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Having trouble understanding

  1. Aug 13, 2011 #1
    Concerning the IVP dy/dx = (1 + y^(2)*sinx)/(y(2cosx - 1)) with y(0) = 1

    Let f(x,y) = (1 + y^(2)*sinx)/(y(2cosx - 1)). Find a rectangular region in the plane, centred at the point (0,1) and on which the two functions f and f_y are continuous. Explain why the problem has a unique solution on some interval containing 0.

    What exactly does this mean? Note that this question comes before the question asking to solve it.
     
  2. jcsd
  3. Aug 14, 2011 #2

    HallsofIvy

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    [tex]f(x,y)= \frac{1+ y^2sin(x)}{y(2cos(x)- 1)}[/tex]
    is continuous as long as the denominator is not 0- that is, as long as y is not 0 and 2cos(x)- 1 is not 0 which is same as saying cos(x) is not 1/2.

    [tex]f_y(x)= \frac{y^2 sin(x)}{y^2(2cos(x)- 1)}[/tex]
    is continuous as long as the denominator is not 0- that is, as long as y is not 0 and cos(x) is not 1/2- the same as the previous condition. Find the largest square, having (0, 1) as center, bounded by y= 0 and x= [itex]\pi/3[/itex] (where cos(x)= 1/2).
     
    Last edited: Aug 14, 2011
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