is nonzero for at least one point in the interval considered, the functions are linearly independent. In the context of differential equation systems, this non-zero condition of the Wronskian ensures that the solution set constructed from these functions is indeed a fundamental set.
Linear Independence
Linear independence is a cornerstone concept in linear algebra and applies prominently to the study of differential equations. A set of functions is linearly independent if none of the functions in the set can be expressed as a linear combination of the others. This concept guarantees that each function brings unique information to the solution set of a differential equation.

To test for linear independence, as mentioned, we use the Wronskian. So, calculating the Wronskian and finding it to be non-zero at some point indicates that our functions are independent, and no one function is merely a 'repeat' of another in the context of the solutions they provide to the system. This aspect is particularly crucial for ensuring we have sufficient information to reconstruct any solution to the system from the basis solutions.
Matrix Determinant
The determinant of a matrix is a scalar value that provides a lot of information about the matrix and its associated linear transformations. For a square matrix, it can tell us whether the matrix is invertible, and in physical terms, it can represent a scale factor for volume when the matrix is considered as a transformation.

In the realm of differential equations, the determinant plays a vital role in the Wronskian. Specifically, the Wronskian involves taking the determinant of a specially constructed matrix filled with the functions of our system and their derivatives. Consequently, a non-zero determinant (Wronskian) establishes that our set of functions are linearly independent, which is precisely the condition needed to assert the existence of a fundamental set of solutions. This interplay between the determinant and the Wronskian highlights the determinant's importance beyond simple matrix arithmetic, extending into the analytical solution of differential equations systems.

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Most popular questions from this chapter

The given matrix \(A\) is diagonalizable. (a) Find \(T\) and \(D\) such that \(T^{-1} A T=D\). (b) Using (12c), determine the exponential matrix \(e^{A t}\).\(A=\left[\begin{array}{rr}0 & 2 \\ -2 & 0\end{array}\right]\)

In each exercise, (a) As in Example 3, rewrite the given scalar initial value problem as an equivalent initial value problem for a first order system. (b) Write the Euler's method algorithm, \(\mathbf{y}_{k+1}=\mathbf{y}_{k}+h\left[P\left(t_{k}\right) \mathbf{y}_{k}+\mathbf{g}\left(t_{k}\right)\right]\), in explicit form for the given problem. Specify the starting values \(t_{0}\) and \(\mathbf{y}_{0}\). (c) Using a calculator and a uniform step size of \(h=0.01\), carry out two steps of Euler's method, finding \(\mathbf{y}_{1}\) and \(\mathbf{y}_{2}\). What are the corresponding numerical approximations to the solution \(y(t)\) at times \(t=0.01\) and \(t=0.02\) ?\(\frac{d}{d t}\left[e^{t} \frac{d y}{d t}\right]+y=2 e^{t}, \quad y(0)=-1, \quad y^{\prime}(0)=1\)

In each exercise, (a) Does \(A B=B A\) ? (b) Calculate the exponential matrices \(e^{A t}, e^{B t}\), and \(e^{(A+B) t} .\) Does \(e^{A t} e^{B t}=e^{(A+B) t} ?$$A=\left[\begin{array}{rr}2 & -1 \\ -1 & 2\end{array}\right], \quad B=\left[\begin{array}{ll}1 & 1 \\ 1 & 1\end{array}\right]\)

The exact solution of the initial value problem \(\mathbf{y}^{\prime}=\left[\begin{array}{cc}0.5 & 1 \\ 1 & 0.5\end{array}\right] \mathbf{y}, \quad \mathbf{y}(0)=\left[\begin{array}{l}1 \\\ 0\end{array}\right] \quad\) is given by \(\quad \mathbf{y}(t)=\frac{1}{2}\left[\begin{array}{c}e^{-t / 2}+e^{3 t / 2} \\\ -e^{-t / 2}+e^{3 t / 2}\end{array}\right] .\) (a) Write a program that applies the Runge-Kutta method (12) to this problem. (b) Run your program on the interval \(0 \leq t \leq 1\), using step size \(h=0.01\). (c) Run your program on the interval \(0 \leq t \leq 1\), using step size \(h=0.005\). (d) Let \(\mathbf{y}_{100}\) and \(\mathbf{y}_{200}\) denote the numerical approximations to \(\mathbf{y}(1)\) computed in parts (b) and (c), respectively. Compute the error vectors \(\mathbf{y}(1)-\mathbf{y}_{100}\) and \(\mathbf{y}(1)-\overline{\mathbf{y}}_{200}\). By roughly what fractional amount is the error reduced when the step size is halved?

Each of the systems of linear differential equations can be expressed in the form \(\mathbf{y}^{\prime}=P(t) \mathbf{y}+\mathbf{g}(t) .\) Determine \(P(t)\) and \(\mathbf{g}(t)\) $$ A^{\prime}(t)=\left[\begin{array}{cc} t^{-1} & 4 t \\ 5 & 3 t^{2} \end{array}\right], \quad A(1)=\left[\begin{array}{rr} 2 & 5 \\ 1 & -2 \end{array}\right] $$

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