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Direction field of y'=cos(πx)

  1. Dec 13, 2013 #1
    This isn't a home work, so only I'm posting here

    When we solve y'=cos(πx) we'll get y=sin(πx)/π +C. But plotting direction field seems little confusing for me

    Since y' is a function of cosine , the value of y' must be within [-1,1].

    If i took y' as 0 and tried to plot it and I'm getting set of points such as (.....-2.5, -1.5, -0.5, 0.5, 1.5, 2.5,....)

    I can't draw a isocline since here it is discontinuois.

    Similarly by setting y' has 1, -1 we'll get points such as (......-4, -2, 0, 2, 4.....) & (......-5, -3, -1, 1, 3, 5....)

    Now how the solution curve should be plotted from the direction field, without isocline (i mean here isocline is just set of points not a curve).......
     
  2. jcsd
  3. Dec 15, 2013 #2
    No one knows where I'm wrong...?
     
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