Okay...this is probably going to come down to simplification, but I am
stumped.  This is directly from the book "Ordinary Differential
Equations" by Tenenbaum & Pollard. (Page 25)

The question is to test whether:

(a) f(x,y) = x^3 + y^3 - 3xy = 0, -infinity < x < infinity

is an implicit solution of 

(b) F(x,y,y') = (y^2 - x)y' - y + x^2 = 0, -infinity < x < infinity

Their solution is to differentiate implicitly to arrive at:

(c) 3x^2 + 3y^2y' - 3xy - 3y = 0

The next sentence begins..."Since (c) now agrees with (b), we know
that the slope of the function..."

I just don't see how the solution (c) agrees with (b).  If I multiply
terms in (b) I get an equation similiar to (c) but without the
coefficient.  What's the deal?

Thanks in advance.

Regards,

Herkdrvr

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Request for Question Clarification by elmarto-ga on 28 Mar 2005 06:49 PST

Hi herkdrvr,
There appears to be a mistake in the equation you wrote... are you
sure (c) shouldn't be

3x^2 + 3y^2y' - 3xy' - 3y = 0

rather than the one you wrote

3x^2 + 3y^2y' - 3xy - 3y = 0 ?

Best regards,
elmarto

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Clarification of Question by herkdrvr-ga on 28 Mar 2005 16:16 PST

Elmarto--

You are correct.  (c) should read as follows:

(c) 3x^2 + 3y^2y' - 3xy' - 3y = 0

Hi herkdrvr!
Actually, (b) and (c) simplify down to the same equation. When
multiplying the terms in (b), you get the following equation:

(b) x^2 + y^2y' - xy' - y = 0

The equation (c) is:

(c) 3x^2 + 3y^2y' - 3xy' - 3y = 0

But taking 3 as common factor, we get:

(c) 3(x^2 + y^2y' - xy' - y) = 0

Now, dividing both sides by 3:

(c) x^2 + y^2y' - xy' - y = 0

So the book is correct when it says that (b) and (c) agree.


I hope this helps!
Best wishes,
elmarto

The light bulb turned on!  Of course...divide by 3!  I'm sure I'll
have more questions as I go through this book.  Thank you for your
swift reply.  Regards,
Herkdrvr