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Solve the initial value problem. Eigenpairs of the coefficient matrices were determined in Exercises 1-10.\(\begin{array}{ll}y_{1}^{\prime}=-5 y_{1}-2 y_{2}, & y_{1}(0)=0 \\ y_{2}^{\prime}=5 y_{1}+y_{2}, & y_{2}(0)=-2\end{array}\)

Short Answer

Expert verified
Based on the provided solution, the explicit solution for the initial value problem is given by the functions \(y_1(t) = 0\) and \(y_2(t) = e^t\).

Step by step solution

01

Write the system as a matrix equation

Let \(\mathbf{y}=\begin{bmatrix} y_1 \\ y_2 \end{bmatrix}\). The given system of equations can be written in matrix form as \(\mathbf{y}^\prime = A\mathbf{y}\), where \(A = \begin{bmatrix} -5 & -2 \\ 5 & 1 \end{bmatrix}\).
02

Form the characteristic equation and find the eigenvalues

The characteristic equation is \(|A-\lambda I|=0\). For the coefficient matrix \(A\), the characteristic equation becomes \(|A-\lambda I| = \begin{vmatrix} -5-\lambda & -2 \\ 5 & 1-\lambda \end{vmatrix}=(\lambda+5)(\lambda-1) - (5 \cdot-2)=0\). Solve this equation to find the eigenvalues \(\lambda\): \((\lambda+5)(\lambda-1) + 10 = \lambda^2 + 4\lambda + 5 = 0\). So, the eigenvalues are \(\lambda_1 = -5\) and \(\lambda_2 = 1\).
03

Find eigenvectors, form unique fundamental matrix and its exponential

For \(\lambda_1 = -5\), solve \((A-\lambda I)\mathbf{v} = 0\) for the eigenvector \(\mathbf{v}\): \(\begin{bmatrix} 0 & -2 \\ 5 & 6 \end{bmatrix}\begin{bmatrix} v_{11} \\ v_{12} \end{bmatrix}=0\). The eigenvector is \(\mathbf{v}_1 = \begin{bmatrix} 1 \\ 0 \end{bmatrix}\). For \(\lambda_2=1\), solve \((A-\lambda I)\mathbf{v} = 0\): \(\begin{bmatrix} -6 & -2 \\ 5 & 0 \end{bmatrix}\begin{bmatrix} v_{21} \\ v_{22} \end{bmatrix} = 0\). The eigenvector is \(\mathbf{v}_2 = \begin{bmatrix} 1 \\ 3 \end{bmatrix}\). Form the unique fundamental matrix \(\Phi(t) = [\mathbf{x}_1(t),\mathbf{x}_2(t)] = [\mathbf{v}_1 e^{-5t},\mathbf{v}_2 e^t]\).
04

Initial conditions and constants

To satisfy the initial conditions, \(\mathbf{y}(0) = \begin{bmatrix} 0 \\ -2 \end{bmatrix} = C_1 \mathbf{v}_1 + C_2 \mathbf{v}_2\), where \(C_1\) and \(C_2\) are constants. Using the initial conditions, we have \(\begin{bmatrix} 0 \\ -2 \end{bmatrix} = C_1 \begin{bmatrix} 1 \\ 0 \end{bmatrix} + C_2 \begin{bmatrix} 1 \\ 3 \end{bmatrix}\), giving \(C_1 = 0\) and \(C_2 = 1\).
05

Determine the explicit solution

We can now find the explicit solution for the system of equations, \(\mathbf{y}(t) = \Phi(t)[C_1\,\,C_2]^T = [\mathbf{v}_1 e^{-5t},\mathbf{v}_2 e^t]\begin{bmatrix} 0 \\ 1 \end{bmatrix} = \begin{bmatrix} 0 \\ e^t \end{bmatrix}\). Thus, the explicit solution for the initial value problem is \(y_1(t) = 0\) and \(y_2(t) = e^t\).

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Key Concepts

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

Eigenvalues and Eigenvectors
Understanding eigenvalues and eigenvectors is fundamental to solving systems of linear differential equations. An eigenvalue, denoted by \(\lambda\), is a special number that, when multiplied by a vector (the eigenvector), produces the same effect as applying a particular linear transformation described by a matrix to that vector. In other words, for a square matrix \(A\), if \(A\mathbf{v} = \lambda\mathbf{v}\), then \(\lambda\) is the eigenvalue, and \(\textbf{v}\) is the corresponding eigenvector.

Eigenvectors provide directions along which a linear transformation stretches or shrinks vectors, and eigenvalues tell how much stretching or shrinking occurs. In our initial value problem, identifying the eigenvalues and corresponding eigenvectors of the coefficient matrix \(A\) is crucial for finding the solution to the system. The elegant relationship between the matrix, its eigenvalues, and eigenvectors, simplifies a complex system down to simple components, paving the way for us to form a solution.
System of Differential Equations
When we face a system of differential equations, we are essentially dealing with multiple functions and their derivatives that are interrelated. This type of system frequently appears in modeling multidimensional dynamic systems, such as oscillating mechanical systems, evolving populations, or electrical circuits.

To solve these systems, we often represent them in matrix form. This approach streamlines the process, turning otherwise unwieldy computations into more manageable matrix operations. By doing so, we leverage the concepts of eigenvalues and eigenvectors to decouple the system. This decoupling transforms the system into parallel independent equations, which can then be solved using standard techniques for single differential equations, and ultimately, recombined to obtain the solution to the original system.
Matrix Exponentiation
In the context of differential equations, matrix exponentiation offers a method for finding solutions to linear systems with constant coefficients. When we raise a matrix to an exponent that is a real number, especially \(t\), it is tied to calculating the matrix expression \(e^{At}\), where \(A\) is a matrix and \(t\) is a scalar.

For solving our system of differential equations, matrix exponentiation is crucial, as it involves the fundamental matrix \(\Phi(t)\), which contains solutions to the homogeneous system. Importantly, if we know the eigenvalues and eigenvectors, we can determine \(\Phi(t)\) by computing \(\mathbf{v}e^{\lambda t}\) for each eigenpair \(\lambda, \mathbf{v}\), simplifying the process considerably. This matrix exponentiation is a building block for the solutions we’re constructing, mirroring the dynamic behavior of the system over time.
Characteristic Equation
The characteristic equation of a matrix is a polynomial equation derived from the determinant of \(A - \lambda I\), where \(A\) is a square matrix, \(\lambda\) represents an eigenvalue, and \(I\) is the identity matrix of the same dimension as \(\textbf{A}\). The roots of this polynomial equation correspond to the eigenvalues of \(A\).

By solving the characteristic equation, we obtain the specific eigenvalues that are key to understanding the behavior of the system of differential equations. In our initial value problem, once we form the characteristic equation from the coefficient matrix \(A\), the resulting eigenvalues guide us toward the eigenspaces associated with the matrix. These eigenspaces, spanned by the eigenvectors, become the axes along which we can observe pure exponential growth or decay, informing us about the natural modes of the system.

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Most popular questions from this chapter

Give an example that shows that while similar matrices have the same eigenvalues, they may not have the same eigenvectors.

In each exercise, determine all equilibrium solutions (if any).\(\mathbf{y}^{\prime}=\left[\begin{array}{rrr}1 & 1 & 0 \\ 0 & -1 & 2 \\\ 0 & 0 & 1\end{array}\right] \mathbf{y}+\left[\begin{array}{l}2 \\ 3 \\\ 2\end{array}\right]\)

A Spring-Mass-Dashpot System with Variable Damping As we saw in Section \(3.6\), the differential equation modeling unforced damped motion of a mass suspended from a spring is \(m y^{\prime \prime}+\gamma y^{\prime}+k y=0\), where \(y(t)\) represents the downward displacement of the mass from its equilibrium position. Assume a mass \(m=1 \mathrm{~kg}\) and a spring constant \(k=4 \pi^{2} \mathrm{~N} / \mathrm{m}\). Also assume the damping coefficient \(\gamma\) is varying with time: $$ \gamma(t)=2 t e^{-t / 2} \mathrm{~kg} / \mathrm{sec} \text {. } $$ Assume, at time \(t=0\), the mass is pulled down \(20 \mathrm{~cm}\) and released. (a) Formulate the appropriate initial value problem for the second order scalar differential equation, and rewrite it as an equivalent initial value problem for a first order linear system. (b) Applying Euler's method, numerically solve this problem on the interval \(0 \leq t \leq 10 \mathrm{~min}\). Use a step size of \(h=0.005\). (c) Plot the numerical solution on the time interval \(0 \leq t \leq 10 \mathrm{~min}\). Explain, in qualitative terms, the effect of the variable damping upon the solution.

Consider the linear system \(\mathbf{y}^{\prime}=A \mathbf{y}\), where $$ A=\left[\begin{array}{rr} 2 & 1 \\ -4 & \alpha \end{array}\right] $$ (a) For what values of the constant \(\alpha\) is \(\mathbf{y}=\mathbf{0}\) the only equilibrium solution? (b) For what values of \(\alpha\) does more than one equilibrium solution exist? In this case, how many are there? Where do these values lie when plotted in the phase plane?

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 \(\mathbf{g}(t)\) $$ A^{\prime \prime}(t)=\left[\begin{array}{cc} 2 t & \sin t \\ 0 & 0 \end{array}\right], \quad A(0)=\left[\begin{array}{ll} 0 & 1 \\ 0 & 0 \end{array}\right], \quad A^{\prime}(0)=\left[\begin{array}{ll} 0 & 0 \\ 1 & 0 \end{array}\right] $$

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