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Math Help - Problem

  1. #1
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    Problem

    Let be a positive constant. Consider the equation


    with boundary conditions and .

    For certain discrete values of , this equation can have non-zero solutions. Find the three smallest values of for which this is the case.

    -----

    I've been trying to do this, but I get nowhere. Maybe I'm doing something wrong... Let me know.

    So the characteristic equation is k^2 + 8k + a = 0.

    The roots are:

    k1 = -4 + sqrt(16-a)
    k2 = -4 - sqrt(16-a)

    General solution:

    y = Ae^[(-4+sqrt(16-a))x] + Be^[(-4-sqrt(16-a))x]

    From the given conditions:

    0 = A+B --> B = -A

    0 = Ae^[-8+2sqrt(16-a)] + Be^[-8-2sqrt(16-a)]

    Substituting -A for B and taking e^-8 out:

    0 = (Ae^(-8))[e^(2sqrt(16-a)) - e^(-2sqrt(16-a)]

    So I thought that for A and B to not equal 0, I need to make the term in the second parantheses 0 and solve for a. However I only get one answer: a=0 which seems to be wrong.

    Maybe I'm completely missing the point here. I would appreciate it if anyone could help me.
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  2. #2
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    Hello, MathTiger!

    I too much be missing something . . . I found only one solution.


    Let a be a positive constant.

    Consider the equation: . \frac{d^2y}{dx^2} + 8\,\frac{dy}{dx} + ay \:=\:0

    with boundary conditions: . y(0) = 0,\:y(2) = 0

    For certain discrete values of a, this equation can have non-zero solutions.
    Find the three smallest values of a for which this is the case.

    Your preliminary work is correct . . .


    So the characteristic equation is: .  k^2 + 8k + a \:= \:0

    The roots are: . \begin{array}{ccc}k_1 &=& -4 +\sqrt{16-a} \\ k_2 &=& -4 - \sqrt{16-a} \end{array}


    General solution: . y \;= \;Ae^{(-4+\sqrt{16-a})x} + Be^{(-4-\sqrt{16-a})x}


    From the given conditions:

    . . 0 \:=\: A+B\quad\Rightarrow\quad B \:=\: -A

    . . 0 \:= \:Ae^{-8+2\sqrt{16-a}} + Be^{-8-2\sqrt{16-a}}

    Substituting -A for B and taking out e^{-8}\!:\;\;Ae^{-8}\,\bigg[e^{2\sqrt{16-a}} - e^{-2\sqrt{16-a}}\bigg] \:=\:0

    So I thought that for A, B \neq 0,
    . . I need to make the second group equal to 0 and solve for a. . Right!

    We have: . e^{2\sqrt{16-a}} - e^{-2\sqrt{16-a}} \;=\;0

    Multiply by e^{2\sqrt{16-a}}\!:\;\;e^{4\sqrt{16-a}} - 1 \:=\:0 \quad\Rightarrow\quad e^{4\sqrt{16-a}} \:=\:1\:=\:e^0

    Then: . 4\sqrt{16-a} \:=\:0 \quad\Rightarrow\quad \sqrt{16-a} \:=\:0 \quad\Rightarrow\quad 16-a \:=\:0 \quad\Rightarrow\quad \boxed{a \:=\:16}

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  3. #3
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    Correction

    Oops, I did mean I only get a=16, not a=0. But the online submission system says that a=16 is neither of these three lowest values of a.
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by MathTiger View Post
    Let be a positive constant. Consider the equation


    with boundary conditions and .

    For certain discrete values of , this equation can have non-zero solutions. Find the three smallest values of for which this is the case.

    -----

    I've been trying to do this, but I get nowhere. Maybe I'm doing something wrong... Let me know.

    So the characteristic equation is k^2 + 8k + a = 0.

    The roots are:

    k1 = -4 + sqrt(16-a)
    k2 = -4 - sqrt(16-a)

    General solution:

    y = Ae^[(-4+sqrt(16-a))x] + Be^[(-4-sqrt(16-a))x]

    From the given conditions:

    0 = A+B --> B = -A

    0 = Ae^[-8+2sqrt(16-a)] + Be^[-8-2sqrt(16-a)]

    Substituting -A for B and taking e^-8 out:

    0 = (Ae^(-8))[e^(2sqrt(16-a)) - e^(-2sqrt(16-a)]

    So I thought that for A and B to not equal 0, I need to make the term in the second parantheses 0 and solve for a. However I only get one answer: a=0 which seems to be wrong.

    Maybe I'm completely missing the point here. I would appreciate it if anyone could help me.
    let \lambda_1 and \lambda_2 be the roots of the charateristic equation.

    Then if A\ne 0 we have:

    e^{2\lambda_1}=e^{2\lambda_2}

    which implies that:

    2\lambda_1 - 2\lambda_2=\text{\bf{i}}\ 2\ n\ \pi

    for some n \in \mathbb{Z}.

    CB
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  5. #5
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    Quote Originally Posted by CaptainBlack View Post
    let \lambda_1 and \lambda_2 be the roots of the charateristic equation.

    Then if A\ne 0 we have:

    e^{2\lambda_1}=e^{2\lambda_2}

    which implies that:

    2\lambda_1 - 2\lambda_2=\text{\bf{i}}\ 2\ n\ \pi

    for some n \in \mathbb{Z}.

    CB
    Sorry... I'm not sure how you get that equation... And even so, how do I choose n to get the lowest values of a?
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  6. #6
    Grand Panjandrum
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    Quote Originally Posted by MathTiger View Post
    Sorry... I'm not sure how you get that equation... And even so, how do I choose n to get the lowest values of a?
    The equation holds because for any integer n :

    e^{a+2 \text{\bf{i}}\ n\ \pi}=e^ae^{2 \text{\bf{i}}\ n\ \pi}=e^a

    So if two powers of e are equal then the powers differ by an integer multiple of 2 \text{\bf{i}}\ \pi

    CB
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