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Math Help - Is this true?

  1. #1
    DCU
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    Is this true?

    If y1 = e^(ax)*Cos(Bx) is a solution of a certain homogenrous second order linear equation with constant coefficents then y2 = e^(ax)*Sin(Bx) is also a solution of this equation?
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  2. #2
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    Quote Originally Posted by DCU View Post
    If y1 = e^(ax)*Cos(Bx) is a solution of a certain homogeneous second order linear equation with constant coefficents then y2 = e^(ax)*Sin(Bx) is also a solution of this equation?
    Saying that e^{ax}Cos(bx) is a solution of a certain homogenreous second order linear equation with constant coefficents implies that a+ Bi is a solution to the characteristic equation.

    IF the differential equation (and therefore characteristic equation) has real coefficients, then a- Bi must also be a solution to the characteristic equation and so e^{ax}Cos(bx) is an independent solution to the differential equation. However, if any of the coefficients is non-real complex, that is not necessarily true.
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  3. #3
    DCU
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    So it's true if the coefficients are real numbers but not if the coefficents aren't?
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