MATH 308 Lecture 34

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Lecture Notes

Section 7.8

To solve a system with 1 eigenvalue with multiplicity 2:

Find space of eigenvectors for eigenvalue $\lambda$.

If there are two linearly independent eigenvectors ${\overrightarrow{v}}_1$ and ${\overrightarrow{v}}_2$, then

$X_1={\mathrm{e}}^{\lambda t}{\overrightarrow{v}}_1$ and $X_2={\mathrm{e}}^{\lambda t}{\overrightarrow{v}}_2$

are two linearly independent solutions

If there is no pair of linearly independent eigenvectors, find an eigenvector ${\overrightarrow{v}}_1$, and then find a second (generalized) eigenvector satisfying

$(A-\lambda I){\overrightarrow{v}}_2={\overrightarrow{v}}_1$

Then

$X_1={\mathrm{e}}^{\lambda t}{\overrightarrow{v}}_1$ and $X_2={\mathrm{e}}^{\lambda t}(t{\overrightarrow{v}}_1+{\overrightarrow{v}}_2)$

Exercise 2

Find general solution to $X^{\prime }=\left[\begin{array}{cc}4& -2\\ 8& -4\end{array}\right]X$.

Eigenvalues: $\lambda \in \{0,0\}$

Eigenvectors: $V=\left[\begin{array}{c}1\\ 2\end{array}\right]$

Therefore, the only solution we have so far is $X_1={\mathrm{e}}^{0t}\left[\begin{array}{c}1\\ 2\end{array}\right]=\left[\begin{array}{c}1\\ 2\end{array}\right]$

Let's find another solution in the form

$X_2=t{\mathrm{e}}^{0t}\left[\begin{array}{c}1\\ 2\end{array}\right]+{\mathrm{e}}^{0t}{\overrightarrow{v}}_2$

Where

${\overrightarrow{v}}_1=\left[\begin{array}{c}1\\ 2\end{array}\right]=(A-0I){\overrightarrow{v}}_2$

We find ${\overrightarrow{v}}_2=\left[\begin{array}{c}\frac{1}{2}\alpha +\frac{1}{2}\\ \alpha \end{array}\right]$

Let $\alpha =1$, and we get the vector ${\overrightarrow{v}}_2=\left[\begin{array}{c}1\\ 1\end{array}\right]$

Therefore, $X_2=t\left[\begin{array}{c}1\\ 2\end{array}\right]+\left[\begin{array}{c}1\\ 1\end{array}\right]$

and the general solution is

$X=c_1\left[\begin{array}{c}1\\ 2\end{array}\right]+c_2(t\left[\begin{array}{c}1\\ 2\end{array}\right]+\left[\begin{array}{c}1\\ 1\end{array}\right])$

Note: We could have taken any value for $\alpha$. The resulting general solution will look different, but the constants $c_1$ and $c_2$ will change to compensate.

Exercise 3

Find particular solution to initioal value problem $X^{\prime }=\left[\begin{array}{cc}1& -4\\ 4& -7\end{array}\right]X$, where $X(0)=\left[\begin{array}{c}3\\ 2\end{array}\right]$

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

Eigenvector: ${\overrightarrow{v}}_1=⟨1,1⟩$

Solve $(A+3I){\overrightarrow{v}}_2=⟨1,1⟩$ for $v_2=⟨\frac{1}{4}+\alpha ,\alpha ⟩$

Let $\alpha =\frac{3}{4}$, then ${\overrightarrow{v}}_2=⟨1,\frac{3}{4}⟩$

Therefore, our general solution is

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

Plug in $t=0$ and solve for $\overrightarrow{c}$:

$\left[\begin{array}{c}3\\ 2\end{array}\right]=c_1\left[\begin{array}{c}1\\ 1\end{array}\right]+c_2\left[\begin{array}{c}1\\ \frac{3}{4}\end{array}\right]$

$\overrightarrow{c}=⟨-1,4⟩$

So our particular solution is

$X=-1{\mathrm{e}}^{-3t}\left[\begin{array}{c}1\\ 1\end{array}\right]+4{\mathrm{e}}^{-3t}(t\left[\begin{array}{c}1\\ 1\end{array}\right]+\left[\begin{array}{c}1\\ \frac{3}{4}\end{array}\right])$