Results 1 to 10 of 10

Math Help - ODE - constant coefficients

  1. #1
    Junior Member
    Joined
    Dec 2008
    From
    Tel Aviv
    Posts
    29

    Unhappy ODE - constant coefficients

    Hello :-)
    We've learned how to solve second order ODE's with constant coefficients by guessing the exponential solution and plugging it into the equation.
    I've tried doing the same here, but it seems impossible to guess the roots of the characteristic polynomial. I've used some website to get the roots and they were pretty ugly (if it's correct), so there must be some other way.
    much help is appreciated with this (see attachment below).
    Attached Thumbnails Attached Thumbnails ODE - constant coefficients-eq.jpg  
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,864
    Thanks
    744
    Hello, zokomoko!

    I can get you started . . .


    \frac{d\;\!^6y}{dx^t} - 4\frac{d\;\!^5y}{dx^5} + 8\frac{d\;\!^4y}{dx^4} - 8\frac{d\;\!^3y}{dx^3} + 4\frac{d\;\!^2y}{dx^2} \:=\:0

    The characteristic equation is: . m^6 - 4m^5 + 8m^4 - 8m^3 + 4m \:=\:0

    . . which factors: . m^2(m^4 - 4m^3 + 8m^2  - 8m + 4) \:=\:0

    . . and factors further: . m^2(m^2-2m + 2)^2 \:=\:0

    The roots are: . 0,\;0,\;1\pm i,\;1\pm i


    Can you write the solution now?

    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    May 2006
    Posts
    244
    Quote Originally Posted by zokomoko View Post
    Hello :-)
    We've learned how to solve second order ODE's with constant coefficients by guessing the exponential solution and plugging it into the equation.
    I've tried doing the same here, but it seems impossible to guess the roots of the characteristic polynomial. I've used some website to get the roots and they were pretty ugly (if it's correct), so there must be some other way.
    much help is appreciated with this (see attachment below).
    The charateristic polynomial is:

    p(\lambda)=\lambda^6-4 \lambda^5+8 \lambda^4-8\lambda^3+4\lambda^2

    which factors:

    p(\lambda)=\lambda^2(\lambda-1+i)^2(\lambda-1-i)^2

    from which you should be able to proceed.

    .
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Dec 2008
    From
    Tel Aviv
    Posts
    29

    Smile :-)

    thank you :-)
    I've one more questions though, how did you factor it ? I've tried various methods and none worked.. how can you tell that it factors into that ?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by zokomoko View Post
    thank you :-)
    I've one more questions though, how did you factor it ? I've tried various methods and none worked.. how can you tell that it factors into that ?
    = \lambda^2 (\lambda^4 - 4 \lambda^3 + 8 \lambda^2 - 8 \lambda + 4)

    Now note that

    \lambda^4 - 4 \lambda^3 + 8 \lambda^2 - 8 \lambda + 4

    = (\lambda^4 + 8 \lambda^2 + 4) - 4 \lambda^3 - 8 \lambda

    = (\lambda^2 + 2)^2 + 4 \lambda^2 - 4 \lambda (\lambda^2 + 2)

    = (\lambda^2 + 2)^2 + 4 \lambda^2 - 2 \lambda (\lambda^2 + 2) - 2 \lambda (\lambda^2 + 2)

    = (\lambda^2 + 2)(\lambda^2 + 2 - 2 \lambda) + 4 \lambda^2 - 2 \lambda (\lambda^2 + 2)

    = (\lambda^2 + 2)(\lambda^2 + 2 - 2 \lambda) - 2 \lambda (-2 \lambda + \lambda^2 + 2)

    = (\lambda^2 + 2 - 2 \lambda)(- 2 \lambda + \lambda^2 + 2)

    = (\lambda^2 - 2 \lambda + 2)(\lambda^2 -  2 \lambda + 2)

    = (\lambda^2 - 2 \lambda + 2)^2.


    It's times like this that having access to a CAS is ....... helpful.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    May 2006
    Posts
    244
    Quote Originally Posted by zokomoko View Post
    thank you :-)
    I've one more questions though, how did you factor it ? I've tried various methods and none worked.. how can you tell that it factors into that ?
    The quartic obviously has no real roots, and so we know it factors into a pair of monic quadratics neither of which has real roots. Writing out the equations that the coefficients of the quadratics must satisfy quickly allows one to find a solution for the coefficients which gives the factorisation given in the earlier posts.

    .
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    Dec 2008
    From
    Tel Aviv
    Posts
    29

    Smile thanks

    thank you very much

    now you have brought it on yourself, I have another question ..
    consider the attached ODE, it has non-constant coefficients. We were asked to solve it using variation of parameters, but I couldn't find the solutions to the associated homogenous equation =/
    Attached Thumbnails Attached Thumbnails ODE - constant coefficients-ode.jpg  
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,387
    Thanks
    52
    Quote Originally Posted by zokomoko View Post
    thank you very much

    now you have brought it on yourself, I have another question ..
    consider the attached ODE, it has non-constant coefficients. We were asked to solve it using variation of parameters, but I couldn't find the solutions to the associated homogenous equation =/
    The homegenous is

    t^3 x''' + t^2 x'' - 2tx' + 2x = 0.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,387
    Thanks
    52
    Quote Originally Posted by danny arrigo View Post
    The homegenous is

    t^3 x''' + t^2 x'' - 2tx' + 2x = 0.
    Substitute the form  x = t^m gives the characteristic equation

    m(m-1)(m-2) + m(m-1) - 2m +2 = 0 which factors giving

    m = -1, 1, 2 The complimentary solution is therefore

    x = \frac{c_1}{t} + c_2 t + c_3 t^2
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Junior Member
    Joined
    Dec 2008
    From
    Tel Aviv
    Posts
    29
    thank you
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Second order ODE with non-constant coefficients
    Posted in the Differential Equations Forum
    Replies: 3
    Last Post: October 21st 2011, 08:39 PM
  2. Homogenous D.E. with Non-Constant Coefficients
    Posted in the Differential Equations Forum
    Replies: 2
    Last Post: October 12th 2011, 09:42 PM
  3. Particular Solution for Constant Coefficients ODE
    Posted in the Differential Equations Forum
    Replies: 3
    Last Post: February 26th 2011, 10:57 AM
  4. 2nd order DE's with constant coefficients.
    Posted in the Differential Equations Forum
    Replies: 6
    Last Post: January 9th 2011, 08:19 AM
  5. Need Help with second order DE with constant coefficients.
    Posted in the Differential Equations Forum
    Replies: 5
    Last Post: December 17th 2010, 08:25 AM

Search Tags


/mathhelpforum @mathhelpforum