Math Problem Statement
Solve (x_1)' = x_1 -5 x_2 (x_2)' = x_1 + 3 x_2
Solution
To solve the system of differential equations:
(x_1)' = x_1 - 5x_2 \\ (x_2)' = x_1 + 3x_2 \end{cases}$$ we can express the system in matrix form: $$\frac{d}{dt} \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = \begin{pmatrix} 1 & -5 \\ 1 & 3 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \end{pmatrix}$$ Let $$\mathbf{x}(t) = \begin{pmatrix} x_1(t) \\ x_2(t) \end{pmatrix}$$, and let $$A = \begin{pmatrix} 1 & -5 \\ 1 & 3 \end{pmatrix}$$. The system can be written as: $$\frac{d\mathbf{x}}{dt} = A\mathbf{x}$$ ### Step 1: Find the Eigenvalues of the Matrix $$A$$ The eigenvalues $$\lambda$$ of the matrix $$A$$ are found by solving the characteristic equation: $$\det(A - \lambda I) = 0$$ where $$I$$ is the identity matrix. Thus, $$\det \begin{pmatrix} 1-\lambda & -5 \\ 1 & 3-\lambda \end{pmatrix} = (1-\lambda)(3-\lambda) - (-5)(1) = 0$$ Expanding the determinant: $$(1-\lambda)(3-\lambda) + 5 = \lambda^2 - 4\lambda + 3 + 5 = \lambda^2 - 4\lambda + 8 = 0$$ The eigenvalues are the solutions to the quadratic equation: $$\lambda^2 - 4\lambda + 8 = 0$$ Using the quadratic formula: $$\lambda = \frac{4 \pm \sqrt{(-4)^2 - 4 \times 1 \times 8}}{2 \times 1} = \frac{4 \pm \sqrt{16 - 32}}{2} = \frac{4 \pm \sqrt{-16}}{2} = \frac{4 \pm 4i}{2} = 2 \pm 2i$$ So the eigenvalues are $$\lambda_1 = 2 + 2i$$ and $$\lambda_2 = 2 - 2i$$. ### Step 2: Find the Eigenvectors For $$\lambda_1 = 2 + 2i$$, we solve: $$(A - \lambda_1 I)\mathbf{v} = 0$$ This leads to: $$\begin{pmatrix} 1-(2+2i) & -5 \\ 1 & 3-(2+2i) \end{pmatrix} \begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = \begin{pmatrix} -1-2i & -5 \\ 1 & 1-2i \end{pmatrix} \begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = 0$$ This simplifies to: $$\begin{cases} (-1 - 2i)v_1 - 5v_2 = 0 \\ v_1 + (1 - 2i)v_2 = 0 \end{cases}$$ This gives the eigenvector $$\mathbf{v}_1$$. We follow similar steps for $$\lambda_2 = 2 - 2i$$ to find $$\mathbf{v}_2$$. ### Step 3: Form the General Solution The general solution is: $$\mathbf{x}(t) = c_1 e^{(2+2i)t} \mathbf{v}_1 + c_2 e^{(2-2i)t} \mathbf{v}_2$$ ### Step 4: Simplify the Solution Using Euler's formula $$e^{i\theta} = \cos(\theta) + i\sin(\theta)$$, the general solution can be expressed in real terms as: $$x_1(t) = e^{2t} \left[ c_1 \cos(2t) - c_2 \sin(2t) \right]$$ $$x_2(t) = e^{2t} \left[ c_1 \cos(2t) - c_2 \sin(2t) \right]$$ where $$c_1$$ and $$c_2$$ are constants determined by initial conditions. Would you like more details or have any questions? Here are 5 related questions: 1. How do you determine the eigenvectors of a matrix for complex eigenvalues? 2. What is the physical interpretation of the solution to this system of differential equations? 3. How do initial conditions affect the constants in the general solution? 4. Can we solve this system using another method, such as diagonalization? 5. What does the presence of complex eigenvalues indicate about the behavior of the system over time? **Tip:** Always check if a matrix is diagonalizable before attempting to solve using eigenvalues and eigenvectors. This can significantly simplify the problem.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Differential Equations
Matrix Operations
Eigenvalues and Eigenvectors
Complex Numbers
Exponential Functions
Formulas
Characteristic equation
Quadratic formula
Euler's formula
Theorems
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Suitable Grade Level
Advanced Undergraduate
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