I was thinking about posting this as a problem of the week question, but it seems non-elementary.

I believe it is time for me to learn the theory of differencial equations.

Q.What does it mean tosolvea differencial equation?

This question is much more complicated than it seems, you cannot simply say a function whose derivative(s) satisfies the conditions, no!

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Consider,

y'=1/sqrt(1-x^2)

We can show that,

y=asin(x) is a solution on the open interval (-1,1)

Then we can show that y=asin(x)+C is too a solution.

Then we can use MVT to prove no other solutions exist.

But that is not true!

What about,

f=asin(x) defined on (-1/2,1/2)

Certainly,

y not = to f

So the question remains what does it mean to solve a differencial equation?

Does it mean a function defined on some open interval that satisfies this differencial equation for all point in the interval? If so, then that can lead to problem that I will soon show.

Or do we need to define an open interval to begin with.

For example,

Solve,

y'=1/sqrt(1-x^2) on interval (-1,1)

Thenallsolutions are,

y=asin(x)+C

I think the latter is much simpler and more elegant.

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To illustrate the elegance of my second defintion is stated in this obvious theorem.

Theorem:

The differencial equation,

y'=f(x)

where f(x) is a countinous function on some open interval

Has infinitely many solutions.

If F(x) is a solution thenallsolutions are {F(x)+C}.

Proof:

It is sufficient to show that at least one function F(x) exists then using MVT we can show (omitted) that all solutions must be the set {F(x)+C}.

To do that we use the second fundamental theorem,

Define the function,

g(x)=INT_{x_0}^x f(x) dx for any point x_0 in the open interval.

Since f(x) is continous on the open interval the derivative of g(x) on any point in the open interval is f(x).

Q.E.D.

Look how simpler it is to state an open interval to solve a solution on it.

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Linear differencial equations do not bother me so much because they have a good formal explanation in my book.

However, seperable equations are really giving me theory problems.

Consider,

y'=yx

Again what does it mean to "solve" this equation?

Do we say a function definied on some open interval that satisfies this equation? (Because that is what the book seems to assume). But leads to major problems, observe.

The way people procede is to divide by y obtaining,

y'/y=x

Thus two functions equal implies integrals equal thus,

INT y'/y = INT x

Using the substitution rule,

1/2y^2=1/2x^2+C

Thus,

y^2=x^2+C

Thus,

y=+\-sqrt(x^2+C)

Here is the obvious problem. We can only divide by y if we know the function is non-vanishing. If we assume that then the solution must take form,

y=+\-sqrt(x^2+C)

We note that C cannot be 0 that leads to |x| which is not everywhere differenciable. And if C<0 then it is differenciable on some finite open interval not the full line, so what doth we do? The only reasonable answer is C>0. If we check this we see it is a solution, good.

But that is only for when y is non-vanishing.

If y is vanishing at some point. We have two possibilities:

1)y is always vanishing.

2)y is non vanishing at some point, thus non vanishing in some neighborhood (property of continous functions).

If we accept statement 1) then,

y=0 is a solution.

Also,

y=0 on any finite or half-finite interval.

If we accept statement 2) then,

[y=+/-sqrt(C+x^2) on neighborhood otherwise y=0]

But leads to problem since there is a transition from non-vanishing to vanishing there is a point where non-differnciabilitiy occurs thus that cannot be a solution.

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I hope you can see how difficult a seemingly simple differencial equation can be if we do not state an open interval.

Thus, the question reamins, what does it mean to "solve" a differencial equation.