# How to decide right solution

The problem that was given:
$${\frac {d}{dx}}y \left( x \right) =2\,{\frac {x}{{x}^{2}\cos \left( y \right) +4\,\sin \left( 2\,y \right) }}$$ after doing some changes we get a bernuli that looks like this
$$2\, \left( {\frac {d}{dy}}x \left( y \right) \right) x \left( y \right) =\cos \left( y \right) \left( x \left( y \right) \right) ^{ 2}+4\,\sin \left( 2\,y \right)$$
with the initial conditions that x=f(y) goes through (1,0) (1=x,0=y)
the solution is
$$\left( x \left( y \right) \right) ^{2}=-8-8\,\sin \left( y \right) + 9\,{{\rm e}^{\sin \left( y \right) }}$$
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 dont know how to choose the final solution for x(y) thus whether is it th possitive root or the negative one

## Answers and Replies

tiny-tim
Homework Helper
with the initial conditions that x=f(y) goes through (1,0) (1=x,0=y)

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

Hi supercali!

But if it goes through (1,0), with x = 1, then it must be the positive solution, mustn't it?

how stupid of me how did i not notie that??????
after solving this creepy thing i must have been too exhausted :-)

tiny-tim