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Math Help - Equation

  1. #1
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    Equation

    Hi everybody,

    We consider the following differential equation:

    (E) : y''+2y'+2y=-(x+1)e^{-x}+2

    g(x)=\int_0^{x} te^{-t} dt is a solution to this equation E

    1)-I must deduct the solutions of the equation (E)

    I know that the solutions are:  e^{-2x}(\alpha cos(x)+ \beta sin(x))

    But how to determine \alpha and \beta??
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  2. #2
    MHF Contributor
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    Quote Originally Posted by lehder View Post
    Hi everybody,

    We consider the following differential equation:

    (E) : y''+2y'+2y=-(x+1)e^{-x}+2

    g(x)=\int_0^{x} te^{-t} dt is a solution to this equation E

    1)-I must deduct the solutions of the equation (E)

    I know that the solutions are:  e^{-2x}(\alpha cos(x)+ \beta sin(x))

    But how to determine \alpha and \beta??
    I'll need initial or boundary conditions to determine those.
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  3. #3
    MHF Contributor chisigma's Avatar
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    The 'incomplete' DE is...

    y^{''} + 2\cdot y^{'} + 2\cdot y=0 (1)

    ... and its 'characteristic equation' is...

    x^{2} + 2\cdot x + 2 =0 (2)

    ... the solution of which are x= -1 \pm i, so that the general integral of the (1) is...

    y= e^{-x}\cdot (\alpha\cdot \cos x + \beta\cdot \sin x) (3)

    ... that is different from the solution of the starting post.

    Now a particular solution of the 'complete' DE...

    y^{''} + 2\cdot y^{'} + 2\cdot y= -(x+1)\cdot e^{-x} + 2 (4)

    ... You can verify is ...

    g(x)= \int_{0}^{x} t\cdot e^{-t}\cdot dt (5)

    ... so that the general integral of (4) is...

    y= e^{-x}\cdot (\alpha\cdot \cos x + \beta\cdot \sin x) + g(x) (6)

    Kind regards

    \chi \sigma
    Last edited by chisigma; February 23rd 2010 at 01:11 AM. Reason: trivial error corrected...
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