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Integrating a polynomial

  1. Jan 13, 2007 #1
    I Just want to know if this is how this should be done..

    [tex]\frac{dy}{dx} = \frac{3y^2}{x}[/tex]

    [tex]dy = \frac{3y^2}{x}dx[/tex]

    [tex]dy = 3^2 x^-^1dx[/tex]

    [tex]\frac {1}{3} \int \frac {dy}{y^2} = \int x^-^1dx[/tex]

    [tex]\frac {1}{3} ln| y^2| = ln | x | + C[/tex]

    because I forget how to integrate the polynomial...
     
  2. jcsd
  3. Jan 13, 2007 #2

    cristo

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    The right hand side is correct. Look again at [tex]\int y^{-2}dy[/tex] How do you integrate this?
     
  4. Jan 13, 2007 #3

    AlephZero

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    [tex]\frac {1}{3} \int \frac {dy}{y^2} = \int x^-^1dx[/tex]

    That's OK, but the integral of 1/y^2 isn't ln(y^2).

    Look up how to integrate polynomials, if you forgot. Or if you know how to differentiate polynominals, work it out from the fact that integration is "anti-differentiation".
     
  5. Jan 13, 2007 #4
    its not [tex]\frac{1}{3y}[/tex] is it?
     
  6. Jan 13, 2007 #5

    cristo

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    Close; it's -1/(3y), since you must divide by the power of y(which is -1)
     
  7. Jan 13, 2007 #6
    oops.. thanks I was thinking that was how you did it originally, I was just confused and did it that other way and forgot ket things... thanks

    EDIT: Hmmm I also need to then solve for Y

    [tex]y^-^1 = \frac {-ln|x|}{3} - \frac {C}{3}[/tex]

    [tex]y = \frac{-3}{ln|x|} - \frac{3}{C}[/tex]
     
    Last edited: Jan 13, 2007
  8. Jan 13, 2007 #7
    that looks wrong to me
     
  9. Jan 13, 2007 #8
    not even close i dont think
     
  10. Jan 13, 2007 #9
    [tex]y=\frac{ln x + C}^-1{3}[/tex]
     
    Last edited by a moderator: Jan 13, 2007
  11. Jan 13, 2007 #10

    cristo

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    [tex]y^{-1} = \frac {-ln|x|}{3} - \frac{C}{3}=\frac{-(ln|x|+C)}{3}[/tex]

    [tex]y=\frac{-3}{(ln|x|+C)}[/tex]

    Can you follow what I've done?

    JJ420; we are here to help, not criticise!
     
  12. Jan 13, 2007 #11
    [tex]y=\frac{ln x + C}^-1{3}[/tex]
    thats what i got but im dumb so dont trust me
     
  13. Jan 13, 2007 #12
    oops that -13 is supposed to be an exponent of the numerator (-1) and 3 in the denominator
     
  14. Jan 13, 2007 #13

    cristo

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    [tex]\frac{(ln x + C)^{-1}}{3}=\frac{1}{3(ln x + C)}[/tex] The 3 should definitely be in the numerator, and we need a - sign too. See my post:

     
    Last edited: Jan 13, 2007
  15. Jan 13, 2007 #14
    no I can't figure it out, common denominator?
     
  16. Jan 13, 2007 #15
    oh.. yes I see

    why is C positive and not negative
     
    Last edited: Jan 13, 2007
  17. Jan 13, 2007 #16

    cristo

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    Yea you just write ln|x| and C as one fraction with denominator 3. Then take the reciprocal of both sides (since the LHS is 1/y) to obtain an expression in terms of y.

    Well, yes. In order to take the reciprocal of both sides, there must only be one fraction on each side.
     
  18. Jan 13, 2007 #17
    looking back to the origonal i don't see how there is a negative
     
  19. Jan 13, 2007 #18

    cristo

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    The minus sign comes about when integrating; integral(y-2)=-y-1 (+C)
     
  20. Jan 13, 2007 #19
    snowJT what does C mean?

    it means any constant...it can be a minus but since we don't know we just write + C
     
  21. Jan 13, 2007 #20
    figured so, there were some examples I just found in my book
     
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