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Math Help - One more DE question

  1. #1
    Junior Member
    Joined
    Jun 2007
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    One more DE question

    Im a bit hazy on my log rules....and this problem isnt helping...

    Original problem: ((e^y+1)^2)e^-ydx+ ((e^x+1)^3)e^-xdy=0

    Where i've gotten: -((e^x+1)^-3)e^xdx= ((e^y+1)^-2))e^y dy

    Answer: ((e^x+1)^-2)+2(e^y+1)^-1)=c

    im not sure how to get rid of those e^x and the e^y.....probaly some log rule im missing, Thanks for any help!
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  2. #2
    Super Member

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    Hello, neven87!

    Problem: . (e^y+1)^2e^{-y}dx + (e^x+1)^3e^{-x}dy\:=\:0

    Answer: . (e^x+1)^{-2} + 2(e^y+1)^{-1}\:=\:C

    Re-arrange the equation: . \frac{(e^y + 1)^2}{e^y}\ dx + \frac{(e^x + 1)^3}{e^x}\ dy \:=\:0\quad\Rightarrow\quad\frac{(e^x + 1)^3}{e^x}\,dy \;=\;-\frac{(e^y + 1)^2}{e^y}

    Separate variables: . \frac{e^y}{(e^y+1)^2}\,dy \;=\;-\frac{e^x}{(e^x+1)^3}\,dx\quad\Rightarrow\quad (e^y+1)^{-2}(e^y\,dy) \;=\;-(e^x+1)^{-3}(e^x\,dx)

    Integrate: . \int(e^y+1)^{-2}(e^y\,dy) \;=\;-\int(e^x+1)^{-3}(e^x\,dx)

    Let u = e^y + 1\quad\Rightarrow\quad du = e^y\,dy
    Let v = e^x + 1\quad\Rightarrow\quad dv= e^x\,dx

    Substitute: . \int u^{-2}du \;=\;-\int v^{-3}\,dv\quad\Rightarrow\quad -u^{-1}\;=\;\frac{v^{-2}}{2} + c

    . . \frac{v^{-2}}{2} + u^{-1}\;=\;c\quad\Rightarrow\quad v^{-2} + 2u^{-1} \;=\;C

    Back-substitute: . (e^x + 1)^{-2} + 2(e^y + 1)^{-1} \;=\;C

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