- #1
supercali
- 53
- 0
The problem that was given:
{\frac {d}{dx}}y \left( x \right) =2\,{\frac {x}{{x}^{2}\cos \left( y<br /> \right) +4\,\sin \left( 2\,y \right) }}<br /> after doing some changes we get a bernuli that looks like this
2\, \left( {\frac {d}{dy}}x \left( y \right) \right) x \left( y<br /> \right) =\cos \left( y \right) \left( x \left( y \right) \right) ^{<br /> 2}+4\,\sin \left( 2\,y \right)
with the initial conditions that x=f(y) goes through (1,0) (1=x,0=y)
the solution is
\left( x \left( y \right) \right) ^{2}=-8-8\,\sin \left( y \right) +<br /> 9\,{{\rm e}^{\sin \left( y \right) }}
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 don't know how to choose the final solution for x(y) thus whether is it th possitive root or the negative one
{\frac {d}{dx}}y \left( x \right) =2\,{\frac {x}{{x}^{2}\cos \left( y<br /> \right) +4\,\sin \left( 2\,y \right) }}<br /> after doing some changes we get a bernuli that looks like this
2\, \left( {\frac {d}{dy}}x \left( y \right) \right) x \left( y<br /> \right) =\cos \left( y \right) \left( x \left( y \right) \right) ^{<br /> 2}+4\,\sin \left( 2\,y \right)
with the initial conditions that x=f(y) goes through (1,0) (1=x,0=y)
the solution is
\left( x \left( y \right) \right) ^{2}=-8-8\,\sin \left( y \right) +<br /> 9\,{{\rm e}^{\sin \left( y \right) }}
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 don't know how to choose the final solution for x(y) thus whether is it th possitive root or the negative one
fragilistic! 
