Results 1 to 3 of 3

Thread: limit to infinity question

  1. #1
    MHF Contributor
    Joined
    Nov 2008
    Posts
    1,401

    limit to infinity question

    $\displaystyle
    y'=\frac{e^{x-y}}{y-1}$
    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.
    $\displaystyle (y-1)dy=e^{x-y}dx$

    $\displaystyle (y-1)dy-e^{x-y}dx=0$
    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; Oct 21st 2010 at 12:34 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Mar 2010
    Posts
    280
    $\displaystyle
    y'=\frac{e^{x-y}}{y-1}
    $

    $\displaystyle
    \frac{dy}{dx}=\frac{e^x}{e^y(y-1)}
    $

    It is seperable
    $\displaystyle
    e^y(y-1) \; dy = e^x \; dx
    $
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Joined
    Nov 2008
    Posts
    1,401
    thanks
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Limit to infinity question
    Posted in the Calculus Forum
    Replies: 9
    Last Post: Oct 12th 2011, 08:51 PM
  2. [SOLVED] confused with limit question.. y does it = infinity?
    Posted in the Calculus Forum
    Replies: 4
    Last Post: Aug 12th 2010, 01:45 PM
  3. Limit approaching infinity question
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Mar 24th 2010, 07:28 PM
  4. Limit to negative infinity question
    Posted in the Calculus Forum
    Replies: 4
    Last Post: Feb 26th 2010, 04:51 AM
  5. question regarding limit to infinity
    Posted in the Differential Geometry Forum
    Replies: 4
    Last Post: Oct 16th 2009, 04:44 AM

Search Tags


/mathhelpforum @mathhelpforum