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Differential equations - interval of existence

  1. Feb 28, 2012 #1
    dy/dx=(sinx)/y Initial condition is y(pi/2)=1
    The solution to the IVP is y=(1-2cosx)^.5
    That I know is correct, but they're saying the interval of existence is when pi/3<x<5*pi/3.
    Is that wrong? I think it should include the π/3 and 5π/3.
     
  2. jcsd
  3. Feb 28, 2012 #2
    No, y=0 at the endpoints. Look at what that does to your original equation.
     
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