Question

Find a fundamental matrix for the system and write the general solution as a matrix. If initial values are given, solve the initial value problem. x1=2x1+x2x3,x2=3x12x2, x3=3x1+x23x3;x1(0)=1,x2(0)=7,x3(0)=3x_{1}^{\prime}=2 x_{1}+x_{2}-x_{3}, x_{2}^{\prime}=3 x_{1}-2 x_{2}, \ x_{3}^{\prime}=3 x_{1}+x_{2}-3 x_{3} ; x_{1}(0)=1, x_{2}(0)=7, x_{3}(0)=3

Solution

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Answered 3 years ago
Answered 3 years ago
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Let

X=(x1x2x3) and X=(x1x2x3)\begin{align*}\mathbf{X}=\begin{pmatrix} x_1\\ x_2\\x_3 \end{pmatrix}\quad \text{ and }\quad \mathbf{X}'=\begin{pmatrix} x_1'\\ x_2'\\x_3' \end{pmatrix} \end{align*}

Let's rewrite the given system in terms of matrices:

X=AX\mathbf{X}'=\mathbf{AX}

with A\mathbf{A} being a real, constant, 3×33\times 3 matrix.

X=(x1x2x3)=(2x1+x2x33x12x23x1+x23x3)=(2ii1132ii03ii13)(x1x2x3)\begin{align*}\mathbf{X}'=\begin{pmatrix} x_1'\\ x_2'\\x_3'\end{pmatrix}&=\mqty(2x_1+x_2-x_3 \\ 3x_1-2x_2\\3x_1+x_2-3x_3)=\mqty (2&\phantom{ii}1&-1\\3&-2&\phantom{ii}0\\ 3&\phantom{ii}1&-3)\begin{pmatrix} x_1\\ x_2\\x_3 \end{pmatrix} \end{align*}

Thus,

X=(2ii1132ii03ii13)AX\begin{align*}\mathbf{X}'=\underbrace{\mqty (2&\phantom{ii}1&-1\\3&-2&\phantom{ii}0\\ 3&\phantom{ii}1&-3)}_{\mathbf{A}} \mathbf{X} \end{align*}

For our example,

A=(2ii1132ii03ii13)\begin{align*}\mathbf{A}=\mqty (2&\phantom{ii}1&-1\\3&-2&\phantom{ii}0\\ 3&\phantom{ii}1&-3)\end{align*}

The solution X\mathbf{X} is given in the form X=Eeλt\mathbf{X=E}e^{\lambda t} where λ\lambda is an eigenvalue of A\mathbf{A} with associated eigenvector E\mathbf{E}.

AE=λE\mathbf{AE=}\lambda \mathbf{E}

Recall, λ\lambda is an eigenvalue\textbf{eigenvalue} of A\mathbf{A} if and only if it's a root of charateristic polynomial of A\mathbf{A}. λ\lambda is obtain by calculating

pA(λ)=λInA=0p_A(\lambda)=|\lambda \mathbf{I_n}-\mathbf{A}|=0

The degree of pA(λ)p_A(\lambda ) determines the number of eigenvalues.Any nontrivial solution E\mathbf{E} of

(λInA)X=0(\lambda \mathbf{I}_n-\mathbf{A})\mathbf{X}=0

is an eigenvalue of A\mathbf{A} associated with λ\lambda.

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