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Thread: Auxillary Equation

  1. #1
    Mar 2009

    Auxillary Equation

    Hi all i need some help with understanding auxillary equation based on boundary values

    An example was:

    y" + 2y' + y

    with y(0) = 1, y(1) = 3

    thus putting it into the auxillary equation and finding the roots:

    r1 = r2 = -1

    putting it into the formula gave

    y = C1e^(-x) + C2xe^(-x) note that C1 is C with subscript one and so on :3

    Heres the part i dont get. In my book it says

    since y(0) = 1 we have 1 = y(0) = C1

    Further, as y(1) = 3 we have

    3 = y(1) = e^(-1) + C2e^(-1)

    So C2 = 3e - 1 and therefore:

    y = e^(-x) + (3e-1)xe

    Can someone help me with this (could you also explain why C1 disappeared)?

    Your contribution is much appreciated.

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  2. #2
    Member Ruun's Avatar
    Mar 2009
    North of Spain
    It didn't dissappear. If $\displaystyle C_1=1$, then $\displaystyle C_1e^{-x}=e^{-x}$

    -edit- forget the explanations:

    $\displaystyle y(0)$ and $\displaystyle y(1)$ are given to know the values of $\displaystyle C_1$ and $\displaystyle C_2$. The solution is fome function $\displaystyle y(x)$ so plugging $\displaystyle x=0$ to obtain $\displaystyle y(0)$ and the same to $\displaystyle x=1$ for $\displaystyle y(1)$, you're able, as you've done, to compute the values of $\displaystyle C_1$ and $\displaystyle C_2$-
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  3. #3
    Mar 2009
    thanks again mate
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