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 function 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 function, 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 let's 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 let's try some really new integrands (studied by the Bernoullis in the 1600's?). let's 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.
