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Math Help - Parametric differential equation

  1. #1
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    Parametric differential equation

    Hello everybody,
    i'm really stuck with this and i hope somebody can help me out.

    I have this differential equation:
    {y}'''-3{y}''+2{y}'+6y=e^{^{hx}}
    sadly there is no further explaination on what i should do with the parameter, so here is how i got through:
    i wrote the omogeneous equation:
    {y}'''-3{y}''+2{y}'+6y=0
    then tried to find its general solution by writing the characteristic equation:
    \lambda^{3}-3\lambda^{2}+2\lambda+6=0
    \left ( \lambda +1 \right )\left ( \lambda^{2}-4\lambda+6 \right )=0
    which has the following zeroes:
    real:
    \lambda_{1}=-1
    and
    complex:
    \lambda_{2,3}=2\pm i\sqrt{2}
    Which leads to the general solution of the omogeneous equation.
    Now, on to the parameter; due to the fact that the x coefficient, correct me if i'm wrong, changes the way my particular solution of the complete equation is wrote, should i study the cases where h equals the characteristic equation zeroes? And if this is the right procedure, when i study the case in which:
    h=2+i\sqrt{2} or h=2-i\sqrt{2}
    my solution will have the complex exponent?
    Such as: A*e^{2+i\sqrt{2x}} ?

    Hope this is a clear enough question, excuse me for my poor english

    and thanks in advance,
    SkyWolf.
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Re: Parametric differential equation

    No problem about a particular solution. According to a well known theorem, if h is not a root of the characteristic equation a particular solution has the form y(x)=Ce^{hx} . If h is a root of the characteristic equation of multiplicity s, a particular solution has the form y(x)=Cx^se^{hx} . Of course if h\in\mathbb{R} you needn't consider the roots 2\pm\sqrt{2}i .
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  3. #3
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    Re: Parametric differential equation

    Thanks a lot man.
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