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Find the roots

  1. Jul 8, 2006 #1
    y''' - 5y'' +7y' - 3y=o

    Can anybody pls teach me how to find the roots of this equation with the steps to compute it

  2. jcsd
  3. Jul 8, 2006 #2


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    Note that 1+7=5+3. :smile:
  4. Jul 8, 2006 #3
    the answer for the roots are 1 ,1 ,3
  5. Jul 8, 2006 #4


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    If you say so. :smile:
  6. Jul 8, 2006 #5
    erm are those primes indicating the power of the y term? the last term being y^0?

  7. Jul 8, 2006 #6


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    No, usually they denote derivatives (y''' is third derivative of y with respect to the independent variable)
  8. Jul 8, 2006 #7
    Solving cubic equations can be long, so a full analysis of the method is tedious. Here are the first steps :
    1. write your equation this way : y^3 - 5y^2 +7y - 3 = 0
    2. get rid of the y^2 term using the binomial thm
    3. solve y^3 + qy + r =0 ( where q and r are real) using this method :

    let y = x + z. Hence x^3 + z^3 + (3xz+q)y + r = 0

    If we further suppose that : 3xz + q = 0 we obtain :

    x^3 + z^3 = -r and (z^3)(x^3) = (q^3)/(-27)

    hence, x^3 and z^3 are the roots of the quadratic t^2 + rt - (q^3)/27 = 0

    solving this equation and putting :

    x^3 = -r/2 + Sqrt((r^2)/4 + (q^3)/27)

    z^3 = -r/2 - Sqrt((r^2)/4 + (q^3)/27)

    Hence we obtain from the relation y = x + z

    x= (-r/2 + Sqrt((r^2)/4 + (q^3)/27))^(1/3) +
    (-r/2 - Sqrt((r^2)/4 + (q^3)/27))^(1/3)

    This solution is known as Cardan's solution but he got it from Tartaglia even if the solution of the cubic seems to have been due originally to Ferreo.

    I hope this helps, but if I were you, I would try factorising (y-1), (y-2), (y-3) from the polynome by an elementary division before going through this harsh method. Maybe an easier one exists, but not that I know of.

    note : enven if this solution appears to have NINE solutions, it is not the case for since the cube roots are to be taken in pairs. A full explanation would require a lot of time and is quite subtle.

    NOTE : arildno has a good point .....
  9. Jul 9, 2006 #8


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    In case you didn't figure out how those roots were found (if you just had the answer key say) and you don't see right off arildno's hint or other roots immediately, you should first forget about using the solution to the general cubic for this problem- it's a last resort and won't be needed here. Start with the oft forgotten rational root theorem:


    it narrows down the possibilites for rational roots. Here, since the lead coefficient is 1, the only candidates to try for rational roots are 1,-1,3, and -3. If you tried them all, you'd see 1 and 3 are roots. Dividing out by the factors x-3 and x-1 will leave x-1 and the 3rd root is 1 (or notice that since the sum of your roots is 5 by looking at the quadratic term, then the third root be another 1, and you get to avoid division).

    They are derivatives as arildno has said. Solving this kind of differential equation can be reduced to solving a polynomial, the roots of this polynomial are what we're after here.
  10. Jul 9, 2006 #9
    Wow, thanks a lot shmoe !
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