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Math Help - Growth and Decay

  1. #1
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    Exclamation Growth and Decay

    1.) xy dx + e^(-x^2)dy=0 ; y(o)=1
    i got that problem down to 1/2 e^(-x^2) = 1/2(y^2) - lny
    but i don't know how to solve that for y


    thanks!!
    Last edited by holly123; January 8th 2009 at 01:10 PM.
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  2. #2
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    Hello, holly123!

    You've made an error somewhere . . .


    1)\;\;xy\,dx + e^{-x^2}dy\:=\:0,\quad y(0)\:=\:1
    Separate variables:
    . . e^{-x^2}dy \:=\:-xy\,dx \quad\Rightarrow\quad \frac{dy}{y} \:=\:-xe^{x^2}\,dx

    Integrate: . \ln y \:=\:-\tfrac{1}{2}e^{x^2} + C .[1]

    . . Since y(0) = 1\!:\;\; \ln1 \:=\:-\tfrac{1}{2}e^0 + C \quad\Rightarrow\quad C \:=\:\tfrac{1}{2}

    Then [1] becomes: . \ln y \:=\:-\tfrac{1}{2}e^{x^2} + \tfrac{1}{2} \quad\Rightarrow\quad \ln y \:=\:\tfrac{1}{2}\left(1 - e^{x^2}\right)


    Therefore: . y \;=\;e^{\frac{1}{2}(1-e^{x2})}


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  3. #3
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    oh im sorry i wrote it wrong! it's
    xy dx + e^(-x^2) (y^2-1) dy=0

    sorry im confused on how to use the symbols and write this all out correctly and i completely forgot to add the (y^2 -1)
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