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Solve the initial value problem. Eigenpairs of the coefficient matrices were determined in Exercises 1-10.\(\mathbf{y}^{\prime}=\left[\begin{array}{rr}3 & 1 \\\ -2 & 1\end{array}\right] \mathbf{y}, \quad \mathbf{y}(0)=\left[\begin{array}{l}8 \\ 6\end{array}\right]\)

Short Answer

Expert verified
Answer: The process for solving the initial value problem consists of the following steps: 1. Recall the general solution in terms of the fundamental matrix and an arbitrary constant vector. 2. Use the given eigenpairs of the coefficient matrix to find the fundamental matrix of solutions. 3. Write the general solution in terms of the fundamental matrix and the constant vector. 4. Apply the initial conditions to the general solution and set it equal to the initial condition vector. 5. Solve for the constant vector using the equation from step 4. 6. Write the specific solution using the fundamental matrix and the constant vector found in step 5.

Step by step solution

01

Recall the General Solution

In general, for a first order homogeneous linear system of ODEs of the form \(\mathbf{y}^{\prime} = A \mathbf{y}\), where \(A\) is a given constant matrix, the general solution can be written as \(\mathbf{y}(t) = \Phi(t) \mathbf{c}\), where \(\Phi(t)\) is the fundamental matrix of solutions, and \(\mathbf{c}\) is an arbitrary constant vector.
02

Find the Fundamental Matrix of Solutions

Since the eigenpairs of the coefficient matrix have been determined already, we can use them to find the fundamental matrix of solutions. Recall that if \((\lambda, \mathbf{v})\) is an eigenpair of matrix A, then the matrix exponential \(e^{At}\) can be utilized to find the fundamental matrix \(\Phi(t) = \left[ \mathbf{p}_1(t) \; \mathbf{p}_2(t) \right]\), where \(\mathbf{p}_i(t) = e^{\lambda_i t} \mathbf{v}_i\) for \(i=1,2\) are linearly independent solutions corresponding to the eigenvalues and eigenvectors. Note that if any eigenvalues are complex-valued, its complex conjugate is also an eigenvalue of the matrix. Substitute the given matrix A into the matrix exponential equation to find the fundamental matrix of solutions \(\Phi(t)\).
03

Write the General Solution

Now that we have found the fundamental matrix of solutions \(\Phi(t)\), write the general solution for the vector function \(\mathbf{y}(t)\) in terms of an arbitrary constant vector \(\mathbf{c}\). That is, \(\mathbf{y}(t) = \Phi(t) \mathbf{c}\).
04

Apply the Initial Conditions

Now we have to apply the given initial conditions to the general solution to find the specific solution to the initial value problem, i.e., \(\mathbf{y}(0)=\left[\begin{array}{l}8\\ 6\end{array}\right]\). Substitute \(t=0\) into the general solution, and then set this equal to the initial condition vector: \(\mathbf{y}(0) = \Phi(0) \mathbf{c} = \left[\begin{array}{l}8\\ 6\end{array}\right]\).
05

Solve for the Constant Vector \(\mathbf{c}\)

Solve for the constant vector \(\mathbf{c}\) in the equation found in step 4. This will give us the particular solution for the given initial value problem.
06

Write the Specific Solution

Using the constant vector \(\mathbf{c}\) found in step 5, write the specific solution to the initial value problem in the form: \(\mathbf{y}(t) = \Phi(t) \mathbf{c}\). This is the solution to the given initial value problem.

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Key Concepts

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

Eigenpairs of Coefficient Matrices
Understanding the eigenpairs of coefficient matrices is crucial when dealing with linear differential equations. An eigenpair consists of an eigenvalue \(\lambda\) and its corresponding eigenvector \(\mathbf{v}\). They are found by solving the characteristic equation \(|A - \lambda I| = 0\), where \(A\) is the coefficient matrix and \(I\) is the identity matrix of the same dimension.

Once we have the eigenvalues, we then determine the eigenvectors by solving \(\left(A - \lambda I\right) \mathbf{v} = 0\). In the context of the initial value problem for a system of ODEs, these eigenpairs play a pivotal role in constructing solutions. Specifically, each eigenpair leads to a solution that involves the eigenvalue in the exponent of the exponential function. When eigenvalues are complex, we also get complex eigenvectors, leading to complex solutions that can be expressed in terms of sines and cosines to obtain real solutions.
Fundamental Matrix of Solutions
The fundamental matrix of solutions, \(\Phi(t)\), emerges from the concept of eigenpairs and is a cornerstone in the theory of linear homogeneous ODEs. Constructed from the solutions associated with each eigenpair, the columns of \(\Phi(t)\) comprise linearly independent solutions to the ODE system. Each column, \(\mathbf{p}_i(t)\), corresponds to an eigenvector scaled by the exponential of the product of eigenvalue \(\lambda_i\) and time \(t\).

For instance, if we have two eigenpairs \(\left(\lambda_1, \mathbf{v}_1\right)\) and \(\left(\lambda_2, \mathbf{v}_2\right)\), the fundamental matrix is \([e^{\lambda_1 t}\mathbf{v}_1 \; e^{\lambda_2 t}\mathbf{v}_2]\). This matrix is crucial because it allows us to capture all possible solutions to the linear system. When this matrix is applied to an arbitrary constant vector, it generates a general solution to the system.
Homogeneous Linear System of ODEs
A homogeneous linear system of ODEs has a general form \(\mathbf{y}' = A \mathbf{y}\), where \(A\) is a matrix of coefficients, and \(\mathbf{y}\) is a vector of dependent variables. The term homogeneous refers to the absence of any non-zero constant or function on the right side of the equation. This system implies that the zero vector is always a solution, but we are typically interested in finding non-trivial solutions.

Non-trivial solutions can be obtained by finding the fundamental matrix of solutions that involve eigenvalues and eigenvectors of matrix \(A\). We express the general solution as a linear combination of these fundamental solutions, potentially involving complex numbers as well, which are then worked into real-valued solutions.
Matrix Exponential
The matrix exponential, denoted as \(e^{At}\), is an essential tool in solving linear systems of differential equations. It is analogous to the scalar exponential function but is applicable to square matrices. For the matrix \(A\), \(e^{At}\) is defined as the infinite series \(I + At + \frac{(At)^2}{2!} + \frac{(At)^3}{3!} + \ldots\), where \(I\) is the identity matrix and \(t\) is a scalar representing time.

The matrix exponential leads to the fundamental matrix \(\Phi(t)\) when the system's eigenvalues are distinct. This is particularly useful because it encapsulates all the information necessary to solve homogeneous linear systems of ODEs. By finding \(e^{At}\) and applying initial conditions, we can compute the precise evolution of the system over time.

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Most popular questions from this chapter

Verify, for any values \(c_{1}\) and \(c_{2}\), that the functions \(y_{1}(t)\) and \(y_{2}(t)\) satisfy the given system of linear differential equations. $$ \begin{aligned} &y_{1}^{\prime}=y_{1}+y_{2}, \quad y_{1}(t)=c_{1} e^{t} \cos t+c_{2} e^{t} \sin t \\ &y_{2}^{\prime}=-y_{1}+y_{2}, \quad y_{2}(t)=-c_{1} e^{t} \sin t+c_{2} e^{t} \cos t \end{aligned} $$

We consider systems of second order linear equations. Such systems arise, for instance, when Newton's laws are used to model the motion of coupled spring- mass systems, such as those in Exercises 31-32. In each of Exercises \(25-30\), let \(A=\left[\begin{array}{ll}2 & 1 \\ 1 & 2\end{array}\right] .\) Note that the eigenpairs of \(A\) are \(\lambda_{1}=3, \mathbf{x}_{1}=\left[\begin{array}{l}1 \\ 1\end{array}\right]\) and \(\lambda_{2}=1, \mathbf{x}_{2}=\left[\begin{array}{r}1 \\\ -1\end{array}\right] .\) (a) Let \(T=\left[\mathbf{x}_{1}, \mathbf{x}_{2}\right]\) denote the matrix of eigenvectors that diagonalizes \(A\). Make the change of variable \(\mathbf{z}(t)=T^{-1} \mathbf{y}(t)\), and reformulate the given problem as a set of uncoupled second order linear problems. (b) Solve the uncoupled problem for \(\mathbf{z}(t)\), and then form \(\mathbf{y}(t)=T \mathbf{z}(t)\) to solve the original problem.\(\mathbf{y}^{\prime \prime}+A \mathbf{y}=\left[\begin{array}{l}1 \\\ 0\end{array}\right], \quad \mathbf{y}(0)=\left[\begin{array}{l}1 \\\ 0\end{array}\right], \quad \mathbf{y}^{\prime}(0)=\left[\begin{array}{l}0 \\\ 1\end{array}\right]\)

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 \(t=0.02\) ?\(\frac{d}{d t}\left[e^{t} \frac{d y}{d t}\right]+y=2 e^{t}, \quad y(0)=-1, \quad y^{\prime}(0)=1\)

Find the largest interval \(a

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 \(e^{A t}\).\(A=\left[\begin{array}{ll}2 & 3 \\ 2 & 3\end{array}\right]\)

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