MATH 308 Lecture 37

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Lecture Notes

Section 7.9

Exercise 1c: Imaginary Eigenvectors

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle X'(t) = \begin{bmatrix}2&-5\\1&-2\end{bmatrix} \, X + \begin{bmatrix} -\cos{t} \\ \sin{t} \end{bmatrix}}

Eigenvalues: Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lambda \in \{ -i, i \}}

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle X_h(t) = c_1 \, \left( \vec{v}_1 \, \cos{t} - \vec{v}_2 \, \sin{t} \right) + c_2 \, \left( \vec{v}_2 \, \cos{t} + \vec{v}_1 \, \sin{t} \right)}

Guess for Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle X_p(t)} :

$X-p(t)=t\,{\vec {a}}\,\cos {t}+t\,{\vec {b}}\,\sin {t}+{\vec {c}}\,\cos {t}+{\vec {d}}\sin {t}$

Exercise 2a: Laplace Transform

$X'(t)={\begin{bmatrix}0&1\\1&0\end{bmatrix}}\,X+{\begin{bmatrix}t\\-1\end{bmatrix}}\quad \quad X(0)={\begin{bmatrix}2\\1\end{bmatrix}}$

${\begin{cases}{\mathcal {L}}\left\{x'\right\}={\mathcal {L}}\left\{y\right\}+{\mathcal {L}}\left\{t\right\}\\{\mathcal {L}}\left\{y'\right\}={\mathcal {L}}\left\{x\right\}+{\mathcal {L}}\left\{-1\right\}\end{cases}}$

${\begin{cases}s\,{\mathcal {L}}\left\{x\right\}-x(0)={\mathcal {L}}\left\{y\right\}+{\frac {1}{s^{2}}}\\s\,{\mathcal {L}}\left\{y\right\}-y(0)={\mathcal {L}}\left\{x\right\}+{\frac {1}{s}}\end{cases}}$

${\begin{cases}s\,{\mathcal {L}}\left\{x\right\}-2={\mathcal {L}}\left\{y\right\}+{\frac {1}{s^{2}}}\\s\,{\mathcal {L}}\left\{y\right\}-1={\mathcal {L}}\left\{x\right\}+{\frac {1}{s}}\end{cases}}$

${\mathcal {L}}\left\{x\right\}=s\,{\mathcal {L}}\left\{y\right\}-1+{\frac {1}{s}}$

${\mathcal {L}}\left\{y\right\}={\frac {1}{s^{2}\,\left(s^{2}-1\right)}}+{\frac {1}{s^{2}-1}}+{\frac {s}{s^{2}-1}}$

${\mathcal {L}}\left\{x\right\}=s\,\left({\frac {1}{s^{2}\,\left(s^{2}-1\right)}}+{\frac {1}{s^{2}-1}}+{\frac {s}{s^{2}-1}}\right)-1+{\frac {1}{s}}$

Take inverse Laplace transforms

Exercise 2b

Using matrix notation,

$X'(t)={\begin{bmatrix}2&-1\\3&-2\end{bmatrix}}\,X(t)+{\begin{bmatrix}1\\-1\end{bmatrix}}\,\mathrm {e} ^{t}\quad \quad X(0)={\begin{bmatrix}3\\0\end{bmatrix}}$

${\begin{aligned}{\mathcal {L}}\left\{X'\right\}&={\begin{bmatrix}2&-1\\3&-2\end{bmatrix}}\,{\mathcal {L}}\left\{X\right\}+{\begin{bmatrix}1,-1\end{bmatrix}}\,{\mathcal {L}}\left\{\mathrm {e} ^{t}\right\}\\s\,{\mathcal {L}}\left\{X\right\}-X(0)&={\begin{bmatrix}2&-1\\3&-2\end{bmatrix}}\,{\mathcal {L}}\left\{X\right\}+{\begin{bmatrix}1\\-1\end{bmatrix}}\,\left({\frac {1}{s-1}}\right)\\\left(s\,I-{\begin{bmatrix}2&-1\\3&-2\end{bmatrix}}\right)\,{\mathcal {L}}\left\{X\right\}&=X(0)+{\frac {1}{s-1}}\,{\begin{bmatrix}1\\-1\end{bmatrix}}\\{\mathcal {L}}\left\{X\right\}&={\begin{bmatrix}s-2&1\\-3&s+2\end{bmatrix}}^{-1}\,\left(X(0)+{\frac {1}{s-1}}\,{\begin{bmatrix}1\\-1\end{bmatrix}}\right)\\{\mathcal {L}}\left\{X\right\}&={\frac {1}{s^{2}-1}}\,{\begin{bmatrix}s+2&-1\\3&s-2\end{bmatrix}}\,{\begin{bmatrix}3+{\frac {1}{s-1}}\\-{\frac {1}{s-1}}\end{bmatrix}}\end{aligned}}$