Homework Week 7 Solutions

pg 359 \#1.

Let $x_{1}=u$, $x_{2}=u^{\prime}$

Then

$x_{1}^{\prime}=x_{2}$,

$x_{2}^{\prime}=u^{\prime\prime}=-0.5u^{\prime}-2u=-0.5x_{2}-2x_{1}=-2x_{1}-0.5x_{2}$.

pg 359 \#5

Let $x_{1}=u$, $x_{2}=u^{\prime}$

Then

$x_{1}^{\prime}=x_{2}$

$x_{2}^{\prime}=-4x_{1}-0.25x_{2}$

$x_{1}(0)=u(0)=1$

$x_{2}(0)=u^{\prime}(0)=-2$

pg 360 \#8

Solving the first equation for $2x_{2}$ we get

$2x_{2}=3x_{1}-x_{1}^{\prime}$

I try to avoid fractions in my solution.

Then $2x_{2}^{\prime}=3x_{1}^{\prime}-x_{1}^{\prime\prime}$

Then $3x_{1}^{\prime}-x_{1}^{\prime\prime}=4x_{1}-2(2x_{2})=4x_{1}-2(3x_{1}-x_{1}^{\prime})=-2x_{1}+2x_{1}^{\prime}$

Then $0=x_{1}^{\prime\prime}-x_{1}^{\prime}-2x_{1}$

Thus the characteristic factors as $(r-2)(r+1)$

I leave the rest for you.

pg 371 \#2

Answer is in the back of the book.

pg 373 \#22

$x^{\prime}=\left(\begin{array}{c}
4\\
2\end{array}\right)2e^{2t}$

$\left(\begin{array}{cc}
3 & -2\\
2 & -2\end{array}\right)\left(\begin{array}{c}
4\\
2\end{array}\right)e^{2t}=\left(\begin{array}{c}
3\cdot4+(-2)\cdot2\\
2\cdot4+(-2)\cdot2\end{array}\right)e^{2t}=\left(\begin{array}{c}
8\\
4\end{array}\right)e^{2t}=\left(\begin{array}{c}
4\\
2\end{array}\right)2e^{2t}$

Great!

pg 384 \#16

$\left(\begin{array}{cc}
5 & -1\\
3 & 1\end{array}\right)-\lambda\left(\begin{array}{cc}
1 & 0\\
0 & 1\end{array}\right)=\left(\begin{array}{cc}
5-\lambda & -1\\
3 & 1-\lambda\end{array}\right)$

$\left|\begin{array}{cc}
5-\lambda & -1\\
3 & 1-\lambda\end{array}\right|=(5-\lambda)(1-\lambda)-(-1)\cdot3=\lambda^{2}-6\lambda+8$

$0=(\lambda-4)(\lambda-2)$

A) $\lambda=4$

$\left(\begin{array}{cc}
1 & -1\\
3 & -3\end{array}\right)\to\left(\begin{array}{cc}
1 & -1\\
0 & 0\end{array}\right)\to x_{1}-x_{2}=0$

The row reduced matrix is rank one, so we have one free variable.
Let $x_{2}=a$. Then $x_{1}=a$. Then we have the eigenspace $\left(\begin{array}{c}
a\\
a\end{array}\right)$. Thus one eigenvector for $\lambda=4$ is $\left(\begin{array}{c}
1\\
1\end{array}\right)$.

B) $\lambda=2$

$\left(\begin{array}{cc}
3 & -1\\
3 & -1\end{array}\right)\to\left(\begin{array}{cc}
3 & -1\\
0 & 0\end{array}\right)\to3x_{1}-x_{2}=0$

As with A) above, we have one free variable. Let $x_{1}=a$. Then
$x_{2}=3a$. Then we have the eigenspace $\left(\begin{array}{c}
a\\
3a\end{array}\right)$. So one eigenvector for $\lambda=2$ is $\left(\begin{array}{c}
1\\
3\end{array}\right)$.

Note with both A and B you can choose the other variable to be the
free variable and you would get the same eigenspace, and same eigenvector
depending on your value for the free variable. Explicitly,

let $x_{2}=b$. Then $x_{1}=\frac{1}{3}b$. Then we have the eigenspace
$\left(\begin{array}{c}
\frac{1}{3}b\\
b\end{array}\right)$. I leave it to you to understand that $\left\{ \left(\begin{array}{c}
\frac{1}{3}b\\
b\end{array}\right)\mid b\in\mathbb{R}\right\} =\left\{ \left(\begin{array}{c}
a\\
3a\end{array}\right)\mid a\in\mathbb{R}\right\} $.

pg 384 \#21

$\det\left(\begin{array}{cc}
-3-\lambda & 3/4\\
-5 & 1-\lambda\end{array}\right)=\lambda^{2}+2\lambda-3+\frac{15}{4}=\lambda^{2}+2\lambda+\frac{3}{4}=0$

This factors into $\left(\lambda+\frac{3}{2}\right)\left(\lambda+\frac{1}{2}\right)$.
You can do this in your head, or use the quadractic formula, or instead
you can multiply by $4$ and work with $4\lambda^{2}+8\lambda+3=(2\lambda+3)(2\lambda+1)$.
In any case the end result is the same. $\lambda=-\frac{3}{2}$ or
$-\frac{1}{2}$.

pg 398 \#1a

Find the eigenvalues to be $\lambda=-1$ and $\lambda=2$.

Find corresponding eigenvectors to be $\left(\begin{array}{c}
1\\
2\end{array}\right)$ and $\left(\begin{array}{c}
2\\
1\end{array}\right)$.

Thus we have a general solution: $c_{1}\left(\begin{array}{c}
1\\
2\end{array}\right)e^{-1\cdot t}+c_{2}\left(\begin{array}{c}
2\\
1\end{array}\right)e^{2t}$

As $t\to\infty$ $e^{-t}\to0$ and so all solutions approach the line
$l(a)=\left(\begin{array}{c}
2\\
1\end{array}\right)a$.

pg 399 \#15

Find the general solution to be $c_{1}\left(\begin{array}{c}
1\\
3\end{array}\right)e^{2\cdot t}+c_{2}\left(\begin{array}{c}
1\\
1\end{array}\right)e^{4\cdot t}$

Let me know if you have difficulty finding the general solution.

Then set $t=0$ and solve for $c_{1}$ and $c_{2}$.

pg 399 \#24

In section 6 I was slightly stuck and I figured out what I did wrong.
Now that I think about it, I might have made a mistake in section
3. In any case, I should have written down the general solution: $c_{1}\left(\begin{array}{c}
-1\\
2\end{array}\right)e^{-t}+c_{2}\left(\begin{array}{c}
1\\
2\end{array}\right)e^{-2t}$ for part a). Then explain that for some $c_{1},c_{2}$ we get a solution
passing through $(2,3)$. If we want, we can assume this happens at
$t=0$ and solve for $c_{1}$and $c_{2}$, but the picture will be
the same as I did it in class.