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Math Help - integrating factor

  1. #1
    Member i_zz_y_ill's Avatar
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    integrating factor

    Why can't I use an integrating factor when asked to find the complementaryt function of this?
    dy/dt + 2y = e^-2t

    I figured I either use an auxiliary eqn or I.F but the I.F doesn't work. Am i missing some conditions conditions to use an I.f or something thnx.

    The I.f rule states in the form dy/dx +Py = Q
    where P and Q are functions of x you can use the I.f
    Since in this example 2 was essentially a function of x(i don't know how i just thought it is a constant) then arguably Q can just be a constant. Does that mean that i can use the I.f method for instance with dy/dx + 2y = 3??thnx
    Last edited by i_zz_y_ill; June 9th 2008 at 05:19 PM. Reason: d
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by i_zz_y_ill View Post
    hy can't I use an integrating factor when asked to find the complementaryt function of this?
    dy/dt + 2y = e^-2t

    I figured I either use an auxiliary eqn or I.F but the I.F doesn't work. Am i missing some conditions conditions to use an I.f or something thnx.
    Why won't it work? The integrating factor is e^{2t}, so your equation turns into
    \left ( e^{2t}y \right ) ^{\prime} = 1

    -Dan
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  3. #3
    Member i_zz_y_ill's Avatar
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    Yeah thanks, I was wondering though since in this stuation 2 is essentially a function of x if I had dy/dx + 3y = 4 couldn't this 4 essentially be a function of x in which case the method of I.f could be used. Giving y = 4/3 + c

    What confuses me is why using an auxiliary eqn and a particular integral gives a more complicated genera solution i.e y = Qx + P + Ae^x

    Furthermore what confuses me is that if 4 is just perceived as a function of x, what's to stop it being a constant if that makes sense function of y so that it can be taken to the LHS and the equation becomes homogenous i need of only a complementary function. In which case a general solution is y = Ae^-3x.

    I've obviously presumed alot which is wrong!!!!I don't understand lol
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by i_zz_y_ill View Post
    Yeah thanks, I was wondering though since in this stuation 2 is essentially a function of x if I had dy/dx + 3y = 4 couldn't this 4 essentially be a function of x in which case the method of I.f could be used. Giving y = 4/3 + c

    What confuses me is why using an auxiliary eqn and a particular integral gives a more complicated genera solution i.e y = Qx + P + Ae^x

    Furthermore what confuses me is that if 4 is just perceived as a function of x, what's to stop it being a constant if that makes sense function of y so that it can be taken to the LHS and the equation becomes homogenous i need of only a complementary function. In which case a general solution is y = Ae^-3x.

    I've obviously presumed alot which is wrong!!!!I don't understand lol
    To solve y' + 2y = e^{-2t} the general solution will be of the form y(t) = Ae^{-2t} + Bte^{-2t}, it does not have a linear term in it, if that's one of the things troubling you. (The A term is the solution to the homogeneous equation and the B term is the particular solution.)

    The "4" can be taken to be a function of anything you desire. In the IF method we take it to be a function of x for convenience.

    -Dan
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