MATH 308 Lecture 39

« previous | Tuesday, April 30, 2013 | next »

End Exam 4 content

Lecture Notes

Review

Exercise 24.1

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' = \begin{bmatrix}1&1\\4&-2\end{bmatrix} \, X + \begin{bmatrix}\mathrm{e}^{-2t}\\-2 \mathrm{e}^{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 \left\{ -3, 2 \right\}}

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 \vec{x}_1 = \left\langle 1,-4 \right\rangle \quad \vec{x}_2 = \left\langle 1,1 \right\rangle}

Geneal Solution to Homogeneous Problem: 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 c_1 \, \mathrm{e}^{-3t} \, \begin{bmatrix}1,-4\end{bmatrix} + c_2 \, \mathrm{e}^{2t} \, \begin{bmatrix}1\\1\end{bmatrix}}

Undetermined Coefficients

Guess for particular solution: $X_{p}(t)={\vec {v}}_{1}\,\mathrm {e} ^{-2t}+{\vec {v}}_{2}\,\mathrm {e} ^{t}$

Note: If the first term would have been $\mathrm {e} ^{2t}$, we would have to multiply it by $t$ and add a corrective term since $\mathrm {e} ^{2t}$ is already a solution

Derivation and Solution:

${\begin{aligned}X'(t)&=-2\mathrm {e} ^{-2t}\,{\vec {v}}_{1}+\mathrm {e} ^{t}{\vec {v}}_{2}\\A\,X(t)+G(t)&={\begin{bmatrix}1&1\\4&-2\end{bmatrix}}\,{\vec {v}}_{1}+\mathrm {e} ^{t}\,{\begin{bmatrix}1&1\\4&-2\end{bmatrix}}\,{\vec {v}}_{2}+{\begin{bmatrix}\mathrm {e} ^{-2t}\\-2\mathrm {e} ^{t}\end{bmatrix}}\\-2\mathrm {e} ^{-2t}\,{\vec {v}}_{1}+\mathrm {e} ^{t}\,{\vec {v}}_{2}&=\mathrm {e} ^{-2t}\,{\begin{bmatrix}a+b\\a-2b\end{bmatrix}}+\mathrm {e} ^{t}\,{\begin{bmatrix}c+d\\4c-2d\end{bmatrix}}+\mathrm {e} ^{-2t}{\begin{bmatrix}1\\0\end{bmatrix}}+\mathrm {e} ^{t}\,{\begin{bmatrix}0\\-2\end{bmatrix}}\\\end{aligned}}$

Coefficient of $\mathrm {e} ^{2t}$:

${\begin{cases}-2a=a+b+1-2b=4a-2b\end{cases}}$

Coefficient of $\mathrm {e} ^{t}$:

${\begin{cases}c=c+dd=4c-2d\end{cases}}$

Variation of Parameters

Recall $c_{1}'\,X_{1}+c_{2}'\,X_{2}=G(t)$

Plugging in our values gives:

$c_{1}'\,\mathrm {e} ^{-3t}\,{\begin{bmatrix}1\\-4\end{bmatrix}}+c_{2}'\,\mathrm {e} ^{2t}\,{\begin{bmatrix}1\\1\end{bmatrix}}={\begin{bmatrix}\mathrm {e} ^{-2t}\\-2\mathrm {e} ^{t}\end{bmatrix}}$

${\begin{cases}c_{1}'\,\mathrm {e} ^{-3t}+c_{2}'\,\mathrm {e} ^{2t}=\mathrm {e} ^{-2t}-4c_{1}'\,\mathrm {e} ^{-3t}+c_{2}'\,\mathrm {e} ^{2t}=-2\mathrm {e} ^{t}\end{cases}}$

${\begin{aligned}c_{1}'&={\frac {\mathrm {e} ^{t}+2\mathrm {e} ^{4t}}{5}}\\c_{2}'&=\left(\mathrm {e} ^{-2t}-\mathrm {e} ^{3t}\,\left({\frac {\mathrm {e} ^{t}+2\mathrm {e} ^{4t}}{5}}\right)\right)\mathrm {e} ^{-2t}\end{aligned}}$

Take the antiderivatives of each function to get the final constants.

Laplace Transform

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 \begin{align} \mathcal{L} \left\{ X'(t) \right\} &= A \, \mathcal{L} \left\{ X(t) \right\} + \mathcal{L} \left\{ G(t) \right\} \\ s \, \mathcal{L} \left\{ X(t) \right\} - X(0) &= A \, \mathcal{L} \left\{ X(t) \right\} + \begin{bmatrix} \frac{1}{s+2} \\ -\frac{2}{s-1} \end{bmatrix} \end{align}}

Let the initial condition 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(0) = \left\langle 0,0 \right\rangle} (we can pick whatever we want)

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 \begin{align} \left( s \, I - A \right) \, \mathcal{L} \left\{ X(t) \right\} &= \begin{bmatrix} \frac{1}{s-2} \\ -\frac{2}{s-1} \end{bmatrix} \\ \begin{bmatrix} s-1 & -1 \\-4 & s+2\end{bmatrix} \, \begin{bmatrix}\mathcal{L} \left\{ x \right\} \\ \mathcal{L} \left\{ y \right\}\end{bmatrix} &= \begin{bmatrix}\frac{1}{s+2} \\ -\frac{2}{s-1}\end{bmatrix} \end{align}}

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 \begin{cases} \mathcal{L} \left\{ y \right\} = \left( s-1 \right) \, \mathcal{L} \left\{ x \right\} - \frac{1}{s+2} \\ -4 \mathcal{L} \left\{ x \right\} + (s+2) \, \left( (s-1) \, \mathcal{L} \left\{ x \right\} - 1 \right) = -\frac{2}{s-1} \end{cases}}

Finally we get 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 \mathcal{L} \left\{ x \right\} = -\frac{2}{(s-1)(s^2 + s - 6} + \frac{1}{s^2+s-6}}

Exercise 20.3

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' = \begin{bmatrix}1&1&1\\2&1&-1\\-8&-5&-3\end{bmatrix} \, X}

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 \{2,-1,-2\}}

…

Exercise 11

Use method of variation of parameters to find a particular solution of 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 y'' +4y'+4y = t^{-2} \, \mathrm{e}^{-2t}}

Solution to Homogeneous equation:

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 \begin{align} 0 &= r^2 + 4r + 4 \\ r &\in \{-2, -2\} \\ y_1 &= \mathrm{e}^{-2t} \\ y_2 &= t \, \mathrm{e}^{-2t} \\ c_1 &= -\int \frac{g(t) \, y_2}{W\{y_1, y_2\}} \, \mathrm{d}t \\ c_2 &= \int \frac{g(t) \, y_1}{W\{y_1,y_2\}} \, \mathrm{d}t \end{align}}

Notice that now we know matrices:

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 \begin{bmatrix}y_1 & y_2 \\ y_1' & y_2' \end{bmatrix} \, \begin{bmatrix}c_1' \\ c_2' \end{bmatrix} = \begin{bmatrix}0 \\ g(t) \end{bmatrix}}