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Math Help - Higher Order Complex Roots

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    Higher Order Complex Roots

    Find the general solution to the given differential equation:

    <br />
y^{(4)} - 8 y' = 0<br />

    <br />
y''' + 5 y'' + 6 y' + 2 y = 0<br />

    <br />
4 y''' + y' + 5 y = 0<br />
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    Quote Originally Posted by fudawala View Post
    Find the general solution to the given differential equation:

    <br />
y^{(4)} - 8 y' = 0<br />
    The characteristic equation is
    m^4 - 8 = 0

    (m^2 + \sqrt{8})(m^2 - \sqrt{8}) = 0

    (m^2 + 2 \sqrt{2})(m^2 - 2 \sqrt{2}) = 0

    The first factor doesn't factorize further over the real numbers, but we can factor it over the complex numbers:
    m^2 + 2sqrt{2} = \left ( m + i \sqrt{2{\sqrt{2}}} \right ) \left ( m - i \sqrt{2{\sqrt{2}}} \right )

    and, of course
    m^2 - 2sqrt{2} = \left ( m + \sqrt{2{\sqrt{2}}} \right ) \left ( m - \sqrt{2{\sqrt{2}}} \right )

    So
    m^4 - 8 = \left ( m + i \sqrt{2{\sqrt{2}}} \right ) \left ( m - i \sqrt{2{\sqrt{2}}} \right ) <br />
\left ( m + \sqrt{2{\sqrt{2}}} \right ) \left ( m - \sqrt{2{\sqrt{2}}} \right ) = 0

    Thus the solution to the differential equation will be of the form:
    y(x) = Ae^{ix \sqrt{2{\sqrt{2}}}} + Be^{-i x \sqrt{2{\sqrt{2}}}} + Ce^{x \sqrt{2{\sqrt{2}}}} + De^{-x \sqrt{2{\sqrt{2}}}}

    or if you prefer:
    y(x) = A^{\prime}~cos \left ( x \sqrt{2{\sqrt{2}}} \right )  + B^{\prime}~sin \left ( x \sqrt{2{\sqrt{2}}} \right ) + Ce^{x \sqrt{2{\sqrt{2}}}} + De^{-x \sqrt{2{\sqrt{2}}}}

    -Dan
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by fudawala View Post
    Find the general solution to the given differential equation:

    <br />
y''' + 5 y'' + 6 y' + 2 y = 0<br />
    The characteristic equation is
    m^3 + 5m^2 + 6m + 2 = 0

    The rational roots theorem says that any rational zeros of this polynomial will be of the form \pm 1, \pm 2. Checking these shows that m = -1 is a solution. Thus:
    m^3 + 5m^2 + 6m + 2 = (m + 1)(m^2 + 4m + 2) = 0

    We use the quadratic formula to get factors of the quadratic factor, giving:
    m^3 + 5m^2 + 6m + 2 = (m + 1)(m^2 + 4m + 2) = (m + 1)(x + 2 + \sqrt{2})(x + 2 - \sqrt{2}) = 0

    which has solutions m = -1, -2 + \sqrt{2}, -2 - \sqrt{2}.

    So the solution of the differential equation will be of the form:
    y(x) = Ae^{-x} + Be^{(-2 + \sqrt{2})x} + Ce^{(-2 - \sqrt{2})x}

    -Dan
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by fudawala View Post
    Find the general solution to the given differential equation:
    <br />
4 y''' + y' + 5 y = 0<br />
    The characteristic equation is
    4m^3 + m +5 = 0

    The possible rational zeros of this polynomial are:
    m = \pm  1, \pm 5, \pm \frac{1}{2}, \pm \frac{5}{2}, \pm \frac{1}{4}, \pm \frac{5}{4}

    Checking these I see that m = -1 is the only rational solution. Thus:
    4m^3 + m +5 = (m + 1)(4m^2 - 4m + 5) = (m + 1) \left ( m -\frac{1}{2} + i \right ) \left ( m -\frac{1}{2} - i \right ) = 0

    You can take it from here.

    -Dan
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