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## Main Question or Discussion Point

**The problem that was given:**

[tex]{\frac {d}{dx}}y \left( x \right) =2\,{\frac {x}{{x}^{2}\cos \left( y

\right) +4\,\sin \left( 2\,y \right) }}

[/tex] after doing some changes we get a bernuli that looks like this

[tex]2\, \left( {\frac {d}{dy}}x \left( y \right) \right) x \left( y

\right) =\cos \left( y \right) \left( x \left( y \right) \right) ^{

2}+4\,\sin \left( 2\,y \right)[/tex]

with the initial conditions that x=f(y) goes through (1,0) (1=x,0=y)

the solution is

[tex]\left( x \left( y \right) \right) ^{2}=-8-8\,\sin \left( y \right) +

9\,{{\rm e}^{\sin \left( y \right) }}[/tex]

for these conditions the boundry is Dy=(arcsin(ln(8/9)),pi/2) which includes the point in the initial conditions the only problem i have is i dont know how to choose the final solution for x(y) thus whether is it th possitive root or the negative one