
Monday, September 26 

It may be useful to understand what this algebraic mess means.
And even more resonance
What if the driving force hits every 2Pi after t=Pi? The mathematical
model is
y´´+Ky´+y=SUM_{j=0}^{infinity}delta(t(2j+1))
with,
again, y(0)=0 and y´(0)=0. Let us analyze this as we've already
done. Of course, we could worry about the infinite sum, which
is sort of a limit, but "Almost every reasonable limit should exist."
I won't worry.
Exactly as before, the Laplace transform is
Y(s)=SUM_{j=0}^{infinity}[e^{(2j+1)Pi·s}]/[s^{2}+1].
This Laplace transform already has built into it the initial
conditions and the "rough" inhomogeneity of the forcing term.
The inverse Laplace transform is
y(t)=SUM_{j=0}^{infinity}sin(t(2j+1)Pi)U(t(2j+1)Pi).
Again please remember that sine is periodic with period 2Pi. Therefore
sin(t(2j+1)Pi)=sin(tPi), and
y(t)=SUM_{j=0}^{infinity}sin(tPi)U(t(2j+1)Pi)=sin(tPi)SUM_{j=0}^{infinity}U(t(2j+1)Pi).
What's left in the sum is not a constant function, but it is a
staircase function (the graph looks more or less like 4.3, #62.
What about the motion? Is it smooth? Where is it smooth? The
graph is a bit difficult to consider for t large, but if we ask
Maple to "zoom in" near a specific t (I think this is t=3Pi)
you can see a kink in the graph. At 3Pi, the slope changes from 1 to
2. And (3+2j)Pi, motion will be continuous (this spring doesn't fly
apart!) but the graph will not be differentiable: slope changes from j
to j+1.
This is a great vocabulary day:
kink
a. a short backward twist in wire or tubing etc. such as may cause
an obstruction.
b. a tight wave in human or animal hair.
jerk
a. a sharp sudden pull, twist, twitch, start, etc.
b. a spasmodic muscular twitch.
c. (in "pl.") [Brit.] [colloq.] exercises ("physical jerks").
d. [sl.] a fool; a stupid person.
And how about limiting the amplitude
Here is a special bonus problem for mechanical
engineering students, and any others who can grasp these concepts. Our
ideal spring y´´+y can have a damping term: y´´+Ky´+y for some real
constant K. This might correspond to the spring vibrating in honey or
10W30 motor oil or ... whatever. Suppose we hit the spring again every
2Pi. I know that the model described above is, indeed, just a
classroom model. It is hard to conceive of Hooke's law applying at the
spring stretches more and more and more (4 light years?). So the
design question: find a value of K so that the spring is limited in
motion (with, say, y(t)<10) for a limited time, say
0<t<200. You must give some supporting evidence that your
candidate for K actually restricts the motion in the required
fashion.
Please note that a math person might actually try to compute the exact
solution, and might then investigate the highest value of y(t),
etc. What could an engineer do? Well, maybe build a spring system
(difficult and perhaps expensive). Or find some other nearly
experimental way to verify that a conjectured value of K works.
Hint After class, I returned to my office, and guessed a value
of K, and got good enough verification within 45 seconds, with some help from ...
QotD
I asked people to find a formula for the Laplace transform, Y(s), of
the solution to the Initial Value Problem:
y´´(t)+3y´(t)+y(t)=t+delta(t)
y(0)=1 and y´(0)=2.
How to do this: I would try to use effectively the dictionary of Laplace transforms displayed on the board. So the equation
y´´(t)+3y´(t)+y(t)=t+delta(t)
becomes:
s^{2}2Y(s)sy(0)y´(0)+3(sY(s)y(0))+Y(s)=1/s+e^{2s}
and, inserting the initial conditions, we get:
s^{2}2Y(s)s+2+3(sY(s)1)+Y(s)=1/s^{2}+e^{2s}
and collecting Y(s) terms:
(s^{2}+3s+1)Y(s)s+23=(1/s^{2})+e^{2s}
so that
Y(s)=[(1/s^{2})+e^{2s}+s+1]/(s^{2}+3s+1).
By the way, amid the general hilarity, I neglected to inform you that my solution which I did half an hour before class, also had an error. Sigh. I then asked the following question, which I believe is useful to think about. Suppose we have this Laplace transform. What sorts of functions should we expect in the inverse transform? It's my hope that you would have a computer algebra system to solve the equation but you should be able to make some rough checks on the answers.
I notice that s^{2}+3s+1 has real roots (the discriminant, 3^{2}4·1·1=5 is positive). Therefore I would not expect sine or cosine. I would expect some sums of exponentials. I would also expect some Heaviside function entry for two reasons: there is a delta in the righthand side, and we can only get that as a derivative of U, and, second, the e^{2s} in the transform, which the second translation theorem will turn into a U. Now let's see what Maple answers, assuming I can type correctly.
> with(inttrans): > invlaplace(((1/s^2)+exp(2*s)+s+1)/(s^2+3*s+1),s,t); / 1/2 1/2 1/2 \ (5 + 3) t 5 3 (5 + 3) t 1/2    (5  3)  / 1/2 \ / 1/2 \   5  5    2   3/2 2   3/2  \ \ 2 / \ 2 / / 1/2 3 + t + 1/5 exp() (10 + 3 5 ) / 1/2 \  5  4   + 3/2 \ 2 / / 1/2 1/2 \ 1/2  (5 + 3) (t  2) (5  3) (t  2)  + 1/5 5 exp( ) + exp() Heaviside(t  2) \ 2 2 / / 1/2 1/2 1/2 \  (5 + 3) t 5 3 (5 + 3) t  1/2     (5  3)  / 1/2 \ / 1/2 \   5   5   2    3/2 2    3/2 \ \ 2 / \ 2 // 1/2  1/5 exp() (10 + 3 5 ) / 1/2 \ 5  4  + 3/2 \ 2 /What a mess, but, qualitatively, the guesses above are verified. I think. Section 4.6 in 4.6 minutes
dx dy 2 +  2x = 1 dt dt dx dy  +  3x 3y = 2 dt dt x(0)=0 and y(0)=0This is a textbook problem, so it will not be very difficult (maybe). And this one was rehearsed (!). You may be able to see various ways to solve this problem, but let me be obedient to chapter 4, and use Laplace transforms.
Here are the Laplace transforms of the equations, with the rather silly initial conditions used.
2sX(s)+sY(s)2X(s)=(1/s) sX(s)+sY(s)3X(s)3Y(s)=(2/s)Let's collect things:
(2s2)X(s) + sY(s) = (1/s) (s3)X(s) + (s3)Y(s) = (2/s)Now this is a system of two linear equations in two unknowns. I will divide the first equation by (2s2):
1X(s) + (s/[2s2])Y(s)=(1/[s[2s2])and now I will multiply the modified first equation by (s3) and subtract it from the second equation. This is resulting second equation:
0X(s) + ((s3)(s3)(s/[2s2]))Y(s)=(2/s)(s3)(1/[s[2s2])Wow! Now I have isolated Y(s) and can divide by its coefficient, so that
^{ }(2/s)(s3)(1/[s[2s2]) Y(s) =  ((s3)(s3)(s/[2s2]))
Maple tells me this:
> invlaplace(((2/s)(s3)*(1/(s*(2*s2))))/((s3)(s3)*(s/(2*s2))),s,t);  1/6 + 8/3 exp(3 t)  5/2 exp(2 t)which is actually the answer in the back of the book. As I mentioned in class, I had to try four times to type the input correctly, matching the parentheses. But let me show you how to find the inverse Laplace transform "by hand".
I will multiply top and bottom by s(2s2). I will also "factor out" the s3 in the bottom.
(2/s)(s3)(1/[s[2s2]) 2(2s2)(s3) Y(s) =  =  ((s3)(s3)(s/[2s2])) (s3)(s(2s2)s^{2})The top: 2(2s2)(s3)=4s4s+3=3s1.
3s1 A B C A(s2)s+B(s3)s+C(s3)(s2) Y(s) =  =  +  +  =  (s3)(s2)s s3 s2 s (s3)(s2)sand 3s1=A(s2)s+B(s3)s+C(s3)(s2). Plug in s=0 to get C=1/6, plug in s=2 to get B=5/2, and plug in s=3 to get A=8/3. The answer will be what the book and Maple predicted.
I could get X(s) by putting the complicated rational function description of Y(s) in the equation 1X(s) + (s/[2s2])Y(s)=(1/[s[2s2]) and then solving for X(s). I am lazy and won't do this. Last year I discussed a simpler example.
What's going on?
Those students who have some background in linear algebra might see
some sort of pattern. I was doing row reduction in the system of
linear equations given for X(s) and Y(s). I was thinking of various
functions involving s as the scalars (!). My goal was to get 1
and 0 as scalars.
Linear algebra ...
Linear algebra is a subject which involves both scalars and
vectors. Let me tell you about both of them, from an
"operational" point of view.
Here are some examples of scalars and vectors.
HOMEWORK
I would like to give an exam in two weeks, on Monday, October 10. The exam would cover
our work on Laplace transforms and some basic linear algebra. Please
check
the draft version of a formula sheet for the first exam and
give me comments.
The exam will cover what we've done on Laplace transforms and two
lectures on linear algebra, so you probably should read ahead about
linear algebra (see the syllabus & textbook
problems.
Further information about the exam will be available soon.

Thursday, September 22 

A simple mechanics problem
I asked people to consider the frictionless eraser.
I tried to slide an eraser along the narrow shelf at the bottom of the
blackboard, asking people to imagine there was no friction. My
mathematical analysis (!) of this problem used F=ma, a wonderful
equation. So here we go: I use F=ma. I will assume that we are
analyzing the motion of a particle on a line. At time t, the particle
is at x(t). I will also assume (since this is a very simple
model!) that m=1 for all time t. If the particle starts from 0 at time
0 with velocity 0, the initial conditions are x(0)=0 and
x´(0)=0. The force will vary with time. What can one expect of such a
problem?
Then setup
Here's an example of a possible force varying with time, t. Initially, the particle would not move. Then, yes, indeed!, it moves as t increase (once t gets to the region where F(t)>0). Since F(t)>0 there, the particle moves to the right, or (if we consider a graph of t, time, versus, x(t), position) "up". Here is what a graph of x(t) might look like, qualitatively. What is happening? For early time, before the bump in F(t), the particle doesn't move at all. After the bump in F(t), the particle does move, and the amount of movement, it turns out, depends only on the total area under the bump. The total area determines the slope. The particle moves at uniform speed because there is no new force acting on it. What doess the total area represent? Well, work is force·distance, but this is essentially force·time, which in this setup is momentum. (Why? Since x´´(t)=F(t), x´(t)=p(t), the momentum (remember here m=1 always). Thus p´(t)=F(t), and the total area under the F(t) is the change in momentum, as Mr. Wolf suggested. Professor M. Kiessling, a wonderful mathematical physicist, helped (forced?) me to agree to this. In this simple setup, I think that work=energy and momentum gained are proportional. In any case, after the bump, the curve showing positive is linear, with a positive slope. The slope is directly proportional to the total net area. 

The example
The setup Suppose the graph of F(t) is as shown. Let's find the motion of the point. Since x´´(t)=F(t) with x´(0)=0 and x(0)=0, and this is Math 421, we should use the Laplace transform. Well, we can write F(t) in a suitable form: F(t)=(1/A)U(t)(1/A)U(tA). Then take the Laplace transform of x´´(t)=F(t) and get, on the lefthand side (with initial conditions built in), s^{2}X(s)sx´(0)x(0). On the right, we will get (1/A)(e^{0·s}/s)(1/A)(e^{As}/s). Therefore we have s^{2}X(s)=(1/A)[1e^{As}]/s and X(s)=(1/A)1/s^{3}(1/A)e^{As}(1/s^{3}). We now take the inverse Laplace transform. You will need to know the table and the second translation theorem! x(t)=(1/A)(1/2)t^{2}(1/A)(1/2)(tA)^{2}U(tA). For 0<t<A, x(t)=(1/A)(1/2)t^{2}. Since (tA)^{2}=t^{2}2At+A^{2}, after some algebra (!!), we see that if t>A, x(t)=(1/A)(1/2)(2AtA^{2})=(1/2)(2tA)=tA/2. It can be checked that x(A)=A/2 for either definition, so the motion is continuous (the eraser didn't suddenly and magically jump!) and actually, the function described is differentiable: the tangent lines agree at the "splices" (t=0 and t=A). A graph of the position is shown.  
Limits of the motion and what we expect 
Dirac delta "function"
Dirac,
a Nobel prizewinning mathematical physicist, worked on quantum
mechanics and relativity. Here is an interesting quote from Dirac: "I
consider that I understand an equation when I can predict the
properties of its solutions, without actually solving it." A contemporary interview with Dirac may
give some idea of his personality.
The Dirac delta function is the limit of these square impulse functions as A>0^{+}. The Dirac delta function, delta(t), is 0 away from 0. Its total integral is 1. You should think about it as an instantaneous bump, a limit of the boxes above as A>0. I computed some integrals. They resembled the following:
I computed
the integral _{300}^{40}delta(t2) dt: this
was 1.
The integral _{300}^{200}delta(t2) dt was
0. The bump or impulse was centered at 2, and 2 is not in the interval
from 300 to 200.
I tried some further computations. I think one was something like _{300}^{40}delta(t2)(t^{2}5t) dt.
Since the delta function is 0 away from 2, only what multiplies
it at (or close to) 2 matters. But close to 2, the function
t^{2}5t is close to 410=6. Therefore this integral is just
6.
A traditional mixing problem
I wanted to make the Chem E's happy. Here is a twopart problem which
I found last year. It is from an
essay by Kurt Bryan of the RoseHulman Institute of Technology:
A salt tank contains 100 liters of pure water at time t=0 when water begins flowing into the tank at 2 liters per second. The incoming liquid contains 1/2 kg of salt per liter. The well stirred liquid lows out of the tank at 2 liters per second. Model the situation with a first order ODE and find the amount of the salt in the tank at any time. 
I expect students had modeled and solved such problems a number of times in various courses, and certainly in Math 244.
Let m(t) be the kgs of salt in the tank at time t in seconds. We are
given information about how y is changing. Indeed, m(t) is increasing
by (1/2)2 kg/sec (mixture coming in) and decreasing by (2/100)m(t),
the part of the salt in the tank at time t leaving each second. So we
have:
m´(t)=1(1/50)m(t) and m(0)=0 since there is initially no salt in the
tank.
Students seemed to realize that the solution curve should look like
this: a curve starting at the origin, concave down, increasing, and
asymptotic to m=50. In the long term, we expect about 50 kgs of salt
in the tank. This is easy to solve by a variety of methods, but in
421 we should use Laplace transforms: so let's look at the Laplace
transform of the equation. We get sM(s)m(0)=(1/s)(1/50)M(s). Use
m(0)=0 and solve for M(s). The result is M(s)=1/[s(s+(1/50))]. This
splits by partial fractions with some guesses to
M(s)=50[(1/s)(1/(s+(1/50)))]. It is easy (it should be easy
for you by now!) to find the inverse Laplace transform and write
m(t)=50(1e^{(1/50)t}) which is certainly the expected
solution.
Slightly more interesting ...
Suppose that at time t=20 seconds 5 kgs of salt is instantaneously dropped into the tank. Modify the ODE from the previous part of the problem and solve it. Plot the solution to make sure it is sensible. 
Students should expect a jump in m(t)
at time 20 of 5, but then the solution should continue to go
asymptotically to 50 when t is large. How should the ODE be modified
to reflect the new chunk of salt? Here is one model:
m´(t)=1(1/50)m(t)+5delta(t20) and m(0)=0.
The delta function at t=20 represents the "instantaneous"
change in m(t) at time 20, an immediate impulsive (?) increase in the
amount of salt present.
It is very reasonable to ask if this is a good model. Please keep this is mind!
Back to solving m´(t)=1(1/50)m(t)+5delta(t20) and m(0)=0. Take the Laplace transform as before, and use the initial condition as before. The result is now sM(s)m(0)=(1/s)(1/50)M(s)+5e^{20s} from which we get M(s)=50[(1/s)(1/(s+(1/50)))]+5[e^{20s}/(s+(1/50))] which has inverse Laplace transform m(t)=50(1e^{(1/50)t})5U(t20)e^{(1/50)(t20)}. Of course I wanted to check my answer, so I used Maple. Here is the command line and here is Maple's response:
> invlaplace(1/(s*(s+(1/50)))+5*exp(20*s)/(s+(1/50)),s,t); t t 50 exp( ) + 50 + 5 Heaviside(t  20) exp(  + 2/5) 50 50so I am happy. I should remark that last year's version of Maple gave this response:
Some pictures Maple made for me
Here's the solution curve for the problem just solved, with an instantaneous increase of 5kg at t=20. The curve jumps up, and then begins to approach 50.  I modified the problem, and dumped in 30 kg of salt at time 60. So you can see the jump above the asymptote, and then the curve trends down towards 50. 
Hitting a spring
Let's look at y´´+y=delta(tPi)
with initial conditions y(0)=0
and y´(0)=0. O.k.: to me, what I hope is interesting for the
students in this course is the physical interpretation of
everything. Then we should "solve" it, and then we should consider the
solution from the physical point of view: if it makes sense, this
helps to validate the method we used.
Well: y´´+y models a vibrating spring with no damping, a rather ideal situation. The righthand side, delta(tPi) is a "forcing function". In fact, here it correspods to hitting the spring instantaneously (!) with a "force" of 1 at time Pi. The initial conditions tell us that the spring is originally in equilibrium. At time Pi, the spring is hit but there is no further force (so far). Probably we should expect that the spring will somehow vibrate in an ideal fashion. Let us try our wonderful Laplace routines. The Laplace transform of y´´+y=delta(tPi) is s^{2}Y(s)sy´(0)y(0)+Y(s)=e^{Pi·s} which turns out to be (s^{2}+1)Y(s)=e^{Pi·s} or Y(s)=e^{Pi·s}/(s^{2}+1). The predicted vibration of the spring is therefore (!) the inverse Laplace transform of e^{Pi·s}/(s^{2}+1). I can read that off from the (mythical?) table using the second translation theorem and the result is y(t)=sin(tPi)U(tPi): the spring, responding to an instantaneous load of 1 unit, then vibrates sinusiodally.
Hitting a spring again
Keep the initial conditions the same, but change the forcing function.
So here look at
y´´+y=delta(tPi)+delta(t2Pi). Now go through the
Laplace transform, using the initial conditions:
s^{2}Y(s)sy´(0)y(0)+Y(s)=e^{Pi·s}+e^{2Pi·s}
becomes
s^{2}Y(s)+Y(s)=e^{Pi·s}+e^{2Pi·s}
and then
(s^{2}+1)Y(s)=e^{Pi·s}+e^{2Pi·s}
so that
Y(s)=[e^{Pi·s}+e^{2Pi·s}]/(s^{2}+1).
The inverse Laplace transform tells me that
y(t)=sin(tPi)U(tPi)+sin(t2Pi)U(t2Pi).
It may be useful to understand what this algebraic mess means.
Confession 
Find the Laplace transform of tU(t7)
Use a version of the second translation theorem: the Laplace transform
of g(t)U(ta) is e^{as} multiplied by the Laplace transform
of g(t+a). So a=7, and g(t)=t, and g(t+a)=t+7 which has Laplace
transform 1/s^{2}+7/s. The answer is e^{7s}( 1/s^{2}+7/s).
Find the Laplace transform of e^{9t}t
We need the first translation theorem: the Laplace transform of
e^{at}f(t) is F(sa). Here a=9 and f(t)=t. So
F(s)=1/s^{2}, and the desired Laplace transform is
1/(s+9)^{2}. The + occurs because sa=s(9)=s+9.
Mr. Clark intelligently remembered a fact. So he did the following: the Laplace transform of e^{9t} is 1/(s+9), so the Laplace transform of te^{9t} is d/ds[1/(s+9)], and this is 1/(s+9)^{2}. I needed to be reminded that there are two minus signs in the result and they cancel. Sigh.
Find the Laplace transform of e^{5t}tU(t2)
This problem is vicious (viscous?). It involves a concatenation of
both translation results.
Today's word concatenation
To connect or link in a series or chain.
Let's do this two ways.
> with(inttrans); [addtable, fourier, fouriercos, fouriersin, hankel, hilbert, invfourier, invhilbert, invlaplace, invmellin, laplace, mellin, savetable] > laplace(exp(5*t)*t*Heaviside(t2),t,s); exp(2 s  10) (2 s + 11)  2 (s + 5)The answer has been made "pretty", and I can't tell how it was computed.
The QotD from last time
Ms. Jones kindly put her solution on
the board. The given information was a mixture of geometric and
algebraic. The graph of a function was drawn. It was 0 for t<1 and
t>3, and was (t1)(t3) for t between 1 and 3. I asked for the
Laplace transform of this function. I emphasized that it would
probably be easier to begin by describing the function using the
Heaviside function.
Function  Its Laplace transform 

t^{2}  2/s^{3} 
t^{3}  6/s^{4} 
(1/60)t^{6}  {6!/60}/s^{7} Notice that 6!/60 is 12. 
e^{At}  1/(sA) 
e^{Bt}  1/(sB) 
[1/(AB)]{e^{At}e^{Bt}}.  [1/(AB)]{1/(sA)1/(sB)}. Notice that, combining fractions, this happens (?) to equal 1/[(sA)(sB)] 
Convolution and Laplace transform
The Laplace transform of the convolution is the product of the Laplace
transforms: the Laplace transform of f(t)_{*}g(t) is F(s)·G(s).
This is Theorem 4.9 of the text (pp.216217) and you should at least glance at the proof there, please. The table above contains two examples of this fact.
What is the convolution of t and t^{2} and t^{3} and t^{4}?
We can compute t_{*}t^{2}_{*}t^{3}t^{4} by
computing the Laplace transforms of each function, multiplying them,
and then taking the inverse Laplace transform. You should judge if
this is "easier" than a direct computation. So:
t becomes 1/s^{2}
t^{2} becomes 2/s^{3}
t^{3} becomes 6/s^{4}
t^{4} becomes
24/s^{5}
and the product of the Laplace transforms is
288/s^{14}, whose inverse Laplace transform is
(288/13!)t^{13}, which is, of course, the desired
convolution. So this is not a very difficult exam question. Please
note that (288/13!) is 1/21,621,600 which I almost would hate
to read on an exam answer  I would not want students to use time
and effort resolving straightforward arithmetic.
5 A Bs+C  =  +  [(s3)(s^{2}+25)] s3 s^{2}+25Now combine the fractions, and let's look at the top of the result:
Therefore we will find the convolution of e^{3t} and sin(5t)
by discovering the inverse Laplace transform of
(5/34)/(s3)(5/34)s/(s^{2}+25)(15/34)/(s^{2}+25).
I can do this by looking things up in my handy table of Laplace transforms. (If I use the table enough, then I won't need it!) So the answer is
(5/34)e^{3t}(5/34)cos(5t)(3/34)sin(5t). (The 15 becomes 3
because the Laplace transform of sin(5t) already has a 5 "on top").
What is the Laplace transform of _{0}^{t}f(tau)dtau?
This is the convolution of f(t) and the function 1 (that is, the
function whose value for any t is 1). Therefore the Laplace transform
is the product of the Laplace transforms of f(t) and 1: that is, just
F(s)/s. This result should be contrasted with what we got for
derivative: the Laplace transform of f´(t) is sF(s)f´(0). The results
aren't exactly opposite since the f' has an initial condition built
in, but then the integral starts at 0 and is therefore equation
to 0 when t=0.
$ and *
Please see here for a discussion of
convolution and the "present value of an income stream"
Step 1
If F(s) is the Laplace transform of s, then we have
F(s)=(1/s)+(2/s^{2})(8/3)(6/s^{4})F(s) because a
convolution turns into a product under Laplace transform.
Partial fractions ... World's quickest practical review. This is an algebraic method of transforming quotients of polynomials (rational functions). Students in calculus see it for the first time as a method allowing (in theory!) any rational function to be antidifferentiated in terms of familiar functions. If the quotient is P(t)/Q(t), then there are a sequence of steps:

> invlaplace((s^3+2*s^2)/(s^4+16),s,t); 1/2 1/2 1/2 1/2 1/2 1/2 1/2 1/2 2 sinh(2 t) cos(2 t) + 1/2 cosh(2 t) (sin(2 t) 2 + 2 cos(2 t))This is consistent with the factorization I got above for s^{4}+16. Huh. But it is not the same as the answer in the back of the book. In fact, the answer in the back of the book is the answer if the minus sign in the problem is changed to a plus sign.
Restated book problem
f(t)=1+t+(8/3)_{0}^{t}(ttau)^{3}f(taudtau.
Now we need the inverse Laplace transform of
[s^{3}+2s^{2}]/[s^{4}16]. I can factor
s^{4}16 and I hope you can also. First, it splits up as
(s^{2}4)(s^{2}+4), and then further as
(s2)(s+2)(s^{2}+4). So we need A, B, C, and D so that
s^{3}+2s^{2} A B Cs+D  =  +   s^{4}+16 s2 s+2 s^{2}+4Now push the fractions together on the right, and look at the tops of both sides. The equation resulting is
If any student sees any errors I made, please tell me. Here is a link to a solution of problem #38 in section 4.4 which I did a year ago. At least I think it did it correctly! It is similar to the problem I attempted here!
Confession 
More about the Heaviside function
We talked a bit more about how the Heaviside function could be used to
write piecewisedefined functions. In particular, I mentioned that
Maple has the name Heaviside reserved for this
function.
I know little about Matlab but Mr. O'Sullivan wrote me the following
message:
Interestingly enough, the Heaviside function is only recently (version 7) supported by default in MATLAB. But, it works as expected. Looks like they chose to just leave U(0) undefined, though. >> x=5:5 x = 5 4 3 2 1 0 1 2 3 4 5 >> heaviside(x) ans = 0 0 0 0 0 NaN 1 1 1 1 1 Trying the function in MATLAB <= 7 or Octave (an open source MATLAB "clone") results in brokenness.
The Second {ShiftingTranslation} Theorem
(Textbook Theorem 4.7, p.208) If the Laplace transform of
f(t) is F(s) and a>0, the Laplace transform of
f(ta)U(ta) is e^{as}F(s).
You don't need to think about this verification again. Just learn how to use it.
Examples
What is the Laplace transform of t^{2}U(t5)?
This is a bit tricky. The "template" of the Second Translation Theorem
is f(ta)U(ta). I guess a should be 5. But then we need to
"recognize" f(ta)=f(t5) as t^{2}. Here is the trick.
t^{2}=([t5]+5)2. This isn't very profound, but it will
work. So ([t5]+5)2=(t5)^{2}+2(t5)+25, and apparently
f(t5)=(t5)^{2}+2(t5)+25 so that
f(Q)=Q^{2}+2Q+25. Therefore the Laplace transform of f is
(2/s^{3})+2/s^{2}+(25/s). THEREFORE
the Laplace transform (according to the Second Translation Theorem)
is e^{5s}[(2/s^{3})+2/s^{2}+(25/s)].
Another version of the Second {ShiftingTranslation}
Theorem
I find an alternate version of the Second Translation Theorem a bit
easier to use. (It's all psychological  is everything
psychological?). Here it is:
The Laplace transform of g(t)U(ta) is e^{as} multiplied by
the Laplace transform of g(t+a).
Examples
I think I tried to find the Laplace transform of U(tPi)sin(t). This
is e^{Pi s} multiplied by the Laplace transform of
sin(t+Pi). Now sin(t+Pi) is sin(t), which has Laplace transform
1/(s^{2}+1)
Weird formula
Here is another formula from the textbook. The Laplace transform of
t·f(t) is F´(s). Why is this?
Therefore the Laplace transform of t·f(t) is (d/ds)F(s). The
textbook remarks that you can repeat this n times, and the result
is:
the Laplace transform of
(1)^{n}t^{n}·f(t) is
(d^{n}/ds^{n})F(s).
For x=0, the partial derivative is 0 (look at the formula!), and the integral is 0.
Convolution
Suppose that f(t) and g(t) are functions defined for nonnegative
numbers, so their domains include [0,infinity). Then the
convolution of f and g will be another function defined on
[0,infinity) defined by
f_{*}g(t)=_{0}^{t}f(tau)g(ttau) dtau.
What's the convolution of t^{2} and t^{3}? This is
_{0}^{t}tau^{2}(ttau)^{3}dtau.
After a huge amount of debate, we decided (please see here
or here if
this is unfamiliar to you) that (ttau)^{3}=
1t^{3}3t^{2}tau^{1}+3t^{1}tau^{2}1tau^{3}
and therefore
_{0}^{t}tau^{2}(ttau)^{3}dtau=
_{0}^{t}tau^{2}[1t^{3}3t^{2}tau^{1}+3t^{1}tau^{2}1tau^{3}]dtau=
_{0}^{t}tau^{2}t^{3}3t^{2}tau^{3}+3t^{1}tau^{4}tau^{5}dtau=
(1/3)tau^{3}t^{3}(3/4)t^{2}tau^{4}+(3/5)t^{1}tau^{5}(1/6)tau^{6}]_{tau=0}^{tau=t}=
(1/3)t^{6}(3/4)t^{6}+(3/5)t^{6}(1/6)t^{6}.
Wow, what a computation. Of course, the result can be
simplified (who could possibly care?) to get
(1/60)t^{6}.
Mr. O'Sullivan suggested the following web page: http://mathworld.wolfram.com/Convolution.html. It has a discussion of convolution and an animation picture of several convolutions.
I need to give more background about this material, and I will.
Confession 
A textbook problem
I daringly attempted problem #36 in section 4.2. 36 is a large number
and even, so the problem must be difficult and there is no
answer in the back of the book!
Section 4.2, Problem #36 Solve the Initial Value Problem:
(ODE) y´´4y´=6e^{3t}3e^{t}
(IC's) y(0)=1,
y´(0)=1
As I remarked, this is a very simple ODE: it is second order, linear, constant coefficient, inhomogeneous. What methods would I expect/hope that students could use on this problem?
Method #1 First solve the "associated homogeneous equation", y´´4y´=0. Plug in e^{rt}, get the characteristic equation, etc. Attempt to get a particular solution using a variety of methods (undetermined coefficients or throwing dice or ...).
Method #2 Let y_{1}=y and y_{2}=y´. Build
a column vector
Y=( y_{1} y_{2})^{t}. Note: the
silly superscript t means transpose (rows go to columns), and I am
using it here (didn't use it in class yet!) because it is harder to
type a column vector in html than a row vector. Sigh. Now if we define
a 2by2 matrix, A, to be this:
(0 1)
(4 0)
then
I hope that the matrix differential equation Y´=AY+S (S=other
stuff I don't want to bother with now) is the same as the original
problem. One can then apply methods of linear algebra (we'll do some
of this later in the course) to solve the matrix DE which in turn will
lead to a solution of the original equation.
Method #3 Use numerical techniques to approximate a solution. For many applications, this is just as good. For some applications such as those requiring longterm asymptotics as a function of some symbolic initial conditions, for example, numerical methods aren't too useful. Errors in ODE "stepping" methods (the most elementary of which is Euler's Method, and one frequently used method is RKF4, a RungeKutta method) tend to accumulta.
Method #4 Use the Laplace transform, which I'll try in a second. However, there is a natural question: why learn another method? Well, for this initial value problem almost every method "works". But each method has its own flavor (?) and gives a different perspective on the ODE. None of these methods will apply perfectly or easily or every problem. Having a variety of methods will allow you to analyze solutions of ODE's better.
So what we do is take the Laplace transform of the entire equation y´´4y´=6e^{3t}3e^{t}. At this time please realize that there was a table of Laplace transforms on the board. So the Laplace transform of the righthand side is easy, using linearity and entries in the table. It is [6/(s3)]+[3/(s+1)]. What about the righthand side? The Laplace transform of the unknown function, y, is usually written, Y. Then what's the Laplace transform of y´´?
Laplace transforms of derivatives
The Laplace transform of y^{(n)} is
s^{n}Ys^{n1}y(0)s^{n2}y´´(0)...y^{(n1)}(0).
> with(inttrans); # loads various useful diff'l equations transforms [addtable, fourier, fouriercos, fouriersin, hankel, hilbert, invfourier, invhilbert, invlaplace,invmellin, laplace, mellin, savetable] > invlaplace(6/((s^24*s)*(s3)),s,t); 3/2 exp(4 t)  2 exp(3 t) + 1/2So I guess we were correct.
Then I gave up, at least computing by hand. But:
> invlaplace(s/(s^24*s)5/(s^24*s)+6/((s^24*s)*(s3))3/((s^24*s)*(s+1)),s,t); 11  exp(4 t) + 5/2  2 exp(3 t)  3/5 exp(t) 10I bet that (11/10)e^{4t}+5/22e^{3t}(3/5)e^{t} is a likely answer. For your amusement, I remark that I had to try three times before I typed in the correct Maple command.
I have experience with several different versions of Maple and they sometimes don't always give the same versions of the answer. For example, some versions package a linear combination of e^{3t} and e^{3t} as an equivalent linear combination of sinh(3t) and cosh(3t). This can be annoying. I've even found a version of Maple (not the current one!) which insisted on using i sinh(i t) instead of sin(t). So using symbolic manipulation software can have its problems as well as benefits!
What kinds of functions theory interlude #1
The classical Laplace transform works with functions that have growth
no bigger than exponential growth. This means that a function f(t)
must satisfy some sort of estimate of the following type:
There are positive constants A and B so that y(t)<Ae^{Bt}
for t>C, where C is some additional positive constant.
There's sort of a picture of the situation to the right. The "Stuff
happens" doesn't sort of matter. What matters is that for t large
enough (t>C), y(t) is "trapped" between Ae^{Bt} and
Ae^{Bt}. The two curves occur because there are absolute
value signs around the y(t). If y(t) is trapped that way, then in the
integral that defines the Laplace transform, _{0}^{infinity}e^{st}y(t) dt,
when s>B, the integral is less than e^{(Bs)t} in absolute
value, and since this is exponential decrease the integral must
converge.
You can find lots of references to this stuff on the web.
Our major present goal is to expand our table of Laplace transforms (and, therefore, inverse Laplace transforms). The two shifting or translation theorems are very important in accomplishing this.
Easy examples
Uhhh, can we find the Laplace transform of e^{5t}sin(3t)? The
table stated that the transform of sin(3t) is 3/(s^{2}+9). But
(take a=5) the Laplace transform of e^{5t}sin(3t) must then be 3/((s5)^{2}+9).
More ofter we'll need to apply this result backwards. Can we find a
function whose Laplace transform is 5/(s+7)^{14}? We look at
the table and see that this seemed to be Constant/s^{14}
shifted. So this should be related to the Laplace transform of
t^{13}. We needed to fix up the multiplicative constant so
that we would get 5. And therefore we multiply t^{13} by 5 and
divided it by 13! so that the Laplace transform of
(5/(13!)t^{13} is 5/s^{14}. Now we need to shift by
7, and conclude that the Laplace transform of
e^{7t}(5/(13!)t^{13} is 5/(s+7)^{14}.
Isn't this cool!
The Heaviside function
I defined the Heaviside or unit step function, called
U(t) in your text (actually with a calligraphic U). Oliver
Heaviside was a brilliant English engineer whose life was,
overall, rather sad. U is the function which is 0 for t<0
and is 1 for t>=0. There's a jump of 1 at 0, and otherwise the
function's graph consists of two half lines.
I think I graphed a few examples, like U(t3) (the jump is moved to 3) and U(t)+4U(t3)2U(t6) (the graph "starts" at 0 with a jump up at 1, jumps up 4 (to 5) at 3, and then down 2 (to 3) at t=6  aside from the jumps the graph just is pieced together from horizontal line segments. 
Suppose y(t) is
the function which is 0 for t<0, is t^{2} for t in the
interval [0,1], is 1 in between 1 and 3, is (t4)^{2} in the interval [3,4], and is 0 for "later" t. I hope I've drawn a picture of y(t) to the left.
What's a formula for y(t)? I sort of work from left to right in t.
First there's nothing. Then we want to "turn on" t^{2} at t=0,
so we need t^{2}U(t). Then we need to turn off t^{2}
at t=1, and turn on a height of 1. Hey, change the formula to
t^{2}U(t)+(t^{2}+1)U(t1). Now let's continue to the
right, where we need to turn of the height of 1, and turn on the
downway parabolic arc. This change is (1+(t4)^{2})U(t3) and
should be added to what we already had. Then, finally, at 4, turn off
(t4)^{2} with (t4)^{2})U(t4). So a formula for this function is
t^{2}U(t)++(t^{2}+1)U(t1)+
(1+(t4)^{2})U(t3) (t4)^{2})U(t4)
Here is what Maple does. The first instruction defines the function, y(t). This uses the builtin function "Heaviside", what we call U(t). The second instruction plots it. And the plot is shown.
>y:=t>t^2*Heaviside(t)+(t^2+1)*Heaviside(t1)+(1+(t4)^2)*Heaviside(t3) (t4)^2*Heaviside(t4): >plot(y(t), t=1..6,thickness=3,scaling=constrained);Now we are lost ...
Return of the entrance exam
I retruned the Entrance Exam. I view this exam as a very useful
diagnostic, which has shown some correlation with final course grades
in Math 421. For me, it is useful because the exam shines a light on
two aspects of students. First, it clearly shows student knowledge of
representative content: everything asked on the exam will be
relevent to parts of the course, and, actually everything asked on the
exam is important in sections of the course. Second, given an
Entrance Exam like this to a group of somewhat sophisticated
educational "consumers" reveals a bit about their psychology: are they
willing to work, to review, to ask questions ... students who do these
things are more likely to learn, and not just barely survive. As I
mentioned in class, this is a fairly advanced undergraduate course,
and the person most resposible for teaching you is ... well, you! I
hope I will be helpful, but stuff will go by very rapidly. I strongly recommend working with one or more other
students in the course on a regular basis, meeting weekly or more
often for an hour or two. This is likely to keep you "on task", and
working together you will increase the odds of success.
HOMEWORK
You should be reading and doing the problems from sections 4.3 and
4.4, please.
Confession 
I hope you will look at this web page. Uhhh ... it is the page you are looking at now. If you aren't looking at it now, then maybe you could look at it some time. Or if ... oh well.
I began by reminding people of the definition of Laplace transform. The Laplace transform of a function y(t) is a function Y(s). Y(s) is _{0}^{infinity}e^{st}y(t) dt. This is also called L(y(t)). Here L means what is usually written by a calligraphic capital L. It is standard notation for the Laplace transform.
Laplace transform is linear
L(y_{1}(t)+y_{2}(t))=_{0}^{infinity}e^{st}[y_{1}(t)+y_{2}(t)] dt=_{0}^{infinity}e^{st}[y_{1}(t)]+e^{st}[y_{2}(t)] dt=_{0}^{infinity}e^{st}y_{1}(t) dt+_{0}^{infinity}e^{st}y_{2}(t) dt=L(y_{1}(t))+L(y_{2}(t)).
Similarly, if c is a constant, then L(cy(t))=cL(y(t)).
Linearity is a tremendously useful property, and we will use it
constantly.
Then I copied most of Theorem 4.2 (on p. 193 of the textbook). I checked that the Laplace transform of e^{at} is 1/(sa) as the theorem stated. Here is what we did.
I wanted to verify the theorem's result for the Laplace transform of cos(kt). I remarked that the definition tells me that I need to compute _{0}^{infty}e^{st}cos(kt) dt. Maybe this really isn't too difficult to compute directly. I would integrate by parts. But I will need to integrate by parts twice, and the chance for error is large. Let me show you how to do the computation another way.
A Laplace transform of one little triangle
One of great powers of the Laplace transform is its ability to deal
with rough functions: functions that may not be differentiable
or may not be continuous (or even, as we will see later, may not be
functions!). These rough functions are probably more accurate models
of many physical situations than the standard functions of calculus.
Here is an example. We will consider the piecewisedefined function
=0 for t<0
y(t)=t for t for t in [0,1]
=0 for t>1
A graph of this function is shown to the right. It certainly is not
continuous at t=1. We will compute its Laplace transform.
_{0}^{1}e^{st}t dt= u dv = uv]  v du u= } { du= dv= } { v=Then I try to make a good choice for u and see what happens. Your experience must guide you, but generally you want to "exchange" the u dv integral for a "better" v du integral.
Here we can use u=t so dv=e^{st} dt. Then du=dt and
v=(1/s)e^{st}. The integration by parts becomes
_{0}^{1}e^{st}t dt=
(t/s)e^{st}]_{0}^{1}_{0}^{1}(1/s)e^{st} dt
Now the "boundary term" (as the uv term is sometimes called) is
(t/s)e^{st}]_{0}^{1} which contributes
e^{s}/s to our result.
_{0}^{1}(1/s)e^{st} dt can be
integrated directly and we get
{{1/s^{2}}}e^{st}]_{t=0}^{t=1} and this is
e^{s}/s^{2} (from t=1) {1/s^{2}} (from
t=0).
The total result is
e^{s}/se^{s}/s^{2}+{1/s^{2}} and, in one fraction, this is
(se^{s}e^{s}+1)/s^{2}
One comment to make is that we have replaced the rough function by a function given by complication combinations of nice smooth functions. Still, it isn't clear that this is a "win" unless further good things occur. They will.
Simple Laplace transform asymptotics
It might be useful to have some simple things that can be checked
about Laplace transforms. Then we might be able to look for errors,
etc. Also, the asymptotic statements I'll make turn out to be what can
be observed. Frequently, for example, in chemical kinetics, we can
only see what happens for large values of the parameter  things
change too quickly.
Suppose our y(t) is some sort of collection of lumps and bumps, sort
of as shown. Well, the Laplace transform formula multiplies this by
e^{st}, and integrates dt. We will only consider positive s,
so the exponential is always decreasing. When s gets larger and larger
positive, the exponential decreases more and more rapidly to 0. Yes,
it stays at 1 when t=0, but the dropoff gets very rapid indeed. The
product of the exponential with y(t) will be some collection
{lb}umps, but the amplitude, the height, will surely >0. So
for the functions we are considering, Y(s)>0 as s>infinity.
The Laplace transform of f´(t)
Here is a major result which will help to answer "Why are we doing all
this?"
If F(s) is the Laplace transform of f(t),
what is the Laplace transform of f´(t)?
Well, let's start with _{0}^{infinity}e^{st}f´(t) dt,
which is the Laplace transform of f´(t). Integration by parts can
be used, with u=e^{st} and dv=f´(t) dt, so that
du=s e^{st}dt and v=f(t). Therefore
_{0}^{infinity}e^{st}f´(t) dt=e^{st}f(t)]_{0}^{infinity}_{0}^{infinity}(s e^{st}f(t) dt.
There is all sorts of sneaky stuff going on here, and we should be
very careful. Let's see. The boundary term is
e^{st}f(t)]_{0}^{infinity}. Now
s>0, so when t>infinity, e^{st}f(t)>0. Technically I
am using the fact that f(t) doesn't grow too fast (in fact, eventually
it is "killed" by exponentials decreasing fast enough, e^{st}
for large s), but let's just try to get the mood of this method. When
t=0, we get f(0) since e^{0}=1. What about the integral term?
Notice that there are two minus signs. They cancel. What's left
is s multiplied by the Laplace transform of f. So now we know:
L(f´)(s)=f(0)+sL(f)(s).
HOMEWORK
Read 4.1 and 4.2, and do the appropriate problems in the
syllabus. Please hand in on Monday: 4.1:10,34 and 4.2: 13,33
Confession 
Special permissions for math courses can only be gotten by applying on the web.
The standard clerical stuff was done. The information discussed, and much more, is available here and here.
I began by reviewing very briefly the idea of initial value problems (IVP's) for ordinary differential equations (ODE's). An ODE is an equation involving an unknown function of one variable and its derivatives. For example, y´=2t. A solution to such an equation is a function which when substituted into the equation makes it "identically" true. For example, in this simple equation y=t^{2}+C (any constant C) solves the equation. An initial condition (IC) is a value, "t_{0}", for which we require the function to have a certain value, "y_{0}". For example, we could ask that the solution to this ODE satisfy the IC (3,17). Then we choose C so that 17=3^{2}+C is true: I guess C=8. The combination of ODE and IC is called an initial value problem. We expect, due to simple physical examples, that IVP's will have unique solutions. The independent variable in this course will frequently be called t for time, so the dependent variable y is a model of some process which depends on time.
Here is one version of the major theoretical result of the subject, the
Existence and Uniqueness Theorem for ODE's
Suppose f(t,y) is a differentiable function in both t and y for
(t,y)'s in a region in R^{2}, and that the point
(t_{0},y_{0}) is in the region. Then the ODE
y´=f(t,y) has a unique solution going through the point
(t_{0},y_{0}).
Another way of writing the IC is y(t_{0})=y_{0}.
This is a wonderful theorem, but its usefulness is definitely limited. You shouldn't read "into" it any more than is already there. So let me show this with some examples.
General disclaimer The examples discussed in class will mostly be very artificial, and chosen so that "hand" computation is practical.
Example 1 y´=e^{(t2)} and y(0)=0. Then the solution is y(t)=_{0}^{t}e^{(w2)}dw. It turns out (and it can be proved!) that this integral can't be simplified or written in terms of any of the standard functions of calculus, using algebraic means of combination, or even using function composition.
Generally, it is impossible to write solutions of a random ODE in terms of familiar functions. Students should know that this is true, even if we use Maple or Matlab or Mathematica or ... and it even be very difficult or impractical to approximate solutions numerically. Sigh. The examples and methods that are shown here and in Math 244 really are a collection of tricks which work on many ODE's modeling physical situations. They are not guaranteed to work on all ODE's, or even all ODE's derived from fairly simple physical models. But the tricks are very useful in many examples. Now, back to work.
Example 2 The solutions to y´=ty^{2} are not difficult to get (the instructor of course made some errors). This is a separable ODE. I separated the variables, integrating, cleared up some algebra, and the solutions seemed to be y=2/(Ct^{2}). I verified this in the most direct way possible, by computing y´ and checking that the result was ty^{2}.
Let's look at some specific solutions. The solution satisfying the IC
(0,100) has C=2/100, so y=2/(.01t^{2}). The natural domain of
this function is (1/10,1/10). As t>1/10^{}, y>infinity:
the solution explodes. If the IC is (0,2·10^{4}) the solution is
y=2/(.0001t^{2}) and it has domain (1/100,1/100). There is
one special case (you need to look critically at the separation of
variables process to spot what goes wrong) but if the initial
condition is (0,0), the solution is y=0 for all t.
The picture to
the right is supposed to an impression of the geometry of the solution
curves to this ODE. Notice, please, that the Existence and Uniqueness
Theorem is totally correct, but really unhelpful. No inspection of
f(t,y)=ty^{2} seems to show anything wrong (!!) with the
function, but many of the solutions don't "live" very long, and they
live for different amounts of time.
The solutions blow up in the most immediate way
(y>infinity) at different values of t. This ODE is nonlinear. We
will mostly concentrate on linear ODE's, where it turns out that such
problems don't occur.
The most routine example is y´´+y=0, which is the ODE modeling an ideal spring using Hooke's law. We can still learn things from this example! This is a second order ODE (order refers to the highest number of derivatives needed to write the equation). Here the trick is to guess a solution: try y=e^{rt}. Then y´´+y=0 becomes magically e^{rt}(r^{2}+1)=0. The exponential function is never 0. Therefore the guess solves the ODE exactly when r^{2}+1=0 (I think this is called the characteristic equation). The roots of this characteristic equation are +/i.
This is a linear ODE. The particular very very important qualitative consequence is that sums of solutions of solutions of the homogeneous equation are solutions, and so are constant multiples of solutions. Why is this true? Look:
Let's use linearity. y´´+y=0 has solutions e^{it} and e^{it}, so if C_{1} and C_{2} are any constants, then C_{1}e^{ix}+C_{2}e^{ix} must be a solution. Let me search for a solution satisfying certain initial conditions. Since y´´+y=0 is a second order equation, the IC's will generally have two parameters. I want a solution satisfying y(0)=1 and y´(0)=0. I will call this an initial position solution, y_{P}. What is it?
If y_{P}(t)=C_{1}e^{it}+C_{2}e^{it} then y_{P}(t)=iC_{1}e^{it}iC_{2}e^{it}. We can get y_{P}(0)=1 by having C_{1}+C_{2}=1. We can get y_{P}(0)=0 by having iC_{1}iC_{2}=0 which is the same as C_{1}C_{2}=0. After some massive amount of thought, we managed to solve this system of linear equations: C_{1}=1/2 and C_{2}=1/2. Therefore y_{P}(t)=[e^{it}+e^{it}]/2.
Similarly, we can solve the initial value problem y´´+y=0 with y(0)=0 and y´(0)=1, which I'll call the initial velocity solution, y_{V}(t). It turns out to be y_{V}(t)=[e^{it}e^{it}]/(2i). You should check this!
Why would one want to know the y_{P} and y_{V} solutions? Well, they are really neat if you want to "solve" lots of initial value problems for y´´+y=0. The pattern of initial conditions (1 and 0, and 0 and 1) allows us to write a solution for the IVP y(0)=7 and y´(0)=13. Here it is: 7y_{P}(t)13y_{V}(t). This is so easy.
Everything I've written is correct, but of course some important things have not been written! First, writing the solutions as complex exponentials conceals important features of the solutions. For example, since y´´+y=0 models simple harmonic motion, the solutions had better be bounded (they shouldn't grow to infinity in any fashion). I think springs don't do that. And maybe my formulation of y_{P} and y_{V} doesn't entirely display this. But we could compare power series or do other stuff and, in some fashion, remember Euler's formula and its consequences. So here read this, please:
e^{it}=cos(t)+i sin(t) and cos(t)=[e^{it}+e^{it}]/2 and sin(t)=[e^{it}e^{it}]/(2i) 

What happens if we just change a sign in the ODE? If y´´y=0, the characteristic equation becomes r^{2}1=0 with roots r=+/1. Solutions are then C_{1}e^{t}+C_{2}e^{t}. The solutions corresponding to the Position and Velocity initial conditions (which I hope I have convinced you are useful, when combined with linearity, in solving IVP's are as follows:
If we now consider a "general" second order, linear, constant coefficient, homogeneous ODE (by the way, each phrase or word I've just written should make some sense to you and if any do not, you must review material from 244  see the syllabus for suggested reading in our text), Ay´´+By´+Cy=0 then I can tell you what to expect about the solutions. If B^{2}4AC<0, probably sines and cosines will appear. If B^{2}4AC>0, the solutions can be written with coshes and sinhes. There will also be some exponential factors (damping or otherwise). What happens if B^{2}4AC=0? It's a mystery, and you should figure it out.
Part I of the course: Escaping the wolves! 

The snow was coming down thicker and the chill, at first merely uncomfortable, was becoming a serious problem. With nightfall, we could hear the howls of the wolves coming closer. It was time to try for the Duke's castle and safety! I grabbed the child, and jumped on my horse. I called to the others in the party, "Get on your brave steeds, and ride rapidly to the castle ..." 
No, no, no!
This is a complicated definition.
Remarks about the definition
Example 2 What is the Laplace transform of f(t)=t? Now we need
to compute
_{0}^{infinity}e^{st}t dt with a
little more care. We need the antiderivative of e^{st}t with
respect to t. This can be done using integration by parts: if u=t,
then everything else is determined:
u=t and dv=e^{st}dt so du=dt and v=(1/s)e^{st}ds.
Now the integral becomes (I am definitely shortening the computation,
eliminating all the limit "stuff"):
The pattern:ORIGINAL INTEGRAL=u dv] v du
_{0}^{infinity}e^{st}dt=(t/s)e^{st}]_{0}^{infinity}_{0}^{infinity}(1/s)e^{st}ds.
Now (t/s)e^{st} when t=0 is surely 0. And (s>0!) when
t"="infinity (this is shorthand for a limit, really) the value is 0
also, because exponentials with negative argument decrease faster than
polynomials grow. So the boundary term in this integration by
parts is 0 (the stuff arising from ]). As for the other, the
(1/s) can be pulled out of the integral, since it is a dt
integral. And the result (two minus signs make a +) is just
(1/s)_{0}^{infinity}e^{st}dt. The integral
was computed in example 1 and had value 1/s. So the Laplace transform
of t is 1/s^{2}.
Example 3 What is the Laplace transform of f(t)=t^{2}?
Philosophical interruption Much of mathematics is pattern
matching, and then attempting to verify the detected patterns. So far
we have:
If n is a nonnegative integer, the Laplace transform of t^{n} is n!/s^{n+1}.
HOMEWORK
Please begin to read the chapter
in the book about Laplace transforms, and begin the problems. Do the
Entrance Exam, and have the result
of that ready to hand in next Thursday.
Maintained by greenfie@math.rutgers.edu and last modified 9/2/2005.