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Math Help - A quick question on complimentary functions

  1. #1
    Super Member craig's Avatar
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    A quick question on complimentary functions

    Hi.

    If the auxiliary equation of a second order differential equation has 2 complex repeated roots, would I still use:

    e^{\alpha x}(A \cos{\beta x} + B\sin{\beta x})

    Just in the form:

    A \cos{\beta x} + B\sin{\beta x}?

    Thanks
    Last edited by craig; June 8th 2009 at 05:38 AM.
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  2. #2
    Super Member craig's Avatar
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    Doesn't matter, found out that you use e^{\alpha x}(A \cos{\beta x} + B\sin{\beta x})
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    Quote Originally Posted by craig View Post
    Hi.

    If the auxiliary equation of a second order differential equation has 2 repeated roots, would I still use:

    e^{\alpha x}(A \cos{\beta x} + B\sin{\beta x})

    Just in the form:

    A \cos{\beta x} + B\sin{\beta x}?

    Thanks
    If the root is repeated then the general solution is y = (Ax + B) e^{\lambda x}.
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  4. #4
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    Quote Originally Posted by craig View Post
    Doesn't matter, found out that you use e^{\alpha x}(A \cos{\beta x} + B\sin{\beta x})
    This is the general solution for when the roots are not real, that is, the roots have the form \lambda = \alpha \pm i \beta.
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  5. #5
    Super Member craig's Avatar
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    Hi thanks for the reply, there was actually a typo in my original post, it was for roots such as \pm 2i.

    Have fixed the typo now.
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    Senior Member Sampras's Avatar
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    Quote Originally Posted by craig View Post
    Hi thanks for the reply, there was actually a typo in my original post, it was for roots such as \pm 2i.

    Have fixed the typo now.
    Assume the coefficients are real. Then the roots come in pairs. I would think the general solution would be  y =  c_{1}e^{ax} \cos bx+c_{2}xe^{ax} \cos bx+c_{3}e^{ax} \sin bx + c_{4}xe^{ax} \sin bx for  \lambda = a \pm bi if degree of multiplicity is  2 .
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  7. #7
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    Quote Originally Posted by Sampras View Post
    Assume the coefficients are real. Then the roots come in pairs. I would think the general solution would be  y =  c_{1}e^{ax} \cos bx+c_{2}xe^{ax} \cos bx+c_{3}e^{ax} \sin bx + c_{4}xe^{ax} \sin bx for  \lambda = a \pm bi if degree of multiplicity is  2 .
    No, that the same as (c_1+ c_3)e^{ax}cos bx+ (c_2+ c_4)e^{ax}sin bx so you have nothing new. For a+ bi a double root of the characteristic equation, you want y= e^{ax}(c_1cos(bx)+ c_2sin(bx))+ xe^{ax}(c_3cos(bx)+ c_4sin(bx)).

    That is, multiply the "base" solution by x as mr. fantastic said.
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