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Posted by on Saturday, July 27, 2013 at 4:33pm.

Solve the DE by using the Laplace transform:

x''+x=cos(3t), x(0)=x'(0)=0

  • calculus - , Saturday, July 27, 2013 at 6:03pm

    L{x"} = s^2*F(s) - s*f(0) - f'(0)
    = s^2*F(s)

    so,

    s^2 F(s) + F(s) = s/(s^2+9)

    F(s) = s/((s^2+9)(s^2+1))
    = 1/8 (s/(s^2+1) - s/(s^2+9))

    f(t) = 1/8 (cos(t) - cos(3t))

    check:
    x' = 1/8 (3sin(3t)-sin(t))
    x" = 1/8 (9cos(3t)-cos(t))

    x'+x = 1/8 (8cos(3t)) = cos(3t)

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