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Diff Equation problem- give me a nudge, please

  1. Jun 26, 2011 #1
    1. The problem statement, all variables and given/known data

    (dy/dx) - cos2(x-y) = 0

    2. Relevant equations



    3. The attempt at a solution

    I'm unsure where to start this. This isn't a liner equation of form (dy/dt) + p(t)y = g(t) ; and it's not separable nor an exact equation.

    Is it illegal to make this (dy/dx) - cos2(x) + cos2y = 0 ???


    Thanks for your time.
     
  2. jcsd
  3. Jun 26, 2011 #2

    ehild

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    It is totally illegal!!:uhh:

    Hint: replace y-x with a new variable.

    ehild
     
  4. Jun 26, 2011 #3
    Yeah, I knew that was illegal. I was just grasping at straws because I had no idea where to begin. And it's early...

    Well, I have not worked with problems like this at all and I'm having trouble finding anything similar in my text, but this is what I have figured out so far:

    (dy/dx) -cos2(x-y) = 0

    (dy/dx) = cos2(x-y) make (x-y) = v

    (dy/dx) = cos2(v)

    v = x - y
    y = x - v
    y' = 1 - v'

    from original : y' = cos2(x - y)
    so: 1 - v' = cos2(v)
    and : v' = 1 - cos2(v)

    Am I heading in the right direction? Another nudge? Thanks again.
     
  5. Jun 26, 2011 #4

    ehild

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    It is all right so far. Is not it a separable de?

    ehild
     
  6. Jun 26, 2011 #5

    Char. Limit

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    Remember that 1-cos2(v) = sin2(v).
     
  7. Jun 26, 2011 #6
    Yes, I should have seen that.

    (dv/dx) = 1-cos2(v)

    dv/1-cos2(v) = dx

    which is : csc^2(v) = dx

    integrate:

    -cot(v) = x + c

    -cot (x - y) = x + c

    Thank you for your help. I was just staring dumbly at it.

    This implicit form should be a suitable answer for this question, yes? I was just told to solve the equation.
     
    Last edited: Jun 26, 2011
  8. Jun 26, 2011 #7

    Char. Limit

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    It should be suitable, but it's a simple process to get the answer in explicit form, if you want it.
     
  9. Jun 26, 2011 #8
    Hmm....

    -cot (x-y) = x + c

    cot (x-y) = -x +c

    (x - y) = arccot (-x + c)

    -y= arccot (-x + c) - x

    y = x - arccot (-x + c) ???
     
    Last edited: Jun 26, 2011
  10. Jun 26, 2011 #9

    Char. Limit

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    Hint: The inverse of the cotangent function is arccotangent, not arctangent.
     
  11. Jun 26, 2011 #10
    hehe, yeah I saw that as soon as I submitted it and then edited it. You're fast :redface:
     
  12. Jun 26, 2011 #11

    Char. Limit

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    Looks good to me. You can plug it in and test it if you want.
     
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