Differential equations - interval of existence

[pyroknife]

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.

 

[alan2]

No, y=0 at the endpoints. Look at what that does to your original equation.