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.
[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).