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Math Help - second order ODE

  1. #1
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    second order ODE

    a,b,c > 0

    ay'' + by' +cy = 0
    and y(0) = c and y'(0) = d

    the equation has either one real root, two real roots or two complex roots, but how do i then show that in the limit x--> +infinity, y(x) will go to zero?

    Appreciate Your help!
    Last edited by sssouljah; September 4th 2009 at 08:54 AM.
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  2. #2
    MHF Contributor chisigma's Avatar
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    If all roots of the polynomial...

    a\cdot t^{2} + b\cdot t + c

    ... have real part <0, then if y(x) is solution of the second order DE you have prosed is...

    \lim_{x \rightarrow \infty} y(x) =0

    Kind regards

    \chi \sigma
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    but how do i show that all the roots are negative
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  4. #4
    MHF Contributor chisigma's Avatar
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    A polynomial whose roots have all real part <0 is called 'Hurwitz polynomial'. If we write a polynomial of degree n in t as p(t)= \sum_{k=0}^{n} a_{k} \cdot t^{k} and it is a_{0}=1, necessary [but not sufficient...] condition for p(t) to be Hurwitz is that  \forall k\ge 1 a_{k}>0. For a suffcient condition see...

    Hurwitz polynomial - Wikipedia, the free encyclopedia

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    \chi \sigma
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  5. #5
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    Quote Originally Posted by sssouljah View Post
    a,b,c > 0

    ay'' + by' +cy = 0
    and y(0) = c and y'(0) = d

    the equation has either one real root, two real roots or two complex roots, but how do i then show that in the limit x--> +infinity, y(x) will go to zero?

    Appreciate Your help!
    Descartes rule of signs tell you that a\lambda^2+b \lambda +c=0 has no positive roots and one negative root when a,\ b,\ c >0. Hence it has one negative root of multiplicity 2 and no complex roots.

    CB
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