1. ## [SOLVED] Re-writing ODE

im stuck on a problem. Show that the ODE $dy/dx=\frac{2x+e^y}{1+x^2}
$
, can be written in the form $(1+2xe^{-y})dx-(1+x^2)e^{-y}dy=0$

Im stuck and i cant figure out what to do or what method to use.
A nudge in the right direction would be good

EDIT: sorry about that, i fixed it up.

Sorry fixed the typo.. :S

2. Hello,

Is it $\frac{2x+e^y}{1+x}$ ?
Is it $2x+\frac{e^y}{1+x}$ ?

In either case are you sure there's not a typo somewhere ? Because I really can't see where 1+x² comes from :s

3. fixed.

4. That's better

$\frac{dy}{dx}=\frac{2x+e^y}{1+x^2} \Rightarrow (1+x^2) dy=(2x+e^y) dx$

Do you agree ? (that's just cross multiplying)

Then you may ask... "How did they get there ?"

Notice that there are $e^{-y}$ and where there was $e^y$, there's 1.
So what you have to think is "I may have to divide the whole equation by $e^y$"

See what this gives :

$\frac{1+x^2}{e^y} \cdot dy=\frac{2x+e^y}{e^y} \cdot dx$

(1/e^y = e^(-y))

$e^{-y} (1+x^2) dy=(2xe^{-y}+1) dx$

Looks better to you ?

5. wow thanks a tonne! yea i did the whole cross multiplying thing already, i just got confused when i saw ${e^-y}$ and that 1.

thanks for that ill take it from there, ill let you know how i go!

6. yea thats the nudge i needed. thanks again, i didnt have to really do anything after that, aside from from minus one side to the other.

thanks again!