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#1
Aug1311, 07:42 PM

P: 14

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
Aug1411, 07:40 AM

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PF Gold
P: 39,565

[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). 


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