Chapter 18: Problem 447
Find the solution of, the following system $$ \begin{aligned} &\dot{\mathrm{x}}_{1}=\mathrm{x}_{2} \\ &\dot{\mathrm{x}}_{2}=-2 \mathrm{x}_{1}-3 \mathrm{X}_{2} \end{aligned} $$
Short Answer
Expert verified
The solution to the given system is:
\[
x_1(t) = (a+b)e^{-t} - be^{-2t},
x_2(t) = 2(a-b)e^{-t} - 2be^{-2t}.
\]
Step by step solution
01
Write the given system as a matrix equation
First, rewrite the given system as a matrix equation Av = v', where A is the coefficient matrix, v is the column vector of unknown functions, and v' is the column vector of their derivatives:
\[
\begin{pmatrix} \dot{x}_1 \\ \dot{x}_2 \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ -2 & -3 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \end{pmatrix}
\]
Next,
\[
v' = Av,
\]
where
\[
A = \begin{pmatrix} 0 & 1 \\ -2 & -3 \end{pmatrix},
v = \begin{pmatrix} x_1 \\ x_2 \end{pmatrix},
v' = \begin{pmatrix} \dot{x}_1 \\ \dot{x}_2 \end{pmatrix}.
\]
02
Find the eigenvalues and eigenvectors of the coefficient matrix A
To find the eigenvalues and eigenvectors of A, we first find the characteristic equation:
\[
\det(A - \lambda I) = 0,
\]
where I is the identity matrix and λ are the eigenvalues. For matrix A, the characteristic equation is:
\[
\det\begin{pmatrix} -\lambda & 1 \\ -2 & -3-\lambda \end{pmatrix} = \lambda^2 + 3\lambda + 2 = (\lambda+1)(\lambda+2) = 0.
\]
The eigenvalues are λ1 = -1 and λ2 = -2.
Next, we find the eigenvectors associated with each eigenvalue by solving the equation:
\[
(A - \lambda I)w = 0,
\]
where w is the eigenvector.
For λ1 = -1, the eigenvector equation is:
\[
\begin{pmatrix} 1 & 1 \\ -2 & -2 \end{pmatrix}\begin{pmatrix} w_1 \\ w_2 \end{pmatrix} = 0.
\]
Choose w1 = 1; then w2 = -1. Thus, w1 = (1, -1).
For λ2 = -2, the eigenvector equation is:
\[
\begin{pmatrix} 2 & 1 \\ -2 & -1 \end{pmatrix}\begin{pmatrix} w_1 \\ w_2 \end{pmatrix} = 0.
\]
Choose w1 = 1; then w2 = 2. Thus, w2 = (1, 2).
03
Find the matrix exponential
Using the eigenvalues and eigenvectors, the matrix exponential can be found following the general solution of a homogenous linear system:
\[
e^{At} = V e^{\Lambda t}V^{-1},
\]
where V is the matrix of eigenvectors, Λ is the diagonal matrix of eigenvalues, and e^{Λt} is the diagonal matrix of exponentials of λt.
In our case,
\[
V = \begin{pmatrix} 1 & 1 \\ -1 & 2 \end{pmatrix}, \Lambda = \begin{pmatrix} -1 & 0 \\ 0 & -2 \end{pmatrix},
\]
and
\[
e^{\Lambda t} = \begin{pmatrix} e^{-t} & 0 \\ 0 & e^{-2t} \end{pmatrix}.
\]
Now, calculate the matrix exponential:
\[
e^{At} = \begin{pmatrix} 1 & 1 \\ -1 & 2 \end{pmatrix}\begin{pmatrix} e^{-t} & 0 \\ 0 & e^{-2t} \end{pmatrix}\begin{pmatrix} 1 & 1 \\ -1 & 2 \end{pmatrix}^{-1} = \begin{pmatrix} e^{-t} - e^{-2t} & e^{-t} + e^{-2t} \\ -2e^{-2t} & 2e^{-t} \end{pmatrix}.
\]
04
Find the solution using initial conditions
To find the solution vector v(t), multiply e^{At} by the initial condition vector v(0):
\[
v(t) = e^{At}v(0).
\]
Assuming the initial conditions are x1(0) = a and x2(0) = b, the solution vector is:
\[
v(t) = \begin{pmatrix} e^{-t} - e^{-2t} & e^{-t} + e^{-2t} \\ -2e^{-2t} & 2e^{-t} \end{pmatrix}\begin{pmatrix} a \\ b \end{pmatrix} = \begin{pmatrix} (a+b)e^{-t} - be^{-2t} \\ 2(a - b)e^{-t} - 2be^{-2t} \end{pmatrix}.
\]
Thus, the solution to the given system is:
\[
x_1(t) = (a+b)e^{-t} - be^{-2t},
x_2(t) = 2(a-b)e^{-t} - 2be^{-2t}.
\]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Matrix Equations
A matrix equation is a way to represent systems of equations or differential equations compactly. It converts a system of linear equations into a more concise form using matrices and vectors. In matrix form, the given differential system:
- \( \dot{x}_1 = x_2 \)
- \( \dot{x}_2 = -2x_1 - 3x_2 \)
Eigenvalues and Eigenvectors
Eigenvalues and eigenvectors are foundational concepts in linear algebra used to analyze linear transformations. For a matrix \( A \), an eigenvalue \( \lambda \) and an eigenvector \( w \) satisfy the equation: \[ (A - \lambda I)w = 0 \] - **Eigenvalues**: These are scalars that provide important insights about the matrix, such as its stability and behaviors of the system it represents. To find them, solve the characteristic equation \( \det(A - \lambda I) = 0 \). For our system's matrix \( A = \begin{pmatrix} 0 & 1 \ -2 & -3 \end{pmatrix} \), the eigenvalues are \( \lambda_1 = -1 \) and \( \lambda_2 = -2 \).- **Eigenvectors**: Corresponding to each eigenvalue, these are vectors that describe directions of stretching or compressing transformations. Solving \( (A - \lambda I)w = 0 \) gives the eigenvectors here: - For \( \lambda_1 = -1 \), \( w_1 = \begin{pmatrix} 1 \ -1 \end{pmatrix} \). - For \( \lambda_2 = -2 \), \( w_2 = \begin{pmatrix} 1 \ 2 \end{pmatrix} \).
Matrix Exponential
The matrix exponential \( e^{At} \) is used to solve systems of linear differential equations effectively. It relates to the expansion of exponential functions for matrices. The expression \( e^{At} \) is given by: \[ e^{At} = V e^{\Lambda t} V^{-1} \]where:
- \(V\) is the matrix of eigenvectors.
- \(\Lambda\) is the diagonal matrix of eigenvalues.
- \(e^{\Lambda t}\) is the diagonal matrix with elements \(e^{\lambda t}\).
Linear Systems
Linear systems involve equations where each term is either a constant or the product of a constant and a single variable. Their solutions provide insight into how variables change together. A linear differential system like our example:
- \( \dot{x}_1 = x_2 \)
- \( \dot{x}_2 = -2x_1 - 3x_2 \)
- \(x_1(t) = (a+b)e^{-t} - be^{-2t}\)
- \(x_2(t) = 2(a-b)e^{-t} - 2be^{-2t}\)