Chapter 8: Problem 34
Find the general solution of the given system. $$ \begin{aligned} &\frac{d x}{d t}=4 x+5 y \\ &\frac{d y}{d t}=-2 x+6 y \end{aligned} $$
Short Answer
Expert verified
The general solution involves eigenvectors and exponentials: solve for constants with initial conditions.
Step by step solution
01
Write System as a Matrix Equation
Given the system of equations, write it in matrix form:\[\begin{bmatrix}\dot{x} \\dot{y}\end{bmatrix} = \begin{bmatrix}4 & 5 \-2 & 6\end{bmatrix} \begin{bmatrix}x \y\end{bmatrix}\]This can also be written as \(\frac{d\mathbf{X}}{dt} = A\mathbf{X}\), where \(A = \begin{bmatrix} 4 & 5 \ -2 & 6 \end{bmatrix}\) and \(\mathbf{X} = \begin{bmatrix} x \ y \end{bmatrix}\).
02
Find Eigenvalues of the Matrix
To find the eigenvalues, solve \(\det(A - \lambda I) = 0\), where \(I\) is the identity matrix:\[\det\begin{bmatrix} 4 - \lambda & 5 \ -2 & 6 - \lambda \end{bmatrix} = (4-\lambda)(6-\lambda) - (5)(-2) = 0\]Calculate:\[(4-\lambda)(6-\lambda) + 10 = \lambda^2 - 10\lambda + 34 = 0\]Solve for \(\lambda\).
03
Solve the Characteristic Equation
The quadratic equation \(\lambda^2 - 10\lambda + 34 = 0\) can be solved using the quadratic formula:\[\lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]where \(a=1\), \(b=-10\), \(c=34\). Calculate:\[\lambda = \frac{10 \pm \sqrt{100 - 136}}{2}\]\[\lambda = \frac{10 \pm \sqrt{-36}}{2}\]\[\lambda = 5 \pm 3i\]The eigenvalues are \(\lambda_1 = 5 + 3i\) and \(\lambda_2 = 5 - 3i\).
04
Find Eigenvectors for Each Eigenvalue
For \(\lambda_1 = 5 + 3i\), substitute into \((A - \lambda I)\mathbf{v} = \mathbf{0}\): \[\begin{bmatrix} -1-3i & 5 \ -2 & 1-3i \end{bmatrix} \begin{bmatrix} v_1 \ v_2 \end{bmatrix} = \begin{bmatrix} 0 \ 0 \end{bmatrix}\]Solve the system to determine the eigenvector associated with \(\lambda_1\) (similar steps for \(\lambda_2\)).
05
Construct the General Solution using Eigenvalues and Eigenvectors
The general solution for the system is given by combining both eigenvalues and eigenvectors:\[\begin{bmatrix} x(t) \ y(t) \end{bmatrix} = c_1 \begin{bmatrix} \text{Re}(v_1) \ \text{Re}(v_2) \end{bmatrix} e^{5t}\cos(3t) + c_1 \begin{bmatrix} -\text{Im}(v_1) \ -\text{Im}(v_2) \end{bmatrix} e^{5t}\sin(3t) + \]\[+c_2 \begin{bmatrix} \text{Re}(v_1) \ \text{Re}(v_2) \end{bmatrix} e^{5t}\sin(3t) + c_2 \begin{bmatrix} \text{Im}(v_1) \ \text{Im}(v_2) \end{bmatrix} e^{5t}\cos(3t)\]where \(c_1\) and \(c_2\) are constants determined by initial conditions.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Eigenvalues and Eigenvectors
When dealing with a system of differential equations, eigenvalues and eigenvectors play a crucial role in determining the behavior of the system. To find them, we first need the matrix form of the system. In our problem, the matrix is given by:
- \(A = \begin{bmatrix} 4 & 5 \ -2 & 6 \end{bmatrix}\)
- Calculate \(\det\begin{bmatrix} 4 - \lambda & 5 \ -2 & 6 - \lambda \end{bmatrix}\)
- Solve \(\lambda^2 - 10\lambda + 34 = 0\) using the quadratic formula, leading to complex roots.
Matrix Equation
Transforming a system of differential equations into a matrix equation helps simplify and systematically solve the problem. In our system:
- \(\frac{d}{dt}\begin{bmatrix}x \ y\end{bmatrix} = A\begin{bmatrix}x \ y\end{bmatrix}\)
- \(A = \begin{bmatrix} 4 & 5 \ -2 & 6 \end{bmatrix}\)
- Eigenvalues (\(\lambda\))
- Eigenvectors (\(\mathbf{v}\))
Complex Solutions
When the characteristic equation yields complex solutions, it indicates oscillatory behavior within the system. For our problem, the characteristic equation yields the complex roots:
- \(\lambda_1 = 5 + 3i\)
- \(\lambda_2 = 5 - 3i\)
- The real part (5) indicates exponential growth, expressed as \(e^{5t}\).
- The imaginary part (\(3i\)) brings the trigonometric functions \(\cos(3t)\) and \(\sin(3t)\) to the solution.
General Solution of Differential Equations
The general solution of a system of differential equations consisting of complex eigenvalues is formed by combining the oscillatory components imparted by these eigenvalues. For our system:
- The eigenvectors create linear combinations based on trigonometric functions.
- Each component in the solution needs both \(\sin(3t)\) and \(\cos(3t)\) terms.
- The exponential term \(e^{5t}\) modulates both trigonometric components.
- Real parts of eigenvectors with \(\cos(3t)\) and \(\sin(3t)\)
- Imaginary parts with the opposite trigonometric functions to maintain orthogonality