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Homework Help: Can anyone check if my answer for this ODE is correct?

  1. Dec 23, 2009 #1
    1. The problem statement, all variables and given/known data
    [tex](x+2y-4)\,dx - (2x-4y)\,dy=0[/tex]




    3. The attempt at a solution
    My answer,
    [tex]\mathrm{ln}[4(y-1)^2+(x-2)^2] + 2\mathrm{arctan}\frac{x-2}{2y-2}=C[/tex]
    Textbook's answer,
    [tex]\mathrm{ln}[4(y-1)^2+(x-2)^2] - 2\mathrm{arctan}\frac{2y-2}{x-2}=C[/tex]
     
    Last edited: Dec 23, 2009
  2. jcsd
  3. Dec 23, 2009 #2

    tiny-tim

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    Hi ronaldor9! :smile:

    tan(a) = 1/tan(π/2 - a), so theyr'e the same, except for a constant :wink:

    what were the initial conditions?
     
  4. Dec 23, 2009 #3
    Hey, sorry i meant to place a constant instead of the zero.
    Does that identity work for arctan?
     
  5. Dec 23, 2009 #4

    tiny-tim

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    oh come on

    if tan(a) = 1/tan(π/2 - a),

    then putting T = tan(a) gives 1/T = tan(π/2 - a),

    so arctanT = a = π/2 - arctan(1/T)
     
  6. Dec 23, 2009 #5
    Thanks! tiny-tim
     
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