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Math Help - Initial Value Problem

  1. #1
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    Initial Value Problem

    y'''' + 2y'' + y = 3t + 4,
    y(0)= y'(0) = 0,
    y''(0) = y'''(0)=1
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  2. #2
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    Quote Originally Posted by justin6mathhelp View Post
    y'''' + 2y'' + y = 3t + 4,
    y(0)= y'(0) = 0,
    y''(0) = y'''(0)=1
    The charachteristic equation is,
    k^4 + 2k^2 + 1 = 0 \implies (k^2 + 1)^2 = 0 \implies (k-i)^2(k+i)^2=0
    Because the roots have multiplicity it means the general solution is,
    y = c_1\sin (t) + c_2 \cos (t) + c_3 t\sin (t) + c_4 t\cos (t)
    However, this only solves the equation y'''' + 2y'' + y = 0.

    To finish the solution look for a particular solution which happens to be (for example) 3t+4.

    Thus, the full solution is, y = c_1\sin (t) + c_2 \cos (t) + c_3 t\sin (t) + c_4 t\cos (t) + 3t+4
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