MATH 308 Lecture 39

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Lecture Notes

Review

Exercise 24.1

$X^{\prime }=\left[\begin{array}{cc}1& 1\\ 4& -2\end{array}\right]X+\left[\begin{array}{c}{\mathrm{e}}^{-2t}\\ -2{\mathrm{e}}^t\end{array}\right]$

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

Eigenvectors: ${\overrightarrow{x}}_1=⟨1,-4⟩{\overrightarrow{x}}_2=⟨1,1⟩$

Geneal Solution to Homogeneous Problem: $c_1{\mathrm{e}}^{-3t}\left[\begin{array}{c}1,-4\end{array}\right]+c_2{\mathrm{e}}^{2t}\left[\begin{array}{c}1\\ 1\end{array}\right]$

Undetermined Coefficients

Guess for particular solution: $X_p(t)={\overrightarrow{v}}_1{\mathrm{e}}^{-2t}+{\overrightarrow{v}}_2{\mathrm{e}}^t$

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:

$\begin{array}{rl}{X}^{\prime }(t)& =-2{\mathrm{e}}^{-2t}{\overrightarrow{v}}_1+{\mathrm{e}}^t{\overrightarrow{v}}_2\\ AX(t)+G(t)& =\left[\begin{array}{cc}1& 1\\ 4& -2\end{array}\right]{\overrightarrow{v}}_1+{\mathrm{e}}^t\left[\begin{array}{cc}1& 1\\ 4& -2\end{array}\right]{\overrightarrow{v}}_2+\left[\begin{array}{c}{\mathrm{e}}^{-2t}\\ -2{\mathrm{e}}^t\end{array}\right]\\ -2{\mathrm{e}}^{-2t}{\overrightarrow{v}}_1+{\mathrm{e}}^t{\overrightarrow{v}}_2& ={\mathrm{e}}^{-2t}\left[\begin{array}{c}a+b\\ a-2b\end{array}\right]+{\mathrm{e}}^t\left[\begin{array}{c}c+d\\ 4c-2d\end{array}\right]+{\mathrm{e}}^{-2t}\left[\begin{array}{c}1\\ 0\end{array}\right]+{\mathrm{e}}^t\left[\begin{array}{c}0\\ -2\end{array}\right]\\ \end{array}$

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

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

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

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

Variation of Parameters

Recall ${c_1}^{\prime }{X}_1+{c_2}^{\prime }{X}_2=G(t)$

Plugging in our values gives:

${c_1}^{\prime }{\mathrm{e}}^{-3t}\left[\begin{array}{c}1\\ -4\end{array}\right]+{c_2}^{\prime }{\mathrm{e}}^{2t}\left[\begin{array}{c}1\\ 1\end{array}\right]=\left[\begin{array}{c}{\mathrm{e}}^{-2t}\\ -2{\mathrm{e}}^t\end{array}\right]$

$\{\begin{array}{l}{c_1}^{\prime }{\mathrm{e}}^{-3t}+{c_2}^{\prime }{\mathrm{e}}^{2t}={\mathrm{e}}^{-2t}-4{c_1}^{\prime }{\mathrm{e}}^{-3t}+{c_2}^{\prime }{\mathrm{e}}^{2t}=-2{\mathrm{e}}^t\end{array}$

$\begin{array}{rl}{c_1}^{\prime }& =\frac{{\mathrm{e}}^t+2{\mathrm{e}}^{4t}}{5}\\ {c_2}^{\prime }& =({\mathrm{e}}^{-2t}-{\mathrm{e}}^{3t}\left(\frac{{\mathrm{e}}^t+2{\mathrm{e}}^{4t}}{5}\right)){\mathrm{e}}^{-2t}\end{array}$

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

Laplace Transform

$\begin{array}{rl}{ℒ}\{X^{\prime }(t)\}& =A{ℒ}\{X(t)\}+{ℒ}\{G(t)\}\\ s{ℒ}\{X(t)\}-X(0)& =A{ℒ}\{X(t)\}+\left[\begin{array}{c}\frac{1}{s+2}\\ -\frac{2}{s-1}\end{array}\right]\end{array}$

Let the initial condition $X(0)=⟨0,0⟩$ (we can pick whatever we want)

$\begin{array}{rl}(sI-A){ℒ}\{X(t)\}& =\left[\begin{array}{c}\frac{1}{s-2}\\ -\frac{2}{s-1}\end{array}\right]\\ \left[\begin{array}{cc}s-1& -1\\ -4& s+2\end{array}\right]\left[\begin{array}{c}{ℒ}\left\{x\right\}\\ {ℒ}\left\{y\right\}\end{array}\right]& =\left[\begin{array}{c}\frac{1}{s+2}\\ -\frac{2}{s-1}\end{array}\right]\end{array}$

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

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

Exercise 20.3

$X^{\prime }=\left[\begin{array}{ccc}1& 1& 1\\ 2& 1& -1\\ -8& -5& -3\end{array}\right]X$

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

…

Exercise 11

Use method of variation of parameters to find a particular solution of $y^{″}+4y^{\prime }+4y=t^{-2}{\mathrm{e}}^{-2t}$

Solution to Homogeneous equation:

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

Notice that now we know matrices:

$\left[\begin{array}{cc}{y}_1& y_2\\ {y_1}^{\prime }& {y_2}^{\prime }\end{array}\right]\left[\begin{array}{c}{c_1}^{\prime }\\ {c_2}^{\prime }\end{array}\right]=\left[\begin{array}{c}0\\ g(t)\end{array}\right]$