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Math Help - Fourth Order Homogeneous Linear Equation with real roots and complex roots.

  1. #1
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    Fourth Order Homogeneous Linear Equation with real roots and complex roots.

    the problem is:

    y(4)-5y''-36y=0

    => m4-5y2-36=0
    => (m2-9)(m2+4)=0

    therefore, m1=3, m2=-3, m3=2i, m4=-2i.

    I have learned how to find the general solution for cases with distinct real roots and for cases with complex roots but I am unsure how to I would go about finding the general solution for a linear equation that has both real and complex roots.

    Would the general solution be the following

    c1e3x+c2e-3x+c3(cosx+(2i)sinx)+c4(cosx+(-2i)sinx)

    by using euler's formula?
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  2. #2
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    Re: Fourth Order Homogeneous Linear Equation with real roots and complex roots.

    Quote Originally Posted by bakinbacon View Post
    the problem is:

    y(4)-5y''-36y=0

    => m4-5y2-36=0
    => (m2-9)(m2+4)=0

    therefore, m1=3, m2=-3, m3=2i, m4=-2i.

    I have learned how to find the general solution for cases with distinct real roots and for cases with complex roots but I am unsure how to I would go about finding the general solution for a linear equation that has both real and complex roots.

    Would the general solution be the following

    c1e3x+c2e-3x+c3(cosx+(2i)sinx)+c4(cosx+(-2i)sinx)

    by using euler's formula?
    The general solution is:

    y(x)=\sum_{k=1}^4 A_ke^{\lambda_k x}

    where \lambda_k, k=1, ... 4 are the roots of the characteristic equation, irrespective of if they are real imaginary or non-trivially complex.

    And for your information: e^{\pm 2 i x}=\cos(2x) \pm i \sin(2x)
    .
    Last edited by zzephod; February 22nd 2013 at 04:13 AM.
    Thanks from jakncoke
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