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Math Help - 2nd order differential eqs

  1. #1
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    2nd order differential eqs

    Ok, so i have the eq

    d^2x/dt^2 + 4x = sin2t

    And have to first find the auxiliary eq (AE) and also the particular soln.
    then obtain the general soln and confirm that no periodic soln exists.

    So far i have found the AE to be m^2 + 4 = 0,

    Which gives the solution Aexp(2jt) + Bexp(-2jt) [where j = -1^0.5]

    Ive then attempted to use a trial solution of the form t^2(asin2t +bcos2t)
    ive differentiated this twice and substituted into the initial equation,

    However this is where i get in a muddle, when i substitute the two terms in and simplify, then by equating like terms i get that for the particular soln:

    (8at +2b)cos2t = 0 and (2a-8bt)sin2t = sin2t

    And i cannot see how to find a and b from here, yet nor can i see where i have gone wrong, as when i tried using t(asin2t + bcos2t) as the trial solution then i ended up with a=b=0
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  2. #2
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    Quote Originally Posted by macabre View Post
    Ok, so i have the eq

    d^2x/dt^2 + 4x = sin2t
    Begin by recognized this is a second order linear differencial equation with constant coefficients.
    Solve, the homogenous equations,
    y''+4t=0
    The characheteris equation is,
    k^2+4=0
    Complex solutions,
    k=+/- 2i
    Thus the general solution is,
    y=C*cos(2t)+K*sin(2t)
    Here is the trick, you cannot use the method of undetermined coefficients and look for a particular solution of the form,
    y=A*sin(2t)+B*cos(2t)
    For that is included in the general solution, thus, multiply by 't' and the theory gaurenttes a solution of the form,
    y=t[A*sin(2t)+B*cos(2t)]
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