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Math Help - showing differential equation solution

  1. #1
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    showing differential equation solution

    for distinct roots (k1, k2) of the equation k^2 + 2Bk + w^2 show that x(t) = Ae^(k1t) + Be^(k2t) is a solution of the following differential equation: (d^2)x/dt^2 + 2B(dx/dt) + (w^2)x = 0
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  2. #2
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    All you need to do is take the general solution and sub it into the d.e. EG

    x = Ae^{k_1t} + Be^{k_2t}

    \frac{dx}{dt} = Ak_1e^{k_1t} + Bk_2e^{k_2t}

    \frac{d^2x}{dt^2} = Ak_1^2e^{k_1t} + Bk_2^2e^{k_2t}

    Substituting into the d.e.

    LHS = Ak_1^2e^{k_1t} + Bk_2^2e^{k_2t} + 2B(Ak_1e^{k_1t} + Bk_2e^{k_2t}) + w^2(Ae^{k_1t} + Be^{k_2t})

    Now collecting like terms

    LHS = Ae^{k_1t}\left(k_1^2 + 2Bk_1 + w^2\right) + Be^{k_2t}\left(k_2^2 +2Bk_2 + w^2\right)

    Now since k_1 and k_2 are roots of (k^2 + 2Bk + w^2) the terms on the LHS are both zero therefore the d.e. is satisfied.
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  3. #3
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    thanks for that i was going about it in such a difficult way, this was much simpler than i had anticipated...
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  4. #4
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    theres another part to this question which i still can't seem to get use the initial conditions x(0) = 1 and x'(0) = 0 to determine the values of A and B, describe how the solution behaves as t -> infinity
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  5. #5
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    Quote Originally Posted by crafty View Post
    theres another part to this question which i still can't seem to get use the initial conditions x(0) = 1 and x'(0) = 0 to determine the values of A and B, describe how the solution behaves as t -> infinity
    What you need to do here is substitute in the boundary conditions to determine A and B so...

    \begin{array}{rcl}<br />
Ae^{k_10}+Be^{k_20} &=& 1\\<br />
A + B &=& 1<br />
\end{array}

    and

    \begin{array}{rcl}<br />
Ak_1e^{k_10}+Bk_2e^{k_20} &=& 0\\<br />
Ak_1 + Bk_2 &=& 0<br />
\end{array}

    You can solve these simultaneous as 2 equations with 2 unknowns (A and B)

    This problem is a bit tricky since throughout the B in the original equation is different from the B in the general solution.

    Regarding the general behavior it depends on the sign of k_1 and k_2. If we assume the B in the original equation is real and positive and the w is real and less than B then the roots are

    k = -B\pm\sqrt{B^2 - w^2}

    which are both definitely negative and in that case the exponential terms go to zero as t goes to infinity.

    Hope this helps.
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  6. #6
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    and how do i solve these simultaneous equations
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  7. #7
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    \begin{array}{rclr}<br />
A + B &=& 1 &(1) \\<br />
k_1A + k_2B &=& 0 &(2)<br />
\end{array}

    Step 1: k_1 \times (1)

    \begin{array}{rclr}<br />
k_1A + k_1B &=& k_1 &(3)<br />
\end{array}

    Step 2: (3) - (2)

    \begin{array}{rclr}<br />
 (k_1-k_2)B &=& k_1 &(4)<br />
 \end{array}

    Step 3: (4) \div (k_1-k_2)

    B = \dfrac{k_1}{k_1-k_2}

    Step 4: Substitute B = \frac{k_1}{k_1-k_2} into (2)

    k_1A + \frac{k_1k_2}{k1-k_2} = 0

    A = \frac{k_2}{k_2 - k_1}
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  8. #8
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    by the way for that part about infinity i forgot to mention that W>B and that means the roots are complex, so what happens then??? sorry about forgetting to mention this earlier
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  9. #9
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    Quote Originally Posted by crafty View Post
    by the way for that part about infinity i forgot to mention that W>B and that means the roots are complex, so what happens then??? sorry about forgetting to mention this earlier
    OK so if W > B and B > 0 then you have the roots equal to

    k_1 = -B + iC and k_2 = -B - iC

    where C = \sqrt{W^2 - B^2}

    Then your solution becomes

    x = \frac{k_1 e^{k_2t}-k_2e^{k_1t}}{k_1 - k_2}

    x = \frac{(-B+iC) e^{(-B-iC)t}-(-B-iC) e^{(-B+iC)t}}{2iC}

    x = e^{-Bt}\left[\frac{1}{2}\left(e^{iCt}+e^{iCt}\right)+\frac{B}{2  iC}\left(e^{iCt}-e^{iCt}\right)\right]

    x = e^{-Bt}\left[\cos(Ct)+\frac{B}{C}\sin(Ct)\right].

    So when t goes to infinity the exponential term sends the function to zero. The other terms just cause the function to wiggle around but the overall trend is x \rightarrow 0.

    If B < 0 then the function x goes to infinity.
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