Results 1 to 4 of 4

Math Help - Integrating factor (ODE)

  1. #1
    Member
    Joined
    Nov 2009
    Posts
    75

    Integrating factor (ODE)

    the problem goes..

    The integrating factor for the ODE dxdt - 2xt = t is the integrating factor e*(-2integral(tdt)) = e^-t^2

    multiplying the ODE by the integrating factor:
    ddt((e^-t^2)*x) = t(e^-t^2)

    integrate : e^-t^2 = -0.5(e^-t^2) + c

    how does one get the integation bit?? i cant understand it at all =( .. especially the -0.5 ( .... ) +c bit

    i tried by parts..but =S.. i end up integrating again and again....and again =| wow
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1
    As you said, The integrating factor is e^{-t^{2}}

    Multiply through:

    \frac{d}{dt}[xe^{-t^{2}}]=te^{-t^{2}}

    Integrating 'negates' the derivative on the left:

    xe^{-t^{2}}=\frac{-1}{2}e^{-t^{2}}+C

    Divide by e^{-t^{2}} and remember the log laws:

    x=Ce^{t^{2}}-\frac{1}{2}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Nov 2009
    Posts
    75
    Quote Originally Posted by galactus View Post
    As you said, The integrating factor is e^{-t^{2}}

    Multiply through:

    \frac{d}{dt}[xe^{-t^{2}}]=te^{-t^{2}}

    Integrating 'negates' the derivative on the left:

    xe^{-t^{2}}=\frac{-1}{2}e^{-t^{2}}+C


    Divide by e^{-t^{2}} and remember the log laws:

    x=Ce^{t^{2}}-\frac{1}{2}



    what do you mean by 'negates'? [sorry if its a stupid simple question!]
    i can see how if i differentiate..i get the original equation.. but i cannt see how to integrate it yet! =(
    if i do it by parts... i have to keep integrating forever...

    do you just see the answer clearly by looking?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,506
    Thanks
    1403
    "Undoes" is probably a better word than "negates".

    Integrating undoes the derivative.

    So if \frac{d}{dt}[xe^{-t^2}] = te^{-t^2}

    \int{\frac{d}{dt}[xe^{-t^2}]\,dt} = \int{te^{-t^2}\,dt}

    The left hand side is xe^{-t^2} + C_1.

    Now use a u substitution on the RHS.

    Let u = -t^2 so that \frac{du}{dt} = -2t.


    So the RHS becomes

    \int{te^{-t^2}\,dt} = -\frac{1}{2}\int{-2te^{-t^2}\,dt}

     = -\frac{1}{2}\int{e^u\,\frac{du}{dt}\,dt}

     = -\frac{1}{2}\int{e^u\,du}

     = -\frac{1}{2}e^u + C_2

     = -\frac{1}{2}e^{-t^2} + C_2.


    Therefore, upon integrating both sides, the DE is

    xe^{-t^2} + C_1 = -\frac{1}{2}e^{-t^2} + C_2

    xe^{-t^2} = -\frac{1}{2}e^{-t^2} + C, where C = C_2 - C_1

    x = -\frac{1}{2} + Ce^{t^2}.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. integrating factor or d.e
    Posted in the Differential Equations Forum
    Replies: 3
    Last Post: June 6th 2010, 10:14 PM
  2. Integrating Factor.
    Posted in the Differential Equations Forum
    Replies: 2
    Last Post: May 27th 2010, 01:28 AM
  3. integrating factor
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: May 1st 2010, 10:00 AM
  4. integrating factor
    Posted in the Differential Equations Forum
    Replies: 3
    Last Post: June 9th 2008, 05:48 PM
  5. Integrating Factor
    Posted in the Calculus Forum
    Replies: 3
    Last Post: March 17th 2008, 07:31 AM

Search Tags


/mathhelpforum @mathhelpforum