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Thread: Initial conditions

  1. #1
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    Initial conditions

    Find the particular solution for the differential equation $\displaystyle \dfrac{d^2x}{dt^2}-\dfrac{dx}{dt}-20x = 0$ that satisfies the initial conditions $\displaystyle x = 0$, $\displaystyle \dfrac{dx}{dt} = 2$ when $\displaystyle t = 0$.

    Substituting $\displaystyle x = Ae^{n_{1}x}+Be^{n_{2}x}$, I got $\displaystyle n = 5$ and $\displaystyle n = -4$, so the general solution is $\displaystyle x = Ae^{5t}+Be^{-4t}.$

    What I don't know is how to make use of the initial conditions to find $\displaystyle A$ and $\displaystyle B$.
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  2. #2
    Junior Member slider142's Avatar
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    Find (dx/dt)(t): $\displaystyle \frac{dx}{dt}(t) = 5Ae^{5t} - 4Be^{-4t}$
    Now you can get two equations in two unknowns using the two initial conditions: (dx/dt)(0) = 2 and x(0) = 0.
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  3. #3
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    Quote Originally Posted by slider142 View Post
    Find (dx/dt)(t): $\displaystyle \frac{dx}{dt}(t) = 5Ae^{5t} - 4Be^{-4t}$
    Now you can get two equations in two unknowns using the two initial conditions: (dx/dt)(0) = 2 and x(0) = 0.
    $\displaystyle 5A-4B = 2$, $\displaystyle 5A = 2+4B$, $\displaystyle A = \dfrac{2+4B}{5}.$

    $\displaystyle A = -B$

    $\displaystyle \dfrac{2+4B}{5} = -B$

    $\displaystyle 2+4B = -5B$

    $\displaystyle B = -\dfrac{2}{9}$

    $\displaystyle \therefore \;\;\; A = \dfrac{9}{2}

    $ and $\displaystyle B = -\dfrac{9}{2}.$

    Therefore the particular equation is $\displaystyle x = -\dfrac{2}{9}e^{5t}+\dfrac{2}{9}e^{-4t}.$
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