Intro. to Differential Equations


by ExtravagantDreams
Tags: differential, equations, intro
Fritz
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#55
Nov2-04, 03:44 AM
P: 75
Quote Quote by ExtravagantDreams
[tex]
\frac {dy} {dt} = ay - b
[/tex]
Here is y the independent variable and a and by the dependent variables?
QuantumTheory
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#56
Dec10-04, 02:14 AM
P: 224
Quote Quote by zeronem
You get rid of the parts of the equation that has more then one differential because more then one differential is just simply too small to have any effect on the whole equation.1 = 0 [/tex] is an implicit solution to ..
YES, YES, YES!!

This is a problem I am having!

I understand that alot of people here dislike the book Calculus For Dummies, for beginners. Alot of people in the real world cannot understand calculus, it is hard. This book really opened my eyes. I understood that a derivative if nothing but a constant rate. I learned the basics to move on.

However, before I got this book, I picked up my first calculus book in my school library. It was made in the very early 1900's, and was about calculus. Beginner calculus.

In this book, he actually had equations of multi-varibles!

For example, [tex]dx^2/dy^2[/tex], or anything with an infinitely small piece of a deriviative squared.

Since a deriviative is an infinitely small section, out of a whole of an infinite amount of infinitely small pieces, how could you have basically a negative infinity squared? This just doesn't make sense. You're already at some negative infinity rate of one [tex]dx[/tex]. So if the derivative is the rate of some curve(therefore the height of the curve), lets call it [tex]f(t)[/tex].

Then,

[tex]f(t)[/tex] = [tex]f(t)dx[/tex]


Does this make sense? I made it up myself..I hope its right.

Basically, I don't understand how you could possibly have some derivative, of some function, squared.

Ugh, anyway, I hate infinity, it confuses me sometimes :(

Weird how an infinite amount of infinitely thin pieces can equal a finite amount, eh?
poolwin2001
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#57
Dec31-04, 06:22 AM
P: 173
The link is not working.
whozum
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#58
Apr12-05, 12:11 AM
P: 2,223
I have to say this guide is amazing. I have one question on post 25:

py'' + qy' + ry = 0

Solving the homogeneous equation will later always provide a way to solve the corresponding nonhomogeneous problem.

I'm not going to proove all this but you can take the kernal of this funtion as

ar2 + br + c = 0

and you can, so to speak, find the roots of this funtion.

r1,2 = (-b √(b2 -4ac))/2a

r1 = (-b + √(b2 -4ac))/2a
r2 = (-b - √(b2 -4ac))/2a

Assuming that these roots are real and different then;
y1(t) = er1t
y2(t) = er2t
How did you get from the above to the last statements?
dextercioby
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#59
Apr12-05, 12:50 AM
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Which one,the first statement?

Daniel.
whozum
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#60
Apr12-05, 03:36 AM
P: 2,223
How does:

Assuming that these roots are real and different then;
y1(t) = er1t
y2(t) = er2t
follow from

py'' + qy' + ry = 0

Solving the homogeneous equation will later always provide a way to solve the corresponding nonhomogeneous problem.

I'm not going to proove all this but you can take the kernal of this funtion as

ar2 + br + c = 0

and you can, so to speak, find the roots of this funtion.

r1,2 = (-b √(b2 -4ac))/2a

r1 = (-b + √(b2 -4ac))/2a
r2 = (-b - √(b2 -4ac))/2a
I think i understand the root analogy and how r1 and r2 come about, but I dont understand why y1 and y2 follow that structure.
dextercioby
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#61
Apr12-05, 10:14 AM
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That's how u got the characteristic equations,by assuming exponential type solutions and plugging in the ODE.

Daniel.
Maxos
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#62
Jun27-05, 02:56 PM
P: 101
Yes, you make the hypothesis:

Y(t)=e^(a*t)

then, if Y(t) must follow the diff.eq. for any real t

e^(a*t)*(p*a^2+q*a+r)=0

from this You get the two (or one) values.

Then, for some algebrical reasons, you demonstrate that the general integral is a linear combination of the found solutions.
jaap de vries
jaap de vries is offline
#63
Aug19-05, 10:47 AM
P: 254
Quote Quote by ExtravagantDreams
Does anyone know an easier way for writing math on the computer and one that looks less confusing. I know I will have difficulty finding some things, especially subscripts and superscripts. Anyone know a better way to denote these?
Just use _ and ^ in LAtex its pretty easy
Ruslan_Sharipov
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#64
Sep3-05, 05:29 AM
P: 95
Quote Quote by ExtravagantDreams
Looking things up and explaining it to others seems to be the best way to learn.
I would like to suggest you something more than to learn a standard textbook on Diff. Equations. Great many of the ODE's you can find in applications are special cases of the following one:
[tex]
y''=P(x,y)+3 Q(x,y) y'+3 R(x,y) (y')^2+S(x,y) (y')^3,
[/tex]
Once you have such an equation, you usually try to reduce it to some standard equation given in a reference book. A change of variables is most often used for this purpose:
[tex]
\tilde x=\tilde x(x,y),
[/tex]
[tex]
\tilde y=\tilde y(x,y).
[/tex]
Such a change of variables is called a point transformation. There is a theory giving some hints how to find proper point transformation for a given equation. It would be best if you learn this theory and explain it to others (including your class instructor). The following links will help you:
http://arxiv.org/abs/math.DG/9802027
http://arxiv.org/abs/solv-int/9706003

Best regards,
Ruslan Sharipov
senthanal
senthanal is offline
#65
Sep11-05, 01:50 AM
P: 1
Please send this docment.
nazgjunk
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#66
Dec9-05, 02:29 PM
P: 83
Hello all, I am pretty new to these forums, so correct me if I do anything wrong.

In maths class, we have just finished a chapter on differential equations. However, we were mainly working on something like this:

[tex]f(x)=2x^4+x^3-0.2x^2+3x[/tex]
*snip*
[tex]f'(x)=8x^3+3x^2-0.4x+3[/tex]

Now I am wondering why I understand so little of what this thread is saying. I figure there are some possibilities:
1: The way of writing down these things is way different from what I have learned

2: I haven't learnt very much yet. At school we have come as far as the multiplication and division rules, eg:
[tex]f(x)= \frac {g}{h}[/tex]

[tex]f'(x)=\frac {g'h-h'g}{h^2}[/tex]


3: I am completely wrong in translating.

Any hints, please? Because I do like to learn.

Thanks in advance,

Nazgjunk
hypermorphism
hypermorphism is offline
#67
Dec9-05, 03:20 PM
P: 509
Hello naz,
The equations you wrote down are derivatives. In other words, given a function, those formulas tell you how to find a derivative.
The study of differential equations is the study of how one can get a set of functions that satisfy a given derivative (It will always be more than one because derivatives annihilate additive constants).
For example, if you know f(x) = - f'(x), we want to find some explicit form of f(x) that will give us that equation as a derivative. In this case, a moment's thought will show f(x) can be of the form f(x) = A*cos(x + B), where A and B are arbitrary constants.
Usually in physics one first sees the differential form of some phenomenon (ie., simple harmonic motion) and one then tries to solve the differential equation(s) to get a more explicit form.
nazgjunk
nazgjunk is offline
#68
Dec9-05, 03:33 PM
P: 83
Ok, thanks. I think I got the basic idea, and it indeed is probably a translation problem. I still think it's weird, though: my dictionary says the Dutch "differentieren" does mean "differentiate", but it seems to be something completely different.
mathwonk
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#69
Jan8-06, 04:49 PM
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boy am mi ticked. i just lost a post that I had been woprking nop fopr over an hour about o.d.e's from the biog picture and various books and their different characteristics, and essential ingredients of a good d.e. cousre etc etc. when i tried to pst it the computer said I was not logged in but when I logged in my post was gone.

this is not the first time this has happened to me.

well good luck for you, bad luck for me.
shmoe
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#70
Jan8-06, 11:53 PM
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Quote Quote by mathwonk
boy am mi ticked. i just lost a post that I had been woprking nop fopr over an hour about o.d.e's from the biog picture and various books and their different characteristics, and essential ingredients of a good d.e. cousre etc etc. when i tried to pst it the computer said I was not logged in but when I logged in my post was gone.

this is not the first time this has happened to me.

well good luck for you, bad luck for me.
I have to say this is a shame. Your "big idea" posts where you run down the main ideas of a topic are some of my favorites.

If I understand what you did, you typed the message, then hit submit but had to log in and then it gave you a blank form (or invalid thread)? This happens to me all the time, hitting the "back" button a couple times usually gets me back to the message I had typed, and it's recovered. Paranoia also makes bme sometimes "copy" a long message before I hit "submit".
Maxwell
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#71
Jan9-06, 02:06 PM
P: 530
Quote Quote by mathwonk
boy am mi ticked. i just lost a post that I had been woprking nop fopr over an hour about o.d.e's from the biog picture and various books and their different characteristics, and essential ingredients of a good d.e. cousre etc etc. when i tried to pst it the computer said I was not logged in but when I logged in my post was gone.
this is not the first time this has happened to me.
well good luck for you, bad luck for me.
When you type up a big post, copy* it before you submit it. I've been burned too many times before to ever let a huge post of mine disappear because of some bad message board voodoo!

*Select your whole post and press CTRL V, incase you didn't know.
mathwonk
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#72
Jan10-06, 03:46 PM
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It happened again, but I SAVED!! Here is todays post:
I am starting to teach o.d.e. and the first thing I am going to do tomorrow (the 2nd day of class) is try to explain why the sort of thing appearing in post #3 here, apparently taken from Boyce and Diprima, (and not to be blamed on the student trying to learn it, my apologies to that student) is completely meaningless nonsense.
I.e. solving the d.e. "dy/dt = g(t)" may or may not be possible, depending on the nature of g.
In particular it makes no sense at all to simply write
"indefinite integral of g = y", and claim to have solved the problem, since the notation "indefinite integral of g" stands for any function whose derivative is g, provided one exists.
So one has made no progress at all in solving the problem in writing this, as one is merely restating it in different notation. Books which say this sort of thing drive me nuts, as they give young students entirely the wrong idea as to what a function is, and what it means to "solve" a d.e.
Assuming "solving" an equation means "finding" a function that satisfies it, what does it mean to "find" such a function? To call out its name? If so, then I have solved dy/dx = 1/x by saying the magic words "natural log". Unfortunately I do not know the value of this function even at the number t = 3.
So I really do not know much about this solution except that it exists.
Actually I don't even know that, I am only claiming it exists, since most students probably cannot really prove the natural log function does exist, and satisfies this equation. But I have been told it does, so I say "natural log" solves this equation.
But wouldn't it make more sense to say I have "found" a function, if I can actually tell you some of its values, or even an arbitrary value, at least to any desired degree of accuracy?
The way to do this with ln(t) is actually to approximate the area under the curve of y = 1/t. Defining ln as that area fucntion at least shows it exists, but I still ought to prove that area function is both differentiable and solves the equation.
So for some g, an antiderivative function exists and for others it does not. For example it does exist for g(t) = 1/t, or g(t) = e^(t^2)), but does not exist for g(t) = 0 for t<0 and g(t) = 1, for t g.e. 0.
The usual sufficient criterion for such an antiderivative to exist is that g be continuous. But these hypotheses are nowhere mentioned, before writing "indefinite integral of g".
Indeed the indefinite integral of 1/t does not exist because ln(t) exists, rather it is the other way around, ln(t) exists for all t>0, because the area function of 1/t exists, because 1/t is continuous.
So the area function (definite integral) is a machine for making differentiable functions out of continuous ones. Every now and then we will learn that some area function equals some other function we have met in a different situation, but so what?
The ones we have met before are no better solutions than the ones we have not. We just understand those better. So books that say: "well, we can solve the d.e. dt/dy = 1/t, but not dy/dt = e^(t^2), because e^(t^2)) cannot be integrated", are lying, and promulgating a false understanding of d.e's, functions, and their solutions.
More correct would be to say: every continuous integrand leads us to a differentiable area function, and to a potentially new and interesting function. A few of these we have met before and given names to.
But in general the integral (area function of) a function is more complicated than the original fucntion, and many of them we have not yet had time to name.
E.g. the integrand g(t) = 1, has area function t+c, and the integrand g(t) = t^r has area function (1/r+1) t^(r+1), except when this makes no sense, namely when r = -1.
In that case the area function (starting at t=1) of 1/t is a function which we shall call ln(t). It happens to be inverse to an exponential function e^t with a base "e" which we would never have met if we were not solving this d.e. But otherwise it resembles exponential functions like 2^t which of course we previously only defined for rational t.
Now this lets us also name the area functions for all fractions, since by partial fractions (using complex numbers) all fractions have integrals which reduce to sums of ones like 1/(t-a), and these are also all natural logs. So spending hours and weeks integrating more and mroe examples of fractions is just repeating the same thing over and over, or if you like, it is practicing algebra. But once you have integrated 1/(t-c) you are not learning anymore about integration by integrating 1/(any polynomial in t).
Now lets try some really new integrands (studied by the Bernoullis in the 1600's?). lets put a square root in the denominator.
e.g. let g(t) = 1/sqrt(1-t^2). this is continuous within the interval -1<t<1, so has a differentiable area function there, which turns out to be the arclength function for a circle, and it already has a name "arcsin" because arclength for circles came up long ago. (Euclid, 1800 BC?)
Moving on we try perhaps 1/sqrt(1-t^3), also continuous on (-1,1) so it too has a nice differentiable area function but most of us do not know any name for it. But Weierstrass studied it in the 1900's and it is called an elliptic integral because it comes up in trying to measure arclength on a lemniscate? Wait a minute, for this story to be any good, it should be arclength on an ellipse. Well maybe it comes up there too.
Now shall we say this integral does not exist? or that we have not solved the d.e. dy/dt = sqrt(1-y^3) ? just because we have not yet chosen a name for this function?
That would be absurd. so we call it maybe frank, as one of my friends says.
anyway, it should be obvious that all the integrands 1/sqrt(1-t^n) do have differentiable area functions, and all deserve names, but we only name the ones we need in our own problems.
frank turns out to have an inverse function which is periodic like sin and cos, but even better, it is not just singly periodic, but doubly periodic in the complex plane. These functions are really wonderful, and not only do exist, but played the main role in Wiles solution of Fermat's last theorem.
So one should never give students the idea that those d.e.'s whose solutions are functions we have not named yet, or whose names we have not heard yet, somehow are not solved, or even not solvable, and yet this is precisely the impression i get from some books.
Solving an equation means producing a function that solves it. Producing a function means defining it, not just hollering out its name (Oh yeah, I know the solution of that equation, it's Harold, or is it Maude? Is that any worse than a student saying the solution of dy/dt = 1/(1+t^2) is arctan, but the student does not know one single value of arctan?).

Defining it can be done by a lot of different processes, usual among these being "taking the area function of a given function" i.e. really integrating it, or inverting a given function.
Now here is the first lesson of o.d.e.:
If g(t) is any continuous function on an interval I, then the area function of g (Riemann's definite integral of g from a to t) is a differentiable function on I which solves the d.e. dy/dt = g(t).
second lesson of o.d.e.:
If g(t) is any continuous function that is never zero on an interval I, then the area function of
1/g(t) is an invertible differentiable function on I, whose inverse function solves the o.d.e. dy/dt = g(y). (That's right the letter in g is y not t, and the g moved from the denominator to the numerator, because that's what happens to the derivative when you take an inverse function.)
This method is usually called "separation of variables", or just "integration".
I.e. the "monkey see monkey do" way of solving the d.e. dy/dt = g(y)
is to multiply by dt and divide by g(y) and get dy/g(y) = dt, thus "separating the variables".
Then, step one, "integrate" both sides, to get G(y) = t+c. (*)
then step 2: invert the function G (which has an inverse because its derivative 1/g(t) was assumed to be zero nowhere), getting a function H,
and
step 3): apply H to both sides of (*) getting: y = H(t+c).
Now none of this makes any sense, unless you understand how the two processes, taking area function, and inverting a function, do in fact transform an appropriate differentiable function, i.e. one satisfying certain hypotheses, into another differentiable function, namely the solution.
ok, I iterate: the student learning this material, or trying to, from the usual books is not to blame for the confusion the books spew everywhere, but is actually the victim.
My point is: one has NOT solved an o.d.e. simple by saying "the name of the solution is frank", but by knowing why the solution exists, what its domain is, what properties it has there, and how to approximate its values there, and sketch the graph.
E.g. one can do all of these things for the solution of dy/dt = e^(t^2)), so it is quite false for books like the one I was reading today, to say "one cannot integrate this d.e. directly".
ok, end of rant. I am just finding out why I never understood this subject, as the books on introductory d.e. are probably the worst in all of elementary mathematics. it is very hard to find a decent, correct, explanation of d.e. the best I have found is the book of v.i. arnol'd, and that is pretty dense going. I admit also to having learned something from Blanchard Devaney and hall, but it is very wordy and there is no theory at all.
Most of the rest are cookbooks of the worst sort, teaching you to spout out the usual names of the solutions to the simplest possible equations,
i.e. ones like y''' + 3y'' + 3y' + y = 0, which are just chosen because they are all (complex) exponentials (i.e. cosines and sines and exponentials).
A d.e. is a racecourse, with speed signs all over, and a solution is a driver in a fast, well handling car, navigating the course at the right speed at every instant, and in the right direction.


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