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IVP applications of second-order ODE

  1. Mar 5, 2012 #1
    1. The problem statement, all variables and given/known data

    Given the equation mx''+cx=cAsin(Ωt) with the initial conditions x(0)=0 and x'(0)=0.
    Solve the initial value problem for the case when Ω < ω and show that |x(t)| < H provided
    A < H(1-(Ω/ω)).


    2. Relevant equations



    3. The attempt at a solution

    For my solution to the equation I got x(t)=(Aω/(ω^2-Ω^2))[ωsin(Ωt)-Ωsin(ωt)]

    So, I'm hoping that this is right which I found using x(t)=x_h+x_p.
    But I'm completely confused about the second part to show that |x(t)| < H provided
    A < H(1-(Ω/ω)). :confused:

    This is my first post here, so apologies if I'm not doing it right. :shy:
    Thanks in advance.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Mar 5, 2012 #2

    LCKurtz

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    Good work so far; you are almost there. Take the absolute value of both sides of your ##x(t)## equatiion and use this:$$
    |\omega \sin(\Omega t) - \Omega \sin(\omega t)|\le
    |\omega| | \sin(\Omega t)| + |\Omega| | \sin(\omega t)|
    \le \omega + \Omega$$and put in the overestimate given for A. See what drops out.
     
  4. Mar 5, 2012 #3
    Ahh Ok, so it's my modulus work that's really poor!

    I'm thinking you're equation changes mine to: Aω/(ω^2-Ω^2)<H(ω+Ω)

    So with the substitution for A I get (ω-Ω)/(ω^2-Ω^2)<(ω+Ω)

    So from this, could I finish by showing this is true? Or have I gone wrong again? I'm thinking that my Aω/(ω^2-Ω^2) should have changed when I was taking the modulus?
     
  5. Mar 5, 2012 #4

    LCKurtz

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    The contraction "you're" is a short form of "you are". The possessive case is "your". Also, while I'm commenting on your post, I should point out that you never told us what ##\omega## stands for. Best not to leave definitions out.

    You will have less trouble if you will write out your full string of inequalities. You have$$
    x(t) = \frac{A\omega}{\omega^2-\Omega^2}(\omega \sin(\Omega t) - \Omega \sin(\omega t))$$Start by taking the absolute value of both sides:$$
    |x(t)|=\left| \frac{A\omega}{\omega^2-\Omega^2} \right| |\omega \sin(\Omega t) - \Omega \sin(\omega t)| \le ...$$Now continue the string on the right using what you know so far.
     
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