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Math Help - limit to infinity question

  1. #1
    MHF Contributor
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    limit to infinity question

     <br />
 y'=\frac{e^{x-y}}{y-1}
    when x goes to minus infinity its solution y goes to zero.
    find y(ln2)?

    i know that y is some sort of exponent of negative power variable or a fraction because for both of them infinity turns the whole thing into zero.
     (y-1)dy=e^{x-y}dx

     (y-1)dy-e^{x-y}dx=0
    there is no way i solve it by integration coefficient method because we have both x and y in the power of the exponent and no defertiation we change that
    Last edited by transgalactic; October 21st 2010 at 01:34 PM.
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  2. #2
    Senior Member
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    <br />
y'=\frac{e^{x-y}}{y-1}<br />

    <br />
\frac{dy}{dx}=\frac{e^x}{e^y(y-1)}<br />

    It is seperable
    <br />
e^y(y-1) \; dy = e^x \; dx<br />
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  3. #3
    MHF Contributor
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    thanks
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