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First Order Equations

  1. Jun 12, 2007 #1
    I have the following equation:


    I was trying to find the integral factor of this to make it a exact differential equation, but ended in a almost imposible integral. do anyone have any idea of how to make this?
  2. jcsd
  3. Jun 13, 2007 #2
    Separate the variables x and y first.
  4. Aug 10, 2007 #3
    It's a total differential equation. It's how i solved it:
    two or three combinations and you get it:
    where C is real

  5. Aug 10, 2007 #4


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    It certainly is NOT an 'exact' equation (total differential) and the solution you give is not correct.
  6. Aug 10, 2007 #5
    I don't know whether there is an easier solution but i would start with
    then divide both sides with x^2 and use u=y/x substitution
  7. Aug 10, 2007 #6
    so where's an error?
  8. Aug 10, 2007 #7
    [itex]P(x,y)=x^2-y^2-y[/itex], and [itex]Q(x,y)=-x^2+y^2+x[/itex], so

    [tex]P_y=-2y-1 \neq -2x+1 = Q_x.[/tex]



    Why don't you show us what you did?
    Last edited: Aug 10, 2007
  9. Aug 10, 2007 #8
    omg I don't want to write 5 more lines.
    but if it's too hard for someone who would read it i can write
  10. Aug 10, 2007 #9
    y and x aren't constant
  11. Aug 10, 2007 #10
    Really? Are you serious?

    Did you even bothered to calculate [itex]\partial P/\partial y[/itex] and [itex]\partial Q/ \partial x[/itex]?
    Last edited: Aug 10, 2007
  12. Aug 11, 2007 #11


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    do you KNOW how to calculate Py and Qx?

    Your remark makes it clear that you don't know how to take a partial derivative!

    It isn't important, of course, that
    makes it clear that this is a troll.
  13. Aug 11, 2007 #12
  14. Aug 11, 2007 #13

    [tex]\frac{\partial P}{\partial y} \neq \frac{\partial Q}{\partial x}[/tex]

    False => False. Your hypothesis is wrong! Go back to the book and read REALLY CAREFULLY this time.

    The least you should do this time is check your answer before claiming it's correct.
  15. Aug 11, 2007 #14
    ok, sorry, no I've understood
  16. Aug 12, 2007 #15


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    So you firmly believe (a quote from your post) that
    -2y- 1= 1- 2x? If so there is not point in trying to change your mind!
  17. Sep 3, 2007 #16
    For a different view on the subject, try this book:

    http://cgi.ebay.com/ws/eBayISAPI.dll?ViewItem&rd=1&item=120156979964&ssPageName=STRK:MESE:IT&ih=002 [Broken]
    Last edited by a moderator: May 3, 2017
  18. Sep 3, 2007 #17
    I couldn't resolve it.

    I suggested integral factor because I founded this equation on an old book in the integral factor chapter. I will give it another try to see what happens.
  19. Sep 3, 2007 #18
    OK, I tried again using the integral factors to make the equation an exact differential equation I end up with nothing again.

    ([tex]\frac{\delta M}{\delta y}[/tex] - [tex]\frac{\delta N}{\delta x}[/tex] )/ N

    I get F(x,y) so this is not helping.

    The other

    ([tex]\frac{\delta N}{\delta x}[/tex] - [tex]\frac{\delta M}{\delta y}[/tex]) / M

    Get G(x,y) nothing

    and the other cases doesn't help at all because multiplying by x or y or X^2 or y^2 wouldn't end up in f(u) u=x/y or u=xy.
    So maybe this book was trying to make headaches on students.

    Anyway thanks for the help to all.
  20. Sep 5, 2007 #19


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    Let y=kx,

    then you find that k=-1

    so, y=-x is the answer.
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