Math Problem Statement

Find the general solution to x′ = Ax for A = 1 −3 1 1

Solution

To find the general solution to the system of differential equations given by x=Ax\mathbf{x}' = A\mathbf{x} where

A=(1311),A = \begin{pmatrix} 1 & -3 \\ 1 & 1 \end{pmatrix},

we follow these steps:

1. Find the Eigenvalues of AA:

The eigenvalues λ\lambda of matrix AA are found by solving the characteristic equation:

det(AλI)=0\det(A - \lambda I) = 0

where II is the identity matrix and det\det denotes the determinant. The characteristic equation is:

det((1311)λ(1001))=det(1λ311λ)=0\det\left(\begin{pmatrix} 1 & -3 \\ 1 & 1 \end{pmatrix} - \lambda \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\right) = \det\begin{pmatrix} 1 - \lambda & -3 \\ 1 & 1 - \lambda \end{pmatrix} = 0

Calculate the determinant:

(1λ)(1λ)(3)(1)=(λ1)2+3=λ22λ+4=0(1 - \lambda)(1 - \lambda) - (-3)(1) = (\lambda - 1)^2 + 3 = \lambda^2 - 2\lambda + 4 = 0

This quadratic equation is:

λ22λ+4=0\lambda^2 - 2\lambda + 4 = 0

2. Solve for Eigenvalues:

To solve the quadratic equation:

λ=2±(2)24(1)(4)2(1)\lambda = \frac{2 \pm \sqrt{(-2)^2 - 4(1)(4)}}{2(1)}

λ=2±4162\lambda = \frac{2 \pm \sqrt{4 - 16}}{2}

λ=2±122\lambda = \frac{2 \pm \sqrt{-12}}{2}

λ=1±i3\lambda = 1 \pm i\sqrt{3}

So the eigenvalues are:

λ1=1+i3,λ2=1i3\lambda_1 = 1 + i\sqrt{3}, \quad \lambda_2 = 1 - i\sqrt{3}

3. Find the Eigenvectors:

For each eigenvalue λi\lambda_i, solve (AλiI)v=0(A - \lambda_i I)\mathbf{v} = 0 to find the corresponding eigenvector vi\mathbf{v}_i.

For λ1=1+i3\lambda_1 = 1 + i\sqrt{3}:

((1311)(1+i3)(1001))v=0\left(\begin{pmatrix} 1 & -3 \\ 1 & 1 \end{pmatrix} - (1 + i\sqrt{3}) \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\right) \mathbf{v} = 0

(i331i3)v=0\begin{pmatrix} -i\sqrt{3} & -3 \\ 1 & -i\sqrt{3} \end{pmatrix} \mathbf{v} = 0

The eigenvector v1\mathbf{v}_1 can be found by solving this system. Similarly, we find v2\mathbf{v}_2 for λ2=1i3\lambda_2 = 1 - i\sqrt{3}.

4. Form the General Solution:

The general solution is given by:

x(t)=c1eλ1tv1+c2eλ2tv2\mathbf{x}(t) = c_1 e^{\lambda_1 t} \mathbf{v}_1 + c_2 e^{\lambda_2 t} \mathbf{v}_2

Given the complex eigenvalues, the general solution can also be written in terms of real functions using Euler's formula:

x(t)=et[c1(Re(v1)Im(v1))cos(3t)+c2(Re(v1)Im(v1))sin(3t)]\mathbf{x}(t) = e^{t} \left[ c_1 \begin{pmatrix} \text{Re}(\mathbf{v}_1) \\ \text{Im}(\mathbf{v}_1) \end{pmatrix} \cos(\sqrt{3}t) + c_2 \begin{pmatrix} \text{Re}(\mathbf{v}_1) \\ \text{Im}(\mathbf{v}_1) \end{pmatrix} \sin(\sqrt{3}t) \right]

Here, Re(v1)\text{Re}(\mathbf{v}_1) and Im(v1)\text{Im}(\mathbf{v}_1) are the real and imaginary parts of the eigenvector v1\mathbf{v}_1.

Final General Solution:

x(t)=et[c1vrcos(3t)+c2visin(3t)]\mathbf{x}(t) = e^t \left[ c_1 \mathbf{v}_r \cos(\sqrt{3}t) + c_2 \mathbf{v}_i \sin(\sqrt{3}t) \right]

where vr\mathbf{v}_r and vi\mathbf{v}_i are real vectors derived from the real and imaginary parts of the eigenvectors.

Would you like more details on the steps, or do you have any questions?


Relative Questions:

  1. How do you find eigenvectors for a given matrix?
  2. What is Euler's formula, and how is it applied in solving differential equations?
  3. Can eigenvalues be real, and what does that imply for the solution?
  4. How do the real and imaginary parts of eigenvalues affect the general solution?
  5. What are the physical interpretations of complex eigenvalues in differential systems?

Tip: When solving systems of differential equations with complex eigenvalues, always express your final solution using real functions (sine and cosine) to make it easier to interpret and apply.

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Math Problem Analysis

Mathematical Concepts

Linear Algebra
Eigenvalues and Eigenvectors
Differential Equations
Complex Numbers

Formulas

Determinant of a 2x2 matrix: det(A) = ad - bc
Quadratic formula: λ = (-b ± √(b² - 4ac)) / 2a
Euler's formula: e^(ix) = cos(x) + i*sin(x)

Theorems

Eigenvalue theorem for solving systems of linear differential equations
Euler's formula for complex exponentials

Suitable Grade Level

College level (Undergraduate)