Page 421 Problem 15
Tuesday, April 5, 2011
Phase Portraits
Alright, so here's a visual aid regarding phase portraits. I included most cases. I omitted the case when one of the eigenvalues is zero. Hopefully your notes on that case are clear.
You can (and probably should) click on the scanned pages to make them bigger (there are two pages).
You can (and probably should) click on the scanned pages to make them bigger (there are two pages).
Monday, April 4, 2011
Relax and Other
Hey All,
So it was brought to my attention the page loads slowly. This was my mistake as I forgot to cut the posts short when they're viewed from the main page. So seven long posts are trying to load. I've added the HTML so that you have to click on the post if you want to see the full post. This will help with load times.
In any case, relax a little before the test and listen to some Demetri Martin (comedian):
And Flight of the Conchords:
So it was brought to my attention the page loads slowly. This was my mistake as I forgot to cut the posts short when they're viewed from the main page. So seven long posts are trying to load. I've added the HTML so that you have to click on the post if you want to see the full post. This will help with load times.
In any case, relax a little before the test and listen to some Demetri Martin (comedian):
And Flight of the Conchords:
Sunday, April 3, 2011
Some Answers and One Question
[This post was missing one of the solutions, it just got updated 12:38 AM. I think I'll call it a night.]
The Laplace transform is a linear operator.
So we can apply the Laplace transform to each term separately.
The first term is the transform of $t^{3}$ but shifted by $c=-2$.
Thus we have $\frac{3!}{(s+2)^{4}}$.
The second term uses the convolution.
The transform of the integral is the product of the transforms.
So $\mathcal{L}\{e^{2t}\}\cdot\mathcal{L}\{\sin3t\}=\frac{1}{s-2}\cdot\frac{3}{s^{2}+9}$
The last term can be done as follows:
$\mathcal{L}\{t^{2}\delta(t-5)\}=\int_{0}^{\infty}t^{2}\delta(t-5)e^{-st}dt=5^{2}e^{-5s}=25e^{-5s}$
The Laplace transform is a linear operator.
So we can apply the Laplace transform to each term separately.
The first term is the transform of $t^{3}$ but shifted by $c=-2$.
Thus we have $\frac{3!}{(s+2)^{4}}$.
The second term uses the convolution.
The transform of the integral is the product of the transforms.
So $\mathcal{L}\{e^{2t}\}\cdot\mathcal{L}\{\sin3t\}=\frac{1}{s-2}\cdot\frac{3}{s^{2}+9}$
The last term can be done as follows:
$\mathcal{L}\{t^{2}\delta(t-5)\}=\int_{0}^{\infty}t^{2}\delta(t-5)e^{-st}dt=5^{2}e^{-5s}=25e^{-5s}$
Page 409 Problem 1
Page 409 Problem 1
$\left(\begin{array}{cc}
3 & -2\\
4 & -1\end{array}\right)$ gives $\lambda^{2}-(2)\lambda+(-3-(-8))=\lambda^{2}-2\lambda+5$
Solving for the roots we have $\lambda=\frac{2\pm\sqrt{4-20}}{2}=1\pm2i$
$\left(\begin{array}{cc}
3 & -2\\
4 & -1\end{array}\right)$ gives $\lambda^{2}-(2)\lambda+(-3-(-8))=\lambda^{2}-2\lambda+5$
Solving for the roots we have $\lambda=\frac{2\pm\sqrt{4-20}}{2}=1\pm2i$
Some Review [Chapter 7.9]
First, in the previous post, I made an error. It has been corrected. The correction is also shown here:
Compute the inverse Laplace transform of
$\frac{e^{-2s}}{s^{1/2}}$
[CORRECTION: It should be -2s instead of 2s]
Compute the inverse Laplace transform of
$\frac{e^{-2s}}{s^{1/2}}$
[CORRECTION: It should be -2s instead of 2s]
Some Questions and Some Review Part 1
I issue the warning to take what I say with a grain of salt.
\textasciitilde{}
Find the Laplace transform of
$f(t)=t^{3}e^{-2t}+\int_{0}^{t}e^{2s}\sin3(t-s)ds+t^{2}\delta(t-5)$
\textasciitilde{}
Compute the inverse Laplace transform of
$\frac{s-3}{(s-3)^{2}+4}\cdot\frac{e^{-2s}}{s^{3}}$
\textasciitilde{}
Compute the inverse Laplace transform of
$\frac{e^{-2s}}{s^{1/2}}$
[CORRECTION: It should be -2s instead of 2s]
\textasciitilde{}
Practice using Laplace transforms to solve initial value problems:
pg 320 \#11-23, pg 343 \#1-12, etc.
\textasciitilde{}
The key relationship regarding the fundamental matrix $\Phi$ with
an arbitrary fundamental matrix $\Psi$ is $\Phi(t)=\Psi(t)\Psi^{-1}(t_{0})$.
Thus if you need to compute $\Phi$, simply compute $\Psi(t_{0})$,
then compute it's inverse $\Psi^{-1}(t_{0})$, and then comptue the
product, $\Psi(t)\Psi^{-1}(t_{0})$.
For a two-by-two matrix, you can remember the formula for the inverse
of a matrix $\left(\begin{array}{cc}
a & b\\
c & d\end{array}\right)$ is $\frac{1}{ad-bc}\left(\begin{array}{cc}
d & -b\\
-c & a\end{array}\right)$.
In general you can consider the augmented matrix and row reduce. For
the two-by-two case we have:
$\left(\begin{array}{ccccc}
a & b & | & 1 & 0\\
c & d & | & 0 & 1\end{array}\right)$ which when we row reduce the left block to the identity we get the
inverse on the right block.
\textasciitilde{}
I don't know if the problem will come up, but if asked, what is $\exp(At)$?
Then the answer will be the fundamental matrix $\Phi$ that satisfies
$\Phi^{\prime}=A\Phi$, $\Phi(0)=I$ (page 417).
Find this $\Phi$by first getting $\Psi$ from the solutions for $x^{\prime}=Ax$.
Then $\Phi(t)=\Psi(t)\Psi^{-1}(0)$
\textasciitilde{}
Let's talk about two-by-two matrices $A$. If $A$ has a repeated
eigenvalue $\lambda$ with two eigenvectors implies $A$ is of the
form $\left(\begin{array}{cc}
\lambda & 0\\
0 & \lambda\end{array}\right)$. For a repeated eigenvalue $\lambda$ with only one eigenvector ,
then we have two solutions to the equation $x^{\prime}=Ax$, one of
the form $\xi e^{2t}$ and the second of the form $\xi te^{2t}+\eta e^{2t}$
where $(A-\lambda I)\xi=0$ and $(A-\lambda I)\eta=\xi$.
Note: $(A-\lambda I)\xi=0$ implies that $(A-\lambda I)\eta=\xi$
has a solution.
Sidenote: What I was thinking about in class was related, but not
useful for computation. $(A-\lambda I)^{2}\xi=(A-\lambda I)\eta=0$.
\textasciitilde{}
I'm gonna be back in about four hours, so until then, hopefully you
can give yourself problems.
\textasciitilde{}
Find the Laplace transform of
$f(t)=t^{3}e^{-2t}+\int_{0}^{t}e^{2s}\sin3(t-s)ds+t^{2}\delta(t-5)$
\textasciitilde{}
Compute the inverse Laplace transform of
$\frac{s-3}{(s-3)^{2}+4}\cdot\frac{e^{-2s}}{s^{3}}$
\textasciitilde{}
Compute the inverse Laplace transform of
$\frac{e^{-2s}}{s^{1/2}}$
[CORRECTION: It should be -2s instead of 2s]
\textasciitilde{}
Practice using Laplace transforms to solve initial value problems:
pg 320 \#11-23, pg 343 \#1-12, etc.
\textasciitilde{}
The key relationship regarding the fundamental matrix $\Phi$ with
an arbitrary fundamental matrix $\Psi$ is $\Phi(t)=\Psi(t)\Psi^{-1}(t_{0})$.
Thus if you need to compute $\Phi$, simply compute $\Psi(t_{0})$,
then compute it's inverse $\Psi^{-1}(t_{0})$, and then comptue the
product, $\Psi(t)\Psi^{-1}(t_{0})$.
For a two-by-two matrix, you can remember the formula for the inverse
of a matrix $\left(\begin{array}{cc}
a & b\\
c & d\end{array}\right)$ is $\frac{1}{ad-bc}\left(\begin{array}{cc}
d & -b\\
-c & a\end{array}\right)$.
In general you can consider the augmented matrix and row reduce. For
the two-by-two case we have:
$\left(\begin{array}{ccccc}
a & b & | & 1 & 0\\
c & d & | & 0 & 1\end{array}\right)$ which when we row reduce the left block to the identity we get the
inverse on the right block.
\textasciitilde{}
I don't know if the problem will come up, but if asked, what is $\exp(At)$?
Then the answer will be the fundamental matrix $\Phi$ that satisfies
$\Phi^{\prime}=A\Phi$, $\Phi(0)=I$ (page 417).
Find this $\Phi$by first getting $\Psi$ from the solutions for $x^{\prime}=Ax$.
Then $\Phi(t)=\Psi(t)\Psi^{-1}(0)$
\textasciitilde{}
Let's talk about two-by-two matrices $A$. If $A$ has a repeated
eigenvalue $\lambda$ with two eigenvectors implies $A$ is of the
form $\left(\begin{array}{cc}
\lambda & 0\\
0 & \lambda\end{array}\right)$. For a repeated eigenvalue $\lambda$ with only one eigenvector ,
then we have two solutions to the equation $x^{\prime}=Ax$, one of
the form $\xi e^{2t}$ and the second of the form $\xi te^{2t}+\eta e^{2t}$
where $(A-\lambda I)\xi=0$ and $(A-\lambda I)\eta=\xi$.
Note: $(A-\lambda I)\xi=0$ implies that $(A-\lambda I)\eta=\xi$
has a solution.
Sidenote: What I was thinking about in class was related, but not
useful for computation. $(A-\lambda I)^{2}\xi=(A-\lambda I)\eta=0$.
\textasciitilde{}
I'm gonna be back in about four hours, so until then, hopefully you
can give yourself problems.
Complex Eigenvalues
Sorry about being a little misleading with the complex eigenvalues.
Let me try again:
Let's say we have eigenvalue $\lambda=a+ib$ and eigenvector$\xi=\left(\begin{array}{c}
u_{1}\\
u_{2}\end{array}\right)+i\left(\begin{array}{c}
v_{1}\\
v_{2}\end{array}\right)$.
Then the solution $\xi e^{\lambda t}=e^{at}\left[\left[\left(\begin{array}{c}
u_{1}\\
u_{2}\end{array}\right)\cos bt-\left(\begin{array}{c}
v_{1}\\
v_{2}\end{array}\right)\sin bt\right]+i\left[\left(\begin{array}{c}
u_{1}\\
u_{2}\end{array}\right)\sin bt+\left(\begin{array}{c}
v_{1}\\
v_{2}\end{array}\right)\cos bt\right]\right]$
Thus we get the two real solutions: $x_{1}(t)=\left[\left(\begin{array}{c}
u_{1}\\
u_{2}\end{array}\right)\cos bt-\left(\begin{array}{c}
v_{1}\\
v_{2}\end{array}\right)\sin bt\right]$
and $x_{2}(t)=e^{at}\left[\left(\begin{array}{c}
u_{1}\\
u_{2}\end{array}\right)\sin bt+\left(\begin{array}{c}
v_{1}\\
v_{2}\end{array}\right)\cos bt\right]$.
You don't need to memorize these formulas, I'm just doing the problem
abstractly.
To determine the direction of the rotation, let's look at $x_{1}$.
$x_{1}(0)=\left(\begin{array}{c}
u_{1}\\
u_{2}\end{array}\right)$
Now let's look at the derivative: $x_{1}^{\prime}(t)=\left[-b\left(\begin{array}{c}
u_{1}\\
u_{2}\end{array}\right)\sin bt-b\left(\begin{array}{c}
v_{1}\\
v_{2}\end{array}\right)\cos bt\right]$
$x_{1}^{\prime}(0)=-b\left(\begin{array}{c}
v_{1}\\
v_{2}\end{array}\right)$. The spiral should be tangent to this vector at $x_{1}(0)$ and be
the direction goes in the direction of this vector.
Let me try again:
Let's say we have eigenvalue $\lambda=a+ib$ and eigenvector$\xi=\left(\begin{array}{c}
u_{1}\\
u_{2}\end{array}\right)+i\left(\begin{array}{c}
v_{1}\\
v_{2}\end{array}\right)$.
Then the solution $\xi e^{\lambda t}=e^{at}\left[\left[\left(\begin{array}{c}
u_{1}\\
u_{2}\end{array}\right)\cos bt-\left(\begin{array}{c}
v_{1}\\
v_{2}\end{array}\right)\sin bt\right]+i\left[\left(\begin{array}{c}
u_{1}\\
u_{2}\end{array}\right)\sin bt+\left(\begin{array}{c}
v_{1}\\
v_{2}\end{array}\right)\cos bt\right]\right]$
Thus we get the two real solutions: $x_{1}(t)=\left[\left(\begin{array}{c}
u_{1}\\
u_{2}\end{array}\right)\cos bt-\left(\begin{array}{c}
v_{1}\\
v_{2}\end{array}\right)\sin bt\right]$
and $x_{2}(t)=e^{at}\left[\left(\begin{array}{c}
u_{1}\\
u_{2}\end{array}\right)\sin bt+\left(\begin{array}{c}
v_{1}\\
v_{2}\end{array}\right)\cos bt\right]$.
You don't need to memorize these formulas, I'm just doing the problem
abstractly.
To determine the direction of the rotation, let's look at $x_{1}$.
$x_{1}(0)=\left(\begin{array}{c}
u_{1}\\
u_{2}\end{array}\right)$
Now let's look at the derivative: $x_{1}^{\prime}(t)=\left[-b\left(\begin{array}{c}
u_{1}\\
u_{2}\end{array}\right)\sin bt-b\left(\begin{array}{c}
v_{1}\\
v_{2}\end{array}\right)\cos bt\right]$
$x_{1}^{\prime}(0)=-b\left(\begin{array}{c}
v_{1}\\
v_{2}\end{array}\right)$. The spiral should be tangent to this vector at $x_{1}(0)$ and be
the direction goes in the direction of this vector.
Saturday, April 2, 2011
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.
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.
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