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anonymous

  • one year ago

check my work (part 2)

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  1. anonymous
    • one year ago
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    Ok so this time I've studied more pls check my work:) \[\frac{d^2 \vec r}{dt^2}+4\vec r=\vec 0\] Let\[\vec r(t)=\vec Ce^{xt}\] The equation is then... \[\vec C x^2e^{xt}+4 \vec Ce^{xt}=\vec 0\]\[\vec C e^{xt}(x^2+4)=\vec 0\]\[\implies x^2+4=0\]\[x=\pm2i\] \[x_{1}=0+2i\]\[x_{2}=0-2i\]\[\lambda=0\]\[\mu=2\] \[\vec r(t)=\vec C_{1}e^{\lambda t}\cos(\mu t)+\vec C_{2}e^{\lambda t}\sin(\mu t)\]\[\vec r(t)=\vec C_{1}e^0\cos(2t)+\vec C_{2}e^0\sin(2t)\]\[\vec r(t)=\vec C_{1}\cos(2t)+\vec C_{2}\sin(2t)\]

  2. Loser66
    • one year ago
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    yup

  3. anonymous
    • one year ago
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    thx :)

  4. Loser66
    • one year ago
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    Question: why didn't you go directly from the original equation? Why did you go from the given solution \(\vec r(t) = \vec C e^{xt}\)

  5. Loser66
    • one year ago
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    What if they don't give you the solution? how can you find it?

  6. anonymous
    • one year ago
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    I don't understand what you mean ?? you just solve the polynomial and make the equations accordingly if variable is real and distinct, real and same and complex right because IT IS the method for solving the equation, I just didn't take exponential function out of the blue :)

  7. Loser66
    • one year ago
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    From r" +4r =0 characteristic equation is \(\lambda ^2 +4 =0\) which gives us \(\lambda = \pm 2i\) hence the general solution is as you get \(r = C_1cos(2t) + C_2 sin(2t)\) it is kind of 2-lines solution. That is what I meant.

  8. anonymous
    • one year ago
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    Oh, But I'm just starting off on this so I'm going slow :)

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