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Question on drawing a diff.equation

  1. May 14, 2008 #1
    i got this equation

    y'=1+x-y

    i got y=x+(1-c)
    for the isocline c=-1
    i got tangent lines which are perpendicular

    for c=0 the are not
    why is that??
    how do i find the slope of each isocline???
     
  2. jcsd
  3. May 14, 2008 #2

    HallsofIvy

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    What do you mean you "got y= x+ (1-c)"?
     
  4. May 15, 2008 #3
    this is a solution of this y'=1+x-y differential is y=x+(1-c)


    for the isocline c=-1
    i got tangent lines which are perpendicular

    for c=0 the are not
    why is that??
    how do i find the slope of each isocline???
     
  5. May 15, 2008 #4

    HallsofIvy

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    No, it isn't. If y= x+ (1-c), then y'= 1, obviously. The general solution to y'= 1+ x- y is
    y= ce-x+ x.


     
  6. May 15, 2008 #5
    ok
    so generally how do i find the slope of each isocline
    (those lines that we draw on the isoc lines)

    ???????/
     
  7. May 16, 2008 #6

    HallsofIvy

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    You don't find the slope of an isocline- you assign it. For example, with dy/dx= 1+ x- y, the isocline ("iso-cline": equal slope) corresponding to slope 0 must include points (x,y) such that 1+ x- y= 0 or y= x+ 1. Draw that line. The isocline corresponding to slope 1 must include all points (x,y) such that 1+ x- y= 1 or y= x. Draw that line. choose whatever values, s, you want for the slope. The corresponding isocline is the line dy/dx= 1+ x- y= s or y= x+ (1-s).
     
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