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Chapter 6: Problem 31

. \(\mathbf{X}^{\prime}=\left(\begin{array}{cc}3 & 2 \\ -5 & 1\end{array}\right) \mathbf{X}+\left(\begin{array}{c}0 \\\ 10\end{array}\right), \mathbf{X}(0)=\left(\begin{array}{c}1 \\\ -2\end{array}\right)\)

Short Answer

Expert verified
The overall solution \( \mathbf{X}(t) = \mathbf{X}_h(t) + \mathbf{X}_p \), with constants determined from initial conditions.

Step by step solution

01

- Identify the components

This is a first-order linear system of differential equations. Identify the coefficient matrix A, the vector function \( \mathbf{X} \), and the inhomogeneous term.
02

- Write down the system

Given \( \mathbf{X}^{\prime}=\left(\begin{array}{cc}3 & 2 \ -5 & 1\end{array}\right) \mathbf{X}+\left(\begin{array}{c}0 \ 10\end{array}\right), \mathbf{X}(0)=\left(\begin{array}{c}1 \ -2\end{array}\right) \), rewrite it as:\( \mathbf{X}^{\prime} = A \mathbf{X} + \mathbf{b} \) where A = \( \left(\begin{array}{cc}3 & 2 \ -5 & 1\end{array}\right) \) and \( \mathbf{b} = \left(\begin{array}{c}0 \ 10\end{array}\right) \).
03

- Find the eigenvalues of A

Compute the eigenvalues \( \lambda \) by solving the characteristic equation \( \text{det}(A - \lambda I) = 0 \).
04

- Compute eigenvalues

The characteristic equation is \[\text{det}\left(\begin{array}{cc}3-\lambda & 2 \ -5 & 1-\lambda\end{array}\right) = (3-\lambda)(1-\lambda) - (-5)(2) = \lambda^2 - 4\lambda + 13 = 0\] Solve for \( \lambda \): \( \lambda = 2 \pm 3i \).
05

- Find the general solution of the homogeneous system

With \( \lambda = 2 \pm 3i \), find the corresponding eigenvectors to construct the general solution of the homogeneous system.
06

- Solve for eigenvectors

For \( \lambda = 2 + 3i \), solve \( (A - (2 + 3i)I)\mathbf{v} = 0 \). Similarly for \( \lambda = 2 - 3i \).
07

- Form the general solution

Combine the solutions obtained from eigenvalues and eigenvectors to form \( \mathbf{X}_{h}(t) = c_1 \mathbf{v}_1 e^{(2 + 3i)t} + c_2 \mathbf{v}_2 e^{(2-3i)t} \).
08

- Incorporate the particular solution

Find the steady-state (particular) solution of the non-homogeneous system: \( \mathbf{X}_p = (A \mathbf{X}_p + \mathbf{b} = 0) \).
09

- Construct the overall solution

Combine the homogeneous and particular solutions: \( \mathbf{X}(t) = \mathbf{X}_{h}(t) + \mathbf{X}_p \).
10

- Apply the initial condition

Use the initial condition \( \mathbf{X}(0) = \left(\begin{array}{c}1 \ -2\end{array}\right) \) to find constants \( c_1 \) and \( c_2 \).

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Key Concepts

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

First-order Linear System
A system of differential equations is referred to as a first-order linear system when it involves the first derivatives of unknown functions and the system can be represented linearly. In our exercise, the system is composed of first-order derivatives, which are expressed as: \( \mathbf{X}^{\prime}=A\mathbf{X}+\mathbf{b} \) , where A is the coefficient matrix, \( \mathbf{X} \) is the vector of unknown functions, and \( \mathbf{b} \) is the inhomogeneous term. This means that the system dynamically changes based on its current state, which makes analysis and solution finding very important. The form provided facilitates the use of matrix operations to solve the system efficiently.
Eigenvalues
Eigenvalues are crucial in solving systems of differential equations, as they provide insight into the system's behavior over time. To find the eigenvalues of the matrix A, we need to solve the characteristic equation: \[\text{det}(A - \lambda I) = 0\]. For our system, the coefficient matrix A is: \(\left(\begin{array}{cc}3 & 2 \ -5 & 1\end{array}\right)\). The characteristic equation is derived as: \[\text{det}\left(\begin{array}{cc}3-\lambda & 2 \ -5 & 1-\lambda\end{array}\right) = (3-\lambda)(1-\lambda) - (-5)(2) = \lambda^2 - 4\lambda + 13 = 0\]. Solving this for \( \lambda \), we find the eigenvalues to be \( 2 \pm 3i \). These eigenvalues indicate a spiral behavior due to the imaginary part, with a growth factor determined by the real part.
Eigenvectors
Eigenvectors are vectors that, when multiplied by matrix A, result in a scaled version of the original vector defined by its respective eigenvalue. To solve our system, we find the eigenvectors corresponding to each eigenvalue. For an eigenvalue \( \lambda \), we solve \( (A - \lambda I)\mathbf{v} = 0 \). For \( \lambda = 2 + 3i \), substracting this from A, we get: \(\left(\begin{array}{cc}1-3i & 2 \ -5 & -1-3i\end{array}\right)\mathbf{v} = 0\). The corresponding eigenvector \( \mathbf{v}_1 \) must be found, and similarly, for \( \lambda = 2 - 3i \), resulting in eigenvector \( \mathbf{v}_2 \). These eigenvectors are used to construct the general solution of our homogeneous system, combining with their exponential components.
Non-homogeneous Systems
A non-homogeneous system includes an inhomogeneous term \( \mathbf{b} \) which adds more complexity to the solution process. In our exercise, it is given as \(\left(\begin{array}{c}0 \ 10\end{array}\right) \). To solve the non-homogeneous system, we need both the homogeneous solution, obtained from eigenvalues and eigenvectors, and a particular solution for the steady state. \( \mathbf{X}_p \) is found such that \( A\mathbf{X}_p + \mathbf{b} = 0 \). Combining \(\mathbf{X}_{h}(t)\) (homogeneous solution) and \(\mathbf{X}_p\) forms the complete solution. Finally, initial conditions allow us to find constants \(c_1\) and \(c_2\) to perfectly match the initial system state, ensuring our overall solution accurately describes the system鈥檚 behavior from start.

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Most popular questions from this chapter

\(x^{\prime}=x \cos x y-\sin 2 y, y^{\prime}=-y \cos x y+\sin 2 x\); \(R=[-\pi, \pi] \times[-\pi, \pi]\)

\(x^{\prime}=-x-2 y, y^{\prime}=x+y\)

(a) Graph the direction field associated with the nonlinear system \(d x / d t=y\), \(d y / d t=-\sin x\) for \(-7 \leq x \leq 7\) and \(-4 \leq y \leq 4\). (b) (i) Approximate the solution to the initial value problem \(x_{1}^{\prime}=y_{1}, y_{1}^{\prime}=\) \(-\sin x_{1}, x_{1}(0)=0, y_{1}(0)=1\). (ii) Graph \(\left\\{x_{1}(t), y_{1}(t)\right\\}\) for \(0 \leq t \leq 7\) and display the graph together with the direction field. Does it appear as though the vectors in the vector field are tangent to the solution curve? (iii) Approximate the solution to the initial value problem \(x_{2}^{\prime}=y_{2}, y_{2}^{\prime}=\) \(-\sin x_{2}, x_{2}(0)=0, y_{2}(0)=2\). (iv) Graph \(\left\\{x_{2}(t), y_{2}(t)\right\\}\) for \(0 \leq t \leq 7\) and display the graph together with the direction field. Does it appear as though the vectors in the vector field are tangent to the solution curve? (v) Graph \(\left\\{x_{1}(t)+x_{2}(t), y_{1}(t)+y_{2}(t)\right\\}\) for \(0 \leq t \leq 7\) and display the graph together with the direction field. Does it appear as though the vectors in the vector field are tangent to the solution curve? (c) Approximate the solution to the initial value problem \(x^{\prime}=y, y^{\prime}=-\sin x\), \(x(0)=0, y(0)=3\) and graph the solution parametrically for \(0 \leq t \leq 7\). Is this the graph of \(\left\\{x_{1}(t)+x_{2}(t), y_{1}(t)+y_{2}(t)\right\\}\) found in (b) (v)? (d) Is the Principle of Superposition valid for nonlinear systems? Explain.

Consider the initial value problem \(\mathbf{X}^{\prime}=\) \(\left(\begin{array}{cc}1 & -7 / 3 \\ -2 & -8 / 3\end{array}\right) \mathbf{X}, \quad x(0)=x_{0}, \quad y(0)=y_{0} .\) (a) Graph the direction field associated with the system \(\mathbf{X}^{\prime}=\left(\begin{array}{cc}1 & -7 / 3 \\ -2 & -8 / 3\end{array}\right) \mathbf{X}\). (b) Find conditions on \(x_{0}\) and \(y_{0}\) so that at least one of \(x(t)\) or \(y(t)\) approaches zero as \(t\) approaches infinity. (c) Is it possible to choose \(x_{0}\) and \(y_{0}\) so that both \(x(t)\) and \(y(t)\) approach zero as \(t\) approaches infinity?

\(x^{\prime \prime}+3 x^{\prime}+2 x=0, x(0)=0, x^{\prime}(0)=-3, t=1\)
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