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Thread: second order differential equations again

  1. #1
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    second order differential equations again

    A particle moves along the x-axis so that the time ts, its displacement xm from the origin satisfies the differential equation

    $\displaystyle \frac{d^2x}{dt^2}+2\frac{dx}{dt}-35x=70t + 31$

    Given that when t=0, the particle is at rest at the origin, find its displacement at time ts.


    I'm struggling with this because i think it involves complex roots, which i haven't done before.i know the C. Function is $\displaystyle e^{\alpha t}(Acos\beta t + \beta sin \beta t)$

    and from the equation i get

    $\displaystyle d^2 + 2d - 35$

    $\displaystyle \frac{-2^+_- \sqrt{4-140}}{2} $

    $\displaystyle \frac{-2^+_- \sqrt{-136}}{2}$

    Im a bit stuck and where to go from here. Can i still use surds when its complex? ie $\displaystyle \sqrt{-136}=2\sqrt{-34}$ then i can cancel the 2's with the denominator?
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  2. #2
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    You are correct that

    $\displaystyle d^2+2d-35=0$

    gives us the auxiliary roots for the homogenous solution of $\displaystyle x''+2x'-35x=70t+31$. To solve this, we factor it into

    $\displaystyle (d+7)(d-5)=0.$

    Using the quadratic formula, we obtain the same answers:

    $\displaystyle \frac{-2\pm\sqrt{2^2-4\cdot1\cdot(-35)}}{2}=\frac{-2\pm\sqrt{144}}{2}=\frac{-2\pm 12}{2}=-1\pm 6=-7\,\mbox{or}\,5.$

    It can be shown that the general solution is of the form

    $\displaystyle x=x_h+x_p$

    where $\displaystyle x_h$ is the homogeneous solution and $\displaystyle x_p$ is a particular solution of $\displaystyle x''+2x'-35x=70t+31$. Since $\displaystyle 70t+31$ is a first-degree polynomial, we may try

    $\displaystyle x_p=A t+B$

    as a solution and then try to figure out what the values of $\displaystyle A$ and $\displaystyle B$ might be.
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  3. #3
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    OK it was just me making a silly mistake! haha thanks
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