MATH307 December 2005
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Question 06 (b)

Consider the differential equation $x'(t)={\begin{bmatrix}1&\alpha \\1&1\end{bmatrix}}x(t)$
(b) Find the matrix exponential $\displaystyle e^{tA}$ when $\displaystyle \alpha =4$

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Solution

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$x'(t)={\begin{bmatrix}1&4\\1&1\end{bmatrix}}x(t)$
Find the eigenpairs to diagonalize the matrix!
${\begin{aligned}{\text{det}}{\begin{bmatrix}\lambda +1&4\\1&\lambda +1\end{bmatrix}}=\lambda ^{2}+2\lambda +1+4\end{aligned}}$
Setting the above to zero and solving yields
${\begin{aligned}&\lambda ={\frac {2\pm {\sqrt {2^{2}4(5)}}}{2}}\\&={\frac {2\pm {\sqrt {16}}}{2}}\\&={\frac {2\pm 4i}{2}}\\&=1\pm 2i\end{aligned}}$
Let $\displaystyle \lambda _{1}=1+2i,\lambda _{2}=12i$
Find corresponding eigenvectors
$(\lambda _{1}IA){\bar {x}}={\bar {0}}$ ${\begin{bmatrix}2i&4\\1&2i\end{bmatrix}}{\begin{bmatrix}x_{1}\\x_{2}\end{bmatrix}}={\begin{bmatrix}0\\0\end{bmatrix}}$
From the second row: $x_{1}=2ix_{2}$ –> corresponding eigenvector
Find corresponding eigenvectors
$(\lambda _{2}IA){\bar {x}}={\bar {0}}$ ${\begin{bmatrix}2i&4\\1&2i\end{bmatrix}}{\begin{bmatrix}x_{1}\\x_{2}\end{bmatrix}}={\begin{bmatrix}0\\0\end{bmatrix}}$
From the second row: $x_{1}=2ix_{2}$ –> corresponding eigenvector
Therefore
$S={\begin{bmatrix}2i&2i\\1&1\end{bmatrix}}$
$D={\begin{bmatrix}1+2i&0\\0&12i\end{bmatrix}}$
$S^{}1=(4i)^{1}{\begin{bmatrix}1&2i\\1&2i\end{bmatrix}}$
So
$e^{tA}=(4i)^{1}{\begin{bmatrix}2i&2i\\1&1\end{bmatrix}}{\begin{bmatrix}e^{t(1+2i)}&0\\0&e^{t(12i)}\end{bmatrix}}{\begin{bmatrix}1&2i\\1&2i\end{bmatrix}}$
NOTE: please ignore the <span>, <>.. not sure how to get rid of them.


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