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Thread: Easiest way to solve

  1. #1
    Newbie khanim's Avatar
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    Easiest way to solve

    Hi

    The section that I am currently busy with is the D operator, I am learning how to find the aux equation and complimentary functions

    Its basic maths but unfortunately I am struggling with the higher DE

    $\displaystyle \frac{d^{3}y}{dx^{3}} + \frac{d^{2}y}{dx^{2}} - 2y = 0$

    I need to know what is the simplest way to solve these higher order DE
    Please help
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  2. #2
    MHF Contributor chisigma's Avatar
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    That's a linear DE with constant coefficients and its 'characteristic equation' is...

    $\displaystyle d^{3} + d^{2} - 2 = 0$ (1)

    But ...

    $\displaystyle d^{3} + d^{2} - 2 = (d^{2} + 2d +2)\cdot (d-1)$ (2)

    ... so that...

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
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  3. #3
    Newbie khanim's Avatar
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    Thx Chi
    how did you get that, I how do see that? is there any special method you used
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  4. #4
    MHF Contributor chisigma's Avatar
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    In this case it is immediate 'discovering' that $\displaystyle d=1$ makes null the polynomial $\displaystyle d^{3} + d^{2} -2$, so that $\displaystyle d-1$ is a factor of that polynomial. In general we aren't so 'lucky' and in such a case we have to 'attack' a third order algebraic equation, that the Italian mathematicians Scipione del Ferro and Niccolò Fontana first successfully resolved between fifteenth and sixteenth century ...

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
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