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Topic: [ap-calculus] Re: 2001 AB-6 question
Replies: 1   Last Post: Mar 19, 2005 12:30 PM

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Richard J Maher

Posts: 80
Registered: 12/6/04
[ap-calculus] Re: 2001 AB-6 question
Posted: Mar 19, 2005 12:30 PM
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Hello everyone,

------------------------
On Fri, 18 Mar 2005, Shawn J. Godack wrote:

After separating variables and integrating you get

-y^(-1) = 6x - x^2 + C

If you substitute the point (3,1/4) in now, you get one value for C.

If you solve the equation first for y and get y = -1 / (6x - x^2) + C,
this C value will be different.

Which way is correct or

------------------------------------------------------------------------
!!!! should the C value remain the in the denominator? !!!!
-------------------------------------------------------------------------
Yes, the C must remain in the denominator. Your solution is

-1/y = 6x - x^2 + C and when you invert and multiply by

minus one, the whole expression goes to the denominator. The constant
remains the same in both cases.

While I don't have the examination, it is clear that the DE involves a
y^2 term. Squaring
(-1/(6x-x^2)) + C and (-1/(6x-x^2 +C))
will give you two entirely different results.

In general, when you obtain a solution that you are not comfortable
with, it is worthwhile to check back and see if it fits the original DE.

Hope this helps.

Dick Maher

Richard J. Maher
Mathematics and Statistics
Loyola University Chicago
6525 N. Sheridan Rd.
Chicago, Illinois 60626
1-773-508-3565
rjm@math.luc.edu




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