Math Problem Statement

Solve the IVP: X0 = AX, where A =   −2 1 0 −5 0 0 0 3 −2   , and X(0) =   −2 −1 2   .

Solution

To solve the initial value problem (IVP) given by dXdt=AX\frac{dX}{dt} = AX, where

A=(210500032)A = \begin{pmatrix} -2 & 1 & 0 \\ -5 & 0 & 0 \\ 0 & 3 & -2 \end{pmatrix}

and X(0)=(212)X(0) = \begin{pmatrix} -2 \\ -1 \\ 2 \end{pmatrix}, we need to find the general solution to this system of differential equations and then use the initial condition to find the specific solution.

Step 1: Find the Eigenvalues of AA

To find the eigenvalues of AA, we solve the characteristic equation:

det(AλI)=0\text{det}(A - \lambda I) = 0

where II is the identity matrix. Thus,

AλI=(2λ105λ0032λ)A - \lambda I = \begin{pmatrix} -2-\lambda & 1 & 0 \\ -5 & -\lambda & 0 \\ 0 & 3 & -2-\lambda \end{pmatrix}

The determinant is computed as:

[ \begin{vmatrix} -2-\lambda & 1 & 0 \ -5 & -\lambda & 0 \ 0 & 3 & -2-\lambda \end{vmatrix} = (-2-\lambda)[(-\lambda)(-2-\lambda) - (0)(3)]

  • 1[-5(0) - (0)(3-\lambda)]
  • 0 ]

This simplifies to:

(2λ)(λ2+2λ)=0(-2-\lambda)(\lambda^2 + 2\lambda) = 0

This characteristic polynomial, λ3+2λ2=0\lambda^3 + 2\lambda^2 = 0, factors to:

λ2(λ+2)=0\lambda^2 (\lambda + 2) = 0

Thus, the eigenvalues are λ=0\lambda = 0 (with algebraic multiplicity 2), and λ=2\lambda = -2.

Step 2: Find the Eigenvectors

For λ=0\lambda = 0:

Solve (A0I)X=0(A - 0I)X = 0:

AX=(210500032)(xyz)=(000)AX = \begin{pmatrix} -2 & 1 & 0 \\ -5 & 0 & 0 \\ 0 & 3 & -2 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}

This gives the system of equations:

2x+y=0-2x + y = 0

5x=0-5x = 0

3y2z=03y - 2z = 0

From 5x=0-5x = 0, we get x=0x = 0. From 2x+y=0-2x + y = 0, with x=0x=0, we get y=0y = 0. From 3y2z=03y - 2z = 0, with y=0y=0, we get z=0z=0.

Thus, corresponding to λ=0\lambda = 0, a single eigenvector v1=(001)\mathbf{v}_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}.

For λ=2\lambda = -2:

Solve (A+2I)X=0(A + 2I)X = 0:

(010520030)(xyz)=(000)\begin{pmatrix} 0 & 1 & 0 \\ -5 & 2 & 0 \\ 0 & 3 & 0 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}

From this, clearly, y=0y = 0. The system reduces to:

5x=0-5x = 0 3y=03y = 0

Setting x=1x = 1, the eigenvector associated is v2=(100)\mathbf{v}_2 = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}.

Step 3: General Solution

The general solution is a combination of matrices and eigenvectors:

X(t)=c1e2t(100)+c2(100)+c3t(001)X(t) = c_1 e^{-2t} \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} + c_2 \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} + c_3 t \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}

With initial condition (212)=c1(100)+c2(100)+c3(001)\begin{pmatrix} -2 \\ -1 \\ 2 \end{pmatrix} = c_1 \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} + c_2 \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} + c_3 \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}.

From matched terms, c2=2c_2 = -2, and c3=2c_3 = 2. This implies c1=0c_1=0.

Thus, X(t)=2(100)+2t(001)X(t) = -2 \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} + 2t \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} gives the complete solution for t0t \geq 0.

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Math Problem Analysis

Mathematical Concepts

Linear Algebra
Differential Equations
Eigenvalues and Eigenvectors
Matrix Exponentiation

Formulas

Matrix Exponential Solution: X(t) = e^(At)X(0)
Characteristic Polynomial: det(A - λI) = 0

Theorems

Eigenvalue Decomposition Theorem
Solution of Linear Differential Equations using Eigenvalues

Suitable Grade Level

Undergraduate - Mathematics or Engineering