Results 1 to 3 of 3

Math Help - Taylor polynomial of ODE

  1. #1
    Junior Member
    Joined
    May 2010
    Posts
    74

    Taylor polynomial of ODE

    dy/dt = [-pi*cos(pi*t)-2ty]/[1+t^2]

    Find the fourth degree polynomial centred about 0

    I know the taylor polynomial goes y(0) + hy'(0) + h^2/2y''(0) + h^3/6y'''(0) + h^4/24y''''(c)

    The problem I am having is not being able to find y'''(0) I know its meant to be h^3[pi(pi^2 + 6)]/6 But i just can't seem to be able to get that answer..Can anyone help out please?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,407
    Thanks
    1294
    The first thing you're going to need to do is solve the DE...

    \frac{dy}{dt} = \frac{-\pi\cos{(\pi t)} - 2ty}{1 + t^2}

    \frac{dy}{dt} = -\frac{\pi \cos{(\pi t)}}{1 + t^2} - \left(\frac{2t}{1 + t^2}\right)y

    \frac{dy}{dt} + \left(\frac{2t}{1 + t^2}\right)y = -\frac{\pi \cos{(\pi t)}}{1 + t^2}.


    This is first order linear, so use the Integrating Factor method.

    The integrating factor is e^{\int{\frac{2t}{1 + t^2}\,dt}} = e^{\ln{(1 + t^2)}} = 1 + t^2.


    Multiplying through by the integrating factor gives

    (1 + t^2)\frac{dy}{dt} + 2t\,y = -\pi \cos{(\pi t)}

    \frac{d}{dt}[(1 + t^2)y] = -\pi \cos{(\pi t)}

    (1 + t^2)y = \int{-\pi \cos{(\pi t)}\,dt}

    (1 + t^2)y = -\sin{(\pi t)} + C

    y = \frac{C - \sin{(\pi t)}}{1 + t^2}.


    Now you should be able to substitute the MacLaurin series for \sin{x}, replacing x with \pi t.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor chisigma's Avatar
    Joined
    Mar 2009
    From
    near Piacenza (Italy)
    Posts
    2,162
    Thanks
    5
    If we suppose that y(t) is analytic in t=0 we can write...

    \displaystyle y(t) = a_{0} + a_{1}\ t + a_{2}\ t^{2} + a_{3}\ t^{3} + a_{4}\ t^{4} + ... (1)

    ... and...

    \displaystyle y^{'} (t) = a_{1} + 2\ a_{2}\ t + 3\ a_{3}\ t^{2} + 4\ a_{4}\ t^{3} + ... (2)

    The DE becomes with a certain number of steps...

    y^{'} (t) = -\{\pi - 2\ a_{0}\ t + (\frac{\pi^{3}}{2}-2\ a_{1} + \pi)\ t^{2} + (2\ a_{0} - 2\ a_{2})\ t^{3} + ...\} \ (1-t^{2} + t^{4} - ...) (3)

    ... and now we are able from (2) and (3), setting a_{0} = y(0) , to derive...

    a_{1} = -\pi

    a_{2} = a_{0}

    3\ a_{3} = \frac{\pi^{3}}{3} -2\ a_{1} + \pi \rightarrow a_{3} = \pi + \frac{\pi^{3}}{9}

    4\ a_{4} = -2\ a_{2} +2\ a_{0}= 0  \rightarrow a_{4} = 0

    Kind regards

    \chi \sigma
    Last edited by chisigma; September 11th 2010 at 10:41 AM. Reason: more general solution obtained...
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Taylor Polynomial
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: January 23rd 2011, 11:26 AM
  2. Taylor polynomial...
    Posted in the Calculus Forum
    Replies: 1
    Last Post: December 4th 2009, 07:40 AM
  3. Help with a Taylor Polynomial
    Posted in the Calculus Forum
    Replies: 0
    Last Post: November 16th 2009, 03:24 AM
  4. Taylor Polynomial
    Posted in the Calculus Forum
    Replies: 1
    Last Post: April 21st 2008, 10:41 AM
  5. Replies: 9
    Last Post: April 3rd 2008, 05:50 PM

Search Tags


/mathhelpforum @mathhelpforum