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Math Help - [SOLVED] Re-writing ODE

  1. #1
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    [SOLVED] Re-writing ODE

    im stuck on a problem. Show that the ODE dy/dx=\frac{2x+e^y}{1+x^2}<br />
, 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
    Last edited by sterps; August 7th 2009 at 11:54 PM.
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  2. #2
    Moo
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    Hello,

    Please put parenthesis...

    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+x comes from :s
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  3. #3
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    fixed.
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  4. #4
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    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 ?
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  5. #5
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    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!
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  6. #6
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    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!
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