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blackrose636 Group Title

find d^2(y)/d(x^2) in terms of x and y (second derivative) 1-xy=x-y

  • one year ago
  • one year ago

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  1. VeritasVosLiberabit Group Title
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    \[1(\frac{ dy }{ dx })-(x)\frac{ dy }{ dx}+\frac{ dx }{dx }y=\frac{ dx }{ dx }-\frac{ dy }{ dx }\] tell me how this looks it's been a while since I differentiated with this method.

    • one year ago
  2. VeritasVosLiberabit Group Title
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    this would be the setup to solve for the first derivative

    • one year ago
  3. mahmit2012 Group Title
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    y=-1 & y"=0

    • one year ago
  4. Algebraic! Group Title
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    http://i.imgur.com/CMgMz.gif

    • one year ago
  5. VeritasVosLiberabit Group Title
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    mother of god...

    • one year ago
  6. blackrose636 Group Title
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    how did u get the second derivative? @mahmit2012

    • one year ago
  7. Algebraic! Group Title
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    -( x*y' + y) = 1 -y'

    • one year ago
  8. Algebraic! Group Title
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    solve for y'

    • one year ago
  9. Algebraic! Group Title
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    differentiate the expression again

    • one year ago
  10. VeritasVosLiberabit Group Title
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    I didn't do the second derivative yet. just do the first then repeat.

    • one year ago
  11. Algebraic! Group Title
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    if you get another y' in your expression for y" sub.s in your result from the first differentiation...

    • one year ago
  12. Algebraic! Group Title
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    Does that make sense?

    • one year ago
  13. mahmit2012 Group Title
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    y=-1 y"=0 why you did not pay attention!

    • one year ago
  14. mahmit2012 Group Title
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    x#1

    • one year ago
  15. blackrose636 Group Title
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    i got y' is -y-1/x-1 then i tried doing quotient rule and thne i got confused

    • one year ago
  16. Algebraic! Group Title
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    you fail at math @mahmit2012

    • one year ago
  17. Algebraic! Group Title
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    k

    • one year ago
  18. Algebraic! Group Title
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    you got y' right

    • one year ago
  19. mahmit2012 Group Title
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    every body sleeping !! cool

    • one year ago
  20. blackrose636 Group Title
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    yeah -y-1/x-1

    • one year ago
  21. Algebraic! Group Title
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    now, f = ( -y-1) g = (x-1) (f/g)' = f'g -fg'/g^2 f' = -y' g' = 1 g^2 = (x-1)^2

    • one year ago
  22. Algebraic! Group Title
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    plug them in and use the result from the first diff. : y' = -y-1/x-1 to sub.s in for y'

    • one year ago
  23. mahmit2012 Group Title
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    find y=-1 it's so easy !!!

    • one year ago
  24. Algebraic! Group Title
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    got it @blackrose636 ?

    • one year ago
  25. blackrose636 Group Title
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    how did u even get y=-1?

    • one year ago
  26. Algebraic! Group Title
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    gl @Callisto

    • one year ago
  27. mahmit2012 Group Title
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    y-xy=x-1 so y=-1 for all x#1

    • one year ago
  28. mahmit2012 Group Title
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    @Algebraic! did you get it !

    • one year ago
  29. Callisto Group Title
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    Ah... 1 - xy = x-y y-xy = x-1 y (1-x) = x-1 y = (x-1) / (1-x) y=-1!

    • one year ago
  30. mahmit2012 Group Title
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    Thank you @Callisto

    • one year ago
  31. mahmit2012 Group Title
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    this is different seeing !

    • one year ago
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