MATH 308 Lecture 39

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Lecture Notes

Review

Exercise 24.1

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

Eigenvalues: $\lambda \in \left\{-3,2\right\}$

Eigenvectors: ${\vec {x}}_{1}=\left\langle 1,-4\right\rangle \quad {\vec {x}}_{2}=\left\langle 1,1\right\rangle$

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

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

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

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

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

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

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

Exercise 20.3

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

Eigenvalues: $\lambda \in \{2,-1,-2\}$

…

Exercise 11

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

Solution to Homogeneous equation:

${\begin{aligned}0&=r^{2}+4r+4\\r&\in \{-2,-2\}\\y_{1}&=\mathrm {e} ^{-2t}\\y_{2}&=t\,\mathrm {e} ^{-2t}\\c_{1}&=-\int {\frac {g(t)\,y_{2}}{W\{y_{1},y_{2}\}}}\,\mathrm {d} t\\c_{2}&=\int {\frac {g(t)\,y_{1}}{W\{y_{1},y_{2}\}}}\,\mathrm {d} t\end{aligned}}$

Notice that now we know matrices:

${\begin{bmatrix}y_{1}&y_{2}\\y_{1}'&y_{2}'\end{bmatrix}}\,{\begin{bmatrix}c_{1}'\\c_{2}'\end{bmatrix}}={\begin{bmatrix}0\\g(t)\end{bmatrix}}$