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Linear Differential Equation of Second Order

  1. May 4, 2012 #1
    1. The problem statement, all variables and given/known data
    I have the equation mx'' + cx' + kx = 0

    where m = 2, c = 12, k = 50, x0 = 0, x'(0) = -8. x is a function of t, and primes denote derivatives w.r.t. t.


    2. Relevant equations



    3. The attempt at a solution

    so the equation is 2x'' + 12x' + 50x = 0, which I simplify to x'' + 6x' + 25x = 0.

    Characteristic equation is r2 + 6r + 25 = 0

    using the quadratic formula, I find roots r = -3 ± 4i

    therefore the general solution is x(t) = e-3t(c1cos(4t) + c2sin(4t))

    I then solve for the unknown constants, c1 and c2:

    x(0) = 0 = c1

    so x(t) = e-3t(c2sin(4t))

    x'(t) = -3e-3t(c2sin(4t) + e-3t(4c2cos(4t))

    x'(0) = -8 = 4c2

    so c2 = -2

    ∴ x(t) = e-3t(-2sin(4t))

    but... the back of the book gives

    x(t) = 2e-3tcos(4t - 3[itex]\pi[/itex]/2)

    as the answer...

    so I must be doing something horribly wrong, but I have no idea what :confused:
     
  2. jcsd
  3. May 4, 2012 #2
    oh whoops, I just realized that, for the set of problems that I'm working on, it asks me to write the function in the form

    C1e-ptcos(ω1t - [itex]\alpha[/itex]1)

    so I guess it's just a matter of converting it
     
  4. May 4, 2012 #3

    sharks

    User Avatar
    Gold Member

    You're correct. :smile:
     
  5. May 4, 2012 #4

    sharks

    User Avatar
    Gold Member

    [tex]x(t)=e^{-3t} (-2\sin 4t)=-2e^{-3t} (\sin 4t)[/tex]
    Now, you try to obtain the above from the given answer, to check for equivalence:
    [tex]x(t)=2e^{-3t} (\cos (4t - \frac{3\pi}{2}))[/tex]
    Comparing the two answers above, what you need is to prove that:
    [tex]\cos (4t - \frac{3\pi}{2})=-\sin 4t[/tex]
    Use the cosine subtraction formula:
    [tex]\cos (4t - \frac{3\pi}{2})=(\cos 4t)(\cos \frac{3\pi}{2})+ (\sin 4t)(\sin \frac{3\pi}{2})=-\sin 4t[/tex]Proved!
     
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