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Math Help - Higher Order Homogeneous Constant Coeefecient

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    Higher Order Homogeneous Constant Coeefecient

    y^(5) - 3y ^ (4) + 3y''' - 3y'' + 2y' = 0

    I know the solution is : y = c1 + c2* e^x + c3*e^2x + c4 cos x + c5 sin x

    I am able to get the c1 solution. After that I am confused as to how the other solutions come about.
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  2. #2
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    Quote Originally Posted by naikn View Post
    y^(5) - 3y ^ (4) + 3y''' - 3y'' + 2y' = 0

    I know the solution is : y = c1 + c2* e^x + c3*e^2x + c4 cos x + c5 sin x

    I am able to get the c1 solution. After that I am confused as to how the other solutions come about.
    Using the ansantz y=e^{mt} gives

    m^5-3m^4+3m^3-3m^2+2m=0

    Now we just need to factor the above polynomial

    m^5-3m^4+3m^3-3m^2+2m=m(m^4-3m^3+3m^2-3m+2)

    By the rational roots theorem if there are other rational roots they must be either \pm1 \text{ or } \pm 2

    now by either long or synthetic division you will find that 1 and 2 are both roots giving the factorization

    m(m-1)(m-2)(m^2+1)

    Putting all of this back in gives the solution

    y=c_1+c_2e^{t}+c_3e^{2t}+c_4\cos(t)+c_5\sin(t)
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