Differential Equations  1

motivated by a discussion on this calculus forum
Here's an interesting problem:
Consider the function
y = x  1 for x ≤ 2
and y = 2x  1 for 2 < x < 0
and y = x  1 for x ≥ 0.
The function looks like Figure 1.
Notice that:
 y(x) is defined for all x
 dy/dx = (1+y)/x = 1 for x < 2
 dy/dx = (1+y)/x = 2 for 2 < x < 0
 dy/dx = (1+y)/x = 1 for x > 0
 Finally, we note that y = 1 when x = 1
So, is this function a "solution" to the problem:
dy/dx = (1+y)/x, y(1) = 1
 Figure 1 
>But your function has no derivative at x = 2 or even at x = 0.
Yes, that's true, so it won't be a solution at those two points because it has no derivative there.
In fact, it's not even continuous at x = 2.
But what about all the other values of x? Is it a solution for all x other than x = 2 and x = 0?
>Well ... uh, I guess so. After all, it satisfies the differential equation, right?
Yes, but does that make it a solution to the problem I posed?
>What do you mean by a "solution"!
Hmmm ... good question.
>If you give me a definition of "solution", I'll tell you if that function is a solution.
Yes, but consider ...
>I'd just check to see if your function satisfies the definition and ...
Okay! So give me a definition.
>Hey! I though you were teaching this stuff.
Okay, let's consider the following function ... similar to the last:
y = Ax  1 for x ≤ 2
and y = 2x  1 for 2 < x < 0
and y = Bx  1 for x ≥ 0
A and B are constants.
Notice that:
 y(x) is defined for all x
 dy/dx = (1+y)/x = A for x < 2
 dy/dx = (1+y)/x = 2 for 2 < x < 0
 dy/dx = (1+y)/x = B for x > 0
 y = 1 when x = 1
Is this function a solution to the problem I posed, above?
>Yeah, so where's your definition of "solution"?
Okay, suppose we do this:
The problem we posed is an "Initial Value Problem".
Our "solution" must pass through (1, 1) and, at that "Initial Value" point, must satisfy the Differential Equation.
Next, we move away from x = 1 and see how far we can get while still having our "solution" satisfy the DE.
If, at any xvalue, our function fails to satify the DE, we stop. The range of xvalues defined in this way provide the domain of the "solution".
How does that sound?
>So, let's see. For that first function we can move from x = 1 back to x = 2 and up to x = 0, right?
So?
>So that first function is y = 2x  1 in 2 < x < 0, but it's not a solution beyond that interval. Right?
Yes, according to our definition.
>Then I'd say the "solution" is y = 2x  1 with domain 2 < x < 0. Am I right?
What about y = 2x  1 for all x < 0?
>Then I'd say it, too, was a "solution" ... according to your definition. Am I right?
So what about the second solution, above?
>But it isn't unique, is it? I mean, you can choose any constants A and B.
So, do you think a "solution" should be unique?
>Sure, why not?
Then how about defining "solution" as a function that passes through the Initial Point and satisfies the DE over as large an interval as possible.
>An interval that contains the initial point, right?
Yes.
>Then y = 2x  1 is our "solution", wouldn't you say?
For x < 0.
In fact, since (1+y)/x isn't even defined for x = 0, our solutioninterval cannot extend from x = 2 to x = 0, so x < 0.
Further, if we attach to our "solution" the largest interval containing x = 1, then we should discard both that first solution and the second solution.
>Unless A = B = 2, right?
Uh ... well, that "B" is attached to our second function for x > 0, so we'd ignore it.
So we'd end up concluding that, for our particular problem:
(1) y = 2x  1 for x < 0.
(2) No solution exists for x ≥ 0.
>So what's our definition of "solution"?
I think we've agreed upon the following:
y(x) is a "solution" to: dy/dx = f(x,y), y(a) = b on an interval I if:
(1) The interval I contains the point x = a.
(2) y(a) = b.
(3) y(x) satisfies the differential equation at every point in the interval I.
(4) The interval I is the largest interval for which (1)  (3) hold true.

So what's the solution to our Initial Value Problem?
>I'd say it's y = 2x  1 on the interval x < 0.
I agree.
Have we learned anything?
>I think we've learned we can make a banana the "solution" ... if we provide the appropriate definition of "solution".
Very funny.
 Multiple solutions 
For a differential equation like dy/dx = f(x,y), it means that the solution which passes through the point, say (a,b), will have slope f(a,b).
That means we can attach a slope to each point in the xy plane ... so long as f(x,y) has a value at that point.
That means ...
>Can you get to the point?
Okay. Here's a spreadsheet where you type in f(x,y) and you get a bunch of those slopes:
Just click on the pretty picture to download the spreadsheet.
NOTE:
The numerical calculation uses RungeKutta, like so:
' start RungeKutta integration, Num1 steps
For i = 10 To 10 + Num1
Cells(i, 35) = x ' stick x into column AI, row i
Cells(i, 36) = y ' stick x into column AJ, row i
Range("C7") = x ' put x into cell C7
Range("D7") = y ' put y into cell D7
k1 = dx * Range("E7") ' cell E7 calculates f(C7,D7) so k1 = f(x,y)*dx is calculated
Range("C7") = x + dx / 2
Range("D7") = y + k1 / 2
k2 = dx * Range("E7") ' k2 = f(x+dx/2,y+k1/2)*dx
Range("C7") = x + dx / 2
Range("D7") = y + k2 / 2
k3 = dx * Range("E7") ' k3 = f(x+dx/2,y+k2/2)*dx
Range("C7") = x + dx
Range("D7") = y + k3
k4 = dx * Range("E7") ' k4 = f(x+dx,y+k3)*dx
x = x + dx ' Take an xstep
y = y + (k1 + 2 * (k2 + k3) + k4) / 6 ' Take a ystep
Next i

