How do I solve this DE method problem with y'cosx = 1-y^2?

[gbacsf]

What method would be used to solve this DE, it look like a Bernoulli but isn't. I'm lost.

y'cosx = 1-y^2

Thanks,

Gab

 

[neutrino]

It looks like you can separate the variables.

 

[Pseudo Statistic]

Treat y' as the limiting ratio dy/dx.
Your aim is to get the xs (and dx) on one side, and the ys (and dy) on the other side and integrate...

 

[gbacsf]

I'm not sure if I can solve it that way. I did some digging and found its in the form of a Riccati eq.

http://en.wikipedia.org/wiki/Riccati_equation

But I'm not sure how to apply it to this problem, any help?

Thanks

 

[FrogPad]

Hint:

\frac{dy}{dx} \,\, \frac{\cos x}{1-y^2}=1

 

[gbacsf]

Well from that I can say that y=sin(x) is a solution.

Then I get:

z' +(-2*tan(x))*z = 1/cos(x)

So then I solve this linear equation:

(sin(x) + C)/(cos(x))^2

 

[J77]

The next step should be to substitute: y=\sin(u) into the equation FrogPad gave.