# How can i solve this separable differential equation with trigonometric function?

The given Problem is separable differential equation:

$$\cos y\ dx + (1+e^{-x})\sin y\ dy = 0$$ $$y(0) = \frac{\pi}{4}$$

My approach was:

this differential equation seems to be in implicit form and its a so called ordinary differential equation of first order. I tried some transformations:

$$\cos y\ dx + (1+e^{-X})\sin y\ dy = 0$$

$$\cos y\ dx = -(1+e^{-X})\sin y\ dy$$

but bow i am stuck, and i am quite confused about the term $\cos y\ dx$ ? maybe you can help?

P.S.: edits were made to improve language and latex

$$\dfrac {1}{(1+e^{−x})} dx=\dfrac{−\sin y}{\cos y} dy$$ $$\dfrac {e^{x}}{(1+e^{x})} dx={−\tan y}\;dy$$ $$\int{\tan y}\;dy=-\int \dfrac {e^{x}}{(1+e^x)} dx$$ $e^x=t\implies e^x=\dfrac {dt}{dx}\implies dx=\dfrac{dt}{e^x}$ $$\log \sec y=-\int \dfrac {t}{(1+t)} \dfrac{dt}{t}$$ $$\log \sec y=-\log (1+t)+\log C$$ $$\log \sec y=-\log (1+e^x)+\log C$$ $y(0)=\dfrac \pi4\;$ $$\log \sqrt2=-\log 2+\log C$$ $$\dfrac32\log 2=\log C\implies C=2\sqrt2$$ $$\log \sec y=-\log (1+e^x)+2\sqrt2$$ $$\log \sec \dfrac \pi4=\log \dfrac {2\sqrt2}{1+e^x}$$ $$\sec y=\dfrac {2\sqrt2}{1+e^x}$$

• thanks alot for the answe, but i'll need some time, to think about it. Do you think it will be possible to get a value/term for y in the end? So is this suggestion correct: $$\sec y=\dfrac {2\sqrt2}{1+e^x}$$ then $$arcsec\left(\dfrac {2\sqrt2}{1+e^x}\right) = y$$ – Toralf Westström Jun 10 '13 at 14:05
• do $\sec^{-1}\;$on other side.but i don't think it will necessary – iostream007 Jun 10 '13 at 14:06

the simplest way to solve this equation is to bring cosy under siny and (1+e^-x) beneath dx and solve the equation.

$\dfrac {1}{(1+e^{−X})} dx=\dfrac{−\sin y}{\cos y} dy$

And now simply you could integrate the equations.

• Is this helpful – Nitesh Sethia Jun 10 '13 at 11:22
• I've correct your dx side – iostream007 Jun 10 '13 at 11:38