Alternating series test (Wikipedia.org)
Definition of an alternating series:
A series of the form $$\sum_{k=0}^\infty (-1)^k a_k$$
where all the $a_k$ are non-negative. The emphasis here is on the non-negative.
We know the absolute value of the remainder $R_n=\sum_{k=n+1}^\infty (-1)^k a_k$ for an alternating series will be less than the next term in the series, $a_{n+1}$.
So what's the point of all this? Well, some power series might look like alternating series, but they'll only be alternating series for certain values of $x$. Thus, you can only estimate the remainder using the above estimate for values of $x$ for which the power series turns into an alternating series.
[Remark: It sort of follows from the above, that a power series that doesn't look like an alternating series, might be an alternating series for certain values of $x$]
Tuesday, December 6, 2011
Answering a student's email questions
For question 6, you should set $x=0$ and solve for $t$. You'll get two values of $t$, but only the one that gives you $y\geq 3$ will work. Then set $y=3$ and solve for $t$. Again you'll get two values of $t$, but only one gives you $x\geq 0$.
For question 7, the furthest distance from the origin is when $r$ is greatest. Thus we are trying to maximize $r$. Then from one variable calculus, just take the derivative with respect to $\theta$ to find critical points. $\frac{dr}{d\theta}=0$. There's a geometric interpretation to this, but it's much easier to explain through pictures than through words. If requested, I'll explain this.
Question 8 is asking for when $r=0$. No derivatives required. Let me know if I'm wrong on this.
For question 10, your mistake might have been that the integral should go from $0$ to $\pi$ and not from $0$ to $2\pi$.
For question 7, the furthest distance from the origin is when $r$ is greatest. Thus we are trying to maximize $r$. Then from one variable calculus, just take the derivative with respect to $\theta$ to find critical points. $\frac{dr}{d\theta}=0$. There's a geometric interpretation to this, but it's much easier to explain through pictures than through words. If requested, I'll explain this.
Question 8 is asking for when $r=0$. No derivatives required. Let me know if I'm wrong on this.
For question 10, your mistake might have been that the integral should go from $0$ to $\pi$ and not from $0$ to $2\pi$.
Taylor Remainder Term
On page 756 (of Stewart's Single Variable Calculus Early Transcendentals), the margin has the following information:
If $f^{(n+1)}$ is continuous on an interval $I$ and $x\in I$, then $$R_n(x)=\frac{1}{n!}\int_a^x(x-t)^n f^{(n+1)}(t)dt$$. This is called the integral form of the remainder term.
Lagrange's form of the remainder term:
$$R_n(x)=\frac{f^{n+1}(z)}{(n+1)!}(x-a)^{n+1})$$
for some $z$ between $x$ and $a$.
Note that Taylor's Inequality follows from Lagrange's form of the remainder term.
Let's do a similar problem to #17 on this the 2010 final exam.
$f(x)=3^x$ and $a=5$.
$f^\prime(x)=\ln(3) 3^x$
$f^{\prime\prime}(x)=\ln(3)^2 3^x$
$f^{(3)}(x)=\ln(3)^3 3^x$
Then $f$ has Taylor series $$\sum_{n=0}^\infty \frac{\ln(3)^n 3^5}{n!}(x-5)^n$$
It's remainder $R_n(x)$ will be $$R_n(x)=\frac{1}{n!}\int_5^x(x-t)^n \ln(3)^{n+1} 3^t dt$$
Note: Using this formula, I would answer the following for part (c): $$R_2(2.1)=\frac{1}{2!}\int_2^{2.1} (2.1-t)^2 \ln(2)^3 2^t dt$$
To answer part (d), just integrate the above by parts and use all the information he gave you to simplify the answer. You might get a different answer since he gave an answer I don't fully understand to part (c). See Wikipedia: Taylor's theorem > Explicit formulae for the remainder
If $f^{(n+1)}$ is continuous on an interval $I$ and $x\in I$, then $$R_n(x)=\frac{1}{n!}\int_a^x(x-t)^n f^{(n+1)}(t)dt$$. This is called the integral form of the remainder term.
Lagrange's form of the remainder term:
$$R_n(x)=\frac{f^{n+1}(z)}{(n+1)!}(x-a)^{n+1})$$
for some $z$ between $x$ and $a$.
Note that Taylor's Inequality follows from Lagrange's form of the remainder term.
Let's do a similar problem to #17 on this the 2010 final exam.
$f(x)=3^x$ and $a=5$.
$f^\prime(x)=\ln(3) 3^x$
$f^{\prime\prime}(x)=\ln(3)^2 3^x$
$f^{(3)}(x)=\ln(3)^3 3^x$
Then $f$ has Taylor series $$\sum_{n=0}^\infty \frac{\ln(3)^n 3^5}{n!}(x-5)^n$$
It's remainder $R_n(x)$ will be $$R_n(x)=\frac{1}{n!}\int_5^x(x-t)^n \ln(3)^{n+1} 3^t dt$$
Note: Using this formula, I would answer the following for part (c): $$R_2(2.1)=\frac{1}{2!}\int_2^{2.1} (2.1-t)^2 \ln(2)^3 2^t dt$$
To answer part (d), just integrate the above by parts and use all the information he gave you to simplify the answer. You might get a different answer since he gave an answer I don't fully understand to part (c). See Wikipedia: Taylor's theorem > Explicit formulae for the remainder
Monday, December 5, 2011
part of page 762
$$\frac{1}{1-x}=\sum_{n=0}^\infty x^n$$
$$e^x=\sum_{n=0}^\infty \frac{x^n}{n!}$$
$$\sin x=\sum_{n=0}^\infty (-1)^n\frac{x^{2n+1}}{(2n+1)!}$$
$$\cos x=\sum_{n=0}^\infty (-1)^n\frac{x^{2n}}{(2n)!}$$
Note: $\sin 0=0$ and $\cos 0=1$
$$\tan^{-1} x=\sum_{n=0}^\infty (-1)^n\frac{x^{2n+1}}{2n+1}$$
$$\ln(1+x)=\sum_{n=1}^\infty (-1)^{n-1}\frac{x^n}{n}$$
$$(1+x)^{k}=\sum_{n=0}^{\infty}\left(\begin{array}{c}
k\\
n\end{array}\right)x^{n}$$
$$\left(\begin{array}{c}
k\\
n\end{array}\right)=\frac{k(k-1)(k-2)\cdots(k-n+1)}{n!}$$
$$e^x=\sum_{n=0}^\infty \frac{x^n}{n!}$$
$$\sin x=\sum_{n=0}^\infty (-1)^n\frac{x^{2n+1}}{(2n+1)!}$$
$$\cos x=\sum_{n=0}^\infty (-1)^n\frac{x^{2n}}{(2n)!}$$
Note: $\sin 0=0$ and $\cos 0=1$
$$\tan^{-1} x=\sum_{n=0}^\infty (-1)^n\frac{x^{2n+1}}{2n+1}$$
$$\ln(1+x)=\sum_{n=1}^\infty (-1)^{n-1}\frac{x^n}{n}$$
$$(1+x)^{k}=\sum_{n=0}^{\infty}\left(\begin{array}{c}
k\\
n\end{array}\right)x^{n}$$
$$\left(\begin{array}{c}
k\\
n\end{array}\right)=\frac{k(k-1)(k-2)\cdots(k-n+1)}{n!}$$
Sunday, December 4, 2011
Tim's Office Status
"Funny" Example:
2011-12-04 12:00 PM Tim entered his office.
2011-12-04 12:01 PM Tim left his office to go the bathroom.
2011-12-04 12:20 PM Tim came back from the bathroom but left immediately to avoid meeting with students.
I'll try to come to the office later this evening. But if not, I'll definitely be there most of Monday and Tuesday with the exception of about two hours on Monday and two hours on Tuesday getting exercise, and an hour each day to eat.
Feel free to send me questions by e-mail. timctran at jhu dot edu. I would start my focus on doing problems which Professor Wilson has worked out in class.
2011-12-05 10:23 AM Tim entered his office.
2011-12-05 11:00 AM Krieger 205
2011-12-05 12:00 PM Krieger 300
2011-12-05 01:30 PM Ended "Review" and Left Krieger 300
Playing touch football, will be back in the office later.
2011-12-05 06:20 PM Tim entered his office. Will be around for another hour.
2011-12-05 07:20 PM Tim left the office to go do back handsprings. Might come back in two hours. Maybe you can come and study while I play games. Muahaha.
Definitely memorize the table on page 762. When allowed, using the summation notation should make the work easier.
Of course you have to be smart about finding the series. Example: Find the Maclaurin series for $e^{x^2+3}$. Some students might be tempted to just plug $x^2+3$ into the series for $e^x$, $\sum_{n=0}^\infty \frac{x^n}{n!}$. However, that wouldn't give you a Maclaurin series. Learn to manipulate your starting expression in the correct way to get what you want. In this case, it's a simple matter of writing $e^{x^2+3}$ as $e^3e^{x^2}$. Then we get $e^3 \sum_{n=0}^\infty \frac{x^2n}{n!}$.
2011-12-05 09:50 PM Tim entered his office.
2011-12-06 02:00 AM Tim left the office to go to sleep.
2011-12-06 02:00 PM Tim entered his office, but didn't update it until 02:30 PM. He'll be gone at 03:00 PM for lunch. He'll be back around 5.
2011-12-06 07:15 PM He came back around 5 and left at 7:15 for modern dance. Check out his recent post. Send him questions by e-mail. He plans to sleep around 11 or 12, so he can get to school at 8:40 AM to get the exams.
A student reminded me of an important tip. Double-check your work when it comes to parameterized curves. This includes the possibility that a curve has already completed itself.
Simple Example: $c(t)=(\cos(t),\sin(t))$ starting at $t=0$ will come back to the starting point $(1,0)$ at $t=2\pi$. $d(t)=(\cos(3t),\sin(3t))$ starting at $t=0$ will come back to the starting point $(1,0)$ at $t=2\pi/3$. If when finding the area or the length you had integrated from $0$ to $2\pi$, then you would get three times the expected answer. And unless the expected answer is 0, then you, my friend, would have the wrong answer. =D
Related to parameterized curves, areas, and lengths, keep in mind that length will always be positive. Area can be negative depending on the context. When asked to find area bounded in a region, it's usually implied that this is total area. In other words, all area is positive, regardless of it's location on the x-y plane. So where does parameterization come in? Well, depending on the direction of the parameterization, you might introduce a negative sign into your answer. So take the absolute value and smile.
Short theoretical explanation of the previous paragraph. When we parametrize a curve, we can choose to integrate from $a$ to $b$, or from $b$ to $a$. And $$\int_a^b f(x)dx=-\int_b^a f(x)dx$$.
2011-12-04 12:00 PM Tim entered his office.
2011-12-04 12:01 PM Tim left his office to go the bathroom.
2011-12-04 12:20 PM Tim came back from the bathroom but left immediately to avoid meeting with students.
I'll try to come to the office later this evening. But if not, I'll definitely be there most of Monday and Tuesday with the exception of about two hours on Monday and two hours on Tuesday getting exercise, and an hour each day to eat.
Feel free to send me questions by e-mail. timctran at jhu dot edu. I would start my focus on doing problems which Professor Wilson has worked out in class.
2011-12-05 10:23 AM Tim entered his office.
2011-12-05 11:00 AM Krieger 205
2011-12-05 12:00 PM Krieger 300
2011-12-05 01:30 PM Ended "Review" and Left Krieger 300
Playing touch football, will be back in the office later.
2011-12-05 06:20 PM Tim entered his office. Will be around for another hour.
2011-12-05 07:20 PM Tim left the office to go do back handsprings. Might come back in two hours. Maybe you can come and study while I play games. Muahaha.
Definitely memorize the table on page 762. When allowed, using the summation notation should make the work easier.
Of course you have to be smart about finding the series. Example: Find the Maclaurin series for $e^{x^2+3}$. Some students might be tempted to just plug $x^2+3$ into the series for $e^x$, $\sum_{n=0}^\infty \frac{x^n}{n!}$. However, that wouldn't give you a Maclaurin series. Learn to manipulate your starting expression in the correct way to get what you want. In this case, it's a simple matter of writing $e^{x^2+3}$ as $e^3e^{x^2}$. Then we get $e^3 \sum_{n=0}^\infty \frac{x^2n}{n!}$.
2011-12-05 09:50 PM Tim entered his office.
2011-12-06 02:00 AM Tim left the office to go to sleep.
2011-12-06 02:00 PM Tim entered his office, but didn't update it until 02:30 PM. He'll be gone at 03:00 PM for lunch. He'll be back around 5.
2011-12-06 07:15 PM He came back around 5 and left at 7:15 for modern dance. Check out his recent post. Send him questions by e-mail. He plans to sleep around 11 or 12, so he can get to school at 8:40 AM to get the exams.
A student reminded me of an important tip. Double-check your work when it comes to parameterized curves. This includes the possibility that a curve has already completed itself.
Simple Example: $c(t)=(\cos(t),\sin(t))$ starting at $t=0$ will come back to the starting point $(1,0)$ at $t=2\pi$. $d(t)=(\cos(3t),\sin(3t))$ starting at $t=0$ will come back to the starting point $(1,0)$ at $t=2\pi/3$. If when finding the area or the length you had integrated from $0$ to $2\pi$, then you would get three times the expected answer. And unless the expected answer is 0, then you, my friend, would have the wrong answer. =D
Related to parameterized curves, areas, and lengths, keep in mind that length will always be positive. Area can be negative depending on the context. When asked to find area bounded in a region, it's usually implied that this is total area. In other words, all area is positive, regardless of it's location on the x-y plane. So where does parameterization come in? Well, depending on the direction of the parameterization, you might introduce a negative sign into your answer. So take the absolute value and smile.
Short theoretical explanation of the previous paragraph. When we parametrize a curve, we can choose to integrate from $a$ to $b$, or from $b$ to $a$. And $$\int_a^b f(x)dx=-\int_b^a f(x)dx$$.
Tuesday, October 18, 2011
Common Values to Know
You're expected to know:
$$\log(1)=0$$
$$\sin(\theta)$$
$$\cos(\theta)$$
$$\tan(\theta)$$
for common values of $\theta$, such as $\theta=0,\pi/6,\pi/4,\pi/3,\pi/2$.
$$\arctan(1)=\pi/4$$
$$\log(1)=0$$
$$\sin(\theta)$$
$$\cos(\theta)$$
$$\tan(\theta)$$
for common values of $\theta$, such as $\theta=0,\pi/6,\pi/4,\pi/3,\pi/2$.
$$\arctan(1)=\pi/4$$
Wednesday, September 28, 2011
Relax a Little
Normally students ask me questions by e-mail and I post some responses here. Unfortunately for them, and fortunately for me, no one asked any questions. I would have been happy to reply. The instructor this semester has rules against explicitly posting answers to various material, but I still could have answered with a method. Anyways, here's a video I wanted to show my 3pm section, but never got around to doing. I just embedded the first in the series:
For multiple reasons, its too late to give any advice I didn't already give. I've already heavily emphasized knowing how to simplify $\sin(\theta)$, $\cos(\theta)$, etc. for common values of $\theta$. I also emphasized various formulas related to trigonometric functions, such as $\sin^2(\theta)+\cos^2(\theta)=1$ and $\cosh^2(\theta)-\sinh^2(\theta)=1$.
There's also
$\int \frac{1}{1+x^2}dx=\arctan(x)+C$
and
$\int \sec(x)dx=\ln \left| (\sec(x)+\tan(x)\right|+C$
Students should also be careful about using Euler's Method.
Suppose we are looking to approximate $y(x)$.
If the step size is $h$ and $x=k\cdot h+x_0$, then the value we seek is $y_k$.
Many students made the ERROR of determining $y_{k+1}$ (ERROR)
Recall some formulas:
$\frac{dy}{dx}=F(x,y)$
$x_n=x_{n-1}+h$
$y_n=y_{n-1}+h\cdot F(x_{n-1},y_{n-1})$
I like to remember the above by (note this is a way to remember the formula, it is not intended to justify or derive anything)
$\Delta y=\Delta x \cdot \frac{\Delta y}{\Delta x}$
$y_k - y_{k-1}=h \cdot \frac{dy}{dx}$
$y_k-y_{k-1}=h\cdot F(x_{k-1},y_{k-1})$
$y_k=y_{k-1}+h\cdot F(x_{k-1},y_{k-1})$
Finally for those hard to solve rate problems, just do your best to break down the problem and determine the rate in and the rate out. Suppose the quantity changing with respect to time is volume. Then let $V$ be the amount of volume. We set up a differential equation:
$\frac{dV}{dt}=$ {rate in} - {rate out}
There's a high chance either the rate in or the rate out will depend on $V$.
After you have completed the problem, remember to solve the initial value problem, if that is what the question is asking for.
For multiple reasons, its too late to give any advice I didn't already give. I've already heavily emphasized knowing how to simplify $\sin(\theta)$, $\cos(\theta)$, etc. for common values of $\theta$. I also emphasized various formulas related to trigonometric functions, such as $\sin^2(\theta)+\cos^2(\theta)=1$ and $\cosh^2(\theta)-\sinh^2(\theta)=1$.
There's also
$\int \frac{1}{1+x^2}dx=\arctan(x)+C$
and
$\int \sec(x)dx=\ln \left| (\sec(x)+\tan(x)\right|+C$
Students should also be careful about using Euler's Method.
Suppose we are looking to approximate $y(x)$.
If the step size is $h$ and $x=k\cdot h+x_0$, then the value we seek is $y_k$.
Many students made the ERROR of determining $y_{k+1}$ (ERROR)
Recall some formulas:
$\frac{dy}{dx}=F(x,y)$
$x_n=x_{n-1}+h$
$y_n=y_{n-1}+h\cdot F(x_{n-1},y_{n-1})$
I like to remember the above by (note this is a way to remember the formula, it is not intended to justify or derive anything)
$\Delta y=\Delta x \cdot \frac{\Delta y}{\Delta x}$
$y_k - y_{k-1}=h \cdot \frac{dy}{dx}$
$y_k-y_{k-1}=h\cdot F(x_{k-1},y_{k-1})$
$y_k=y_{k-1}+h\cdot F(x_{k-1},y_{k-1})$
Finally for those hard to solve rate problems, just do your best to break down the problem and determine the rate in and the rate out. Suppose the quantity changing with respect to time is volume. Then let $V$ be the amount of volume. We set up a differential equation:
$\frac{dV}{dt}=$ {rate in} - {rate out}
There's a high chance either the rate in or the rate out will depend on $V$.
After you have completed the problem, remember to solve the initial value problem, if that is what the question is asking for.
Tuesday, September 20, 2011
Trig Functions and Hyperbolic Trig Functions
$$\cos (x) = \frac{e^{ix}+e^{-ix}}{2}$$
$$\sin (x) = \frac{e^{ix}-e^{-ix}}{2i}$$
$$\cos^2(x)+\sin^2(x)=1$$
From this we get:
$$\cos^2(x)+\csc^{-2}(x)=1$$
$$1+\tan^2(x)=\cos^{-2}(x)$$
$$1+\tan^2(x)=\sec^2(x)$$
$$\cosh (x) = \frac{e^{x}+e^{-x}}{2}$$
$$\sinh (x) = \frac{e^{x}-e^{-x}}{2}$$
$$\cosh^2(x)-\sinh^2(x)=1$$
At the moment, this post is incomplete.
$$\sin (x) = \frac{e^{ix}-e^{-ix}}{2i}$$
$$\cos^2(x)+\sin^2(x)=1$$
From this we get:
$$\cos^2(x)+\csc^{-2}(x)=1$$
$$1+\tan^2(x)=\cos^{-2}(x)$$
$$1+\tan^2(x)=\sec^2(x)$$
$$\cosh (x) = \frac{e^{x}+e^{-x}}{2}$$
$$\sinh (x) = \frac{e^{x}-e^{-x}}{2}$$
$$\cosh^2(x)-\sinh^2(x)=1$$
At the moment, this post is incomplete.
Monday, September 12, 2011
Problems on First Homework
Mathematical Errors:
Forgetting the constant C for an indefinite integral.
Adding a constant C for a definite integral.
Sign errors. I sometimes marked off for this. One sign error comes up when making substitutions. Another sign error happens when using integration by parts.
Stylistic Errors:
Doing problems out of order.
Using a lot of paper, possibly only one side
Writing messy
Being concise can work wonders.
Forgetting the constant C for an indefinite integral.
Adding a constant C for a definite integral.
Sign errors. I sometimes marked off for this. One sign error comes up when making substitutions. Another sign error happens when using integration by parts.
Stylistic Errors:
Doing problems out of order.
Using a lot of paper, possibly only one side
Writing messy
Being concise can work wonders.
Monday, May 9, 2011
Substitution for $t^2 y^{\prime \prime}+t y^{\prime}+ y =0$
$t^2 y^{\prime \prime}+t y^{\prime}+ y =0$
Make the substitution $u=\ln t$
Then compute $\frac{dy}{dt}=\frac{dy}{du}\frac{du}{dt}$ and
compute $\frac{d^2y}{dt^2}=\frac{d^2y}{d^2u}(\frac{du}{dt})^2+\frac{dy}{du}\frac{d^2u}{dt^2}$
I'll get the page reference soon. Just wanted to get this up because I said I would.
Make the substitution $u=\ln t$
Then compute $\frac{dy}{dt}=\frac{dy}{du}\frac{du}{dt}$ and
compute $\frac{d^2y}{dt^2}=\frac{d^2y}{d^2u}(\frac{du}{dt})^2+\frac{dy}{du}\frac{d^2u}{dt^2}$
I'll get the page reference soon. Just wanted to get this up because I said I would.
Sunday, May 8, 2011
Phase Portraits Update
As I said in discussion section, instead of using the solutions to determine the directions of the spiral or loop, we can use the original system of equations. Let me know if this is confusing.
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.
Thursday, March 10, 2011
page 173 number 27
page 173 number 27
This is one of the situations where starting the problem abstractly is beneficial.
$Ay^{\prime\prime}+By^\prime+Cy=0$
$y_2=y_1v$
$y_2^\prime=y_1^\prime v+y_1 v^\prime$
$y_2^{\prime\prime}=y_1^{\prime\prime}v+2y_1^\prime v^\prime+y_1 v^{\prime\prime}$
Terms that survive:
$A(2y_1^\prime v^\prime+y_1 v^{\prime\prime})+By_1 v^\prime=0$
Which is:
$Ay_1 v^{\prime\prime}+(2Ay_1^\prime+By_1)v^\prime=0$
Let $w=v^\prime$ to obtain:
$Ay_1 w^\prime+(2Ay_1^\prime+By_1)w=0$
So then:
$\int \frac{1}{w} dw=\int \left[-(2\frac{y_1^\prime}{y_1}+\frac{B}{A})\right] dx$
Note at this point I just enter the coefficients $A$ and $B$:
$\ln w=-2\ln(y_1)-\int \left[\frac{-1}{x}\right] dx$
$v^\prime=e^{\ln(y_1)^{-2}+\ln(x)+C}=C_2\cdot x(y_1)^{-2}$
I originally forgot the constant here. You should remember it.
$v=C_2\int(\frac{x}{\sin^2 x^2})dx=C_2\frac{1}{2}\int(\csc^2(x^2))dx^2$
$=-C_2\frac{1}{2}(\cot x^2+C)=C_3\cot x^2+C_4$
I made the error of forgetting the minus sign. You should not make this error. That is:
$\frac{d}{dx}\cot u=-\csc^2 u \frac{du}{dx}$
Though in this problem the error doesn't matter. We pick $C_3=1$ and $C_4=0$ and get
$y_2=\cot x^2 \sin x^2 =\cos x^2$
This is one of the situations where starting the problem abstractly is beneficial.
$Ay^{\prime\prime}+By^\prime+Cy=0$
$y_2=y_1v$
$y_2^\prime=y_1^\prime v+y_1 v^\prime$
$y_2^{\prime\prime}=y_1^{\prime\prime}v+2y_1^\prime v^\prime+y_1 v^{\prime\prime}$
Terms that survive:
$A(2y_1^\prime v^\prime+y_1 v^{\prime\prime})+By_1 v^\prime=0$
Which is:
$Ay_1 v^{\prime\prime}+(2Ay_1^\prime+By_1)v^\prime=0$
Let $w=v^\prime$ to obtain:
$Ay_1 w^\prime+(2Ay_1^\prime+By_1)w=0$
So then:
$\int \frac{1}{w} dw=\int \left[-(2\frac{y_1^\prime}{y_1}+\frac{B}{A})\right] dx$
Note at this point I just enter the coefficients $A$ and $B$:
$\ln w=-2\ln(y_1)-\int \left[\frac{-1}{x}\right] dx$
$v^\prime=e^{\ln(y_1)^{-2}+\ln(x)+C}=C_2\cdot x(y_1)^{-2}$
I originally forgot the constant here. You should remember it.
$v=C_2\int(\frac{x}{\sin^2 x^2})dx=C_2\frac{1}{2}\int(\csc^2(x^2))dx^2$
$=-C_2\frac{1}{2}(\cot x^2+C)=C_3\cot x^2+C_4$
I made the error of forgetting the minus sign. You should not make this error. That is:
$\frac{d}{dx}\cot u=-\csc^2 u \frac{du}{dx}$
Though in this problem the error doesn't matter. We pick $C_3=1$ and $C_4=0$ and get
$y_2=\cot x^2 \sin x^2 =\cos x^2$
Friday, February 25, 2011
Friday, February 18, 2011
Friday, February 11, 2011
Homework Week 1 Solutions
P8: 22
P24: 1, 5; P25: 7, 9, 15, 17.
P39: 13, 17; P40: 30
P47: 1, 3; P48: 7, 8; P50: 31, 36
P77: 31.
Select solutions have been posted 2/24/2011. Problems with solutions have hyperlinks to them. If you want to see a solution which is not posted, please let me know. You may also request solutions to problems outside the homework set as well.
P24: 1, 5; P25: 7, 9, 15, 17.
P39: 13, 17; P40: 30
P47: 1, 3; P48: 7, 8; P50: 31, 36
P77: 31.
Select solutions have been posted 2/24/2011. Problems with solutions have hyperlinks to them. If you want to see a solution which is not posted, please let me know. You may also request solutions to problems outside the homework set as well.
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