Having trouble understanding

  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. HallsofIvy

    HallsofIvy 40,221
    Staff Emeritus
    Science Advisor

    [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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