Results 1 to 6 of 6

Math Help - Reduction of Order

  1. #1
    Newbie
    Joined
    Mar 2011
    Posts
    4

    Reduction of Order

    Hello,

    I have the equation t^2y''-2y=0, where t>0, and I know that
    y_1(t)=t^2.

    I would then like to use reduction of order to find some v(t), but I am getting a bit stuck, so I will just show my workings and hopefully someone will pick up on what I am doing wrong, as I remain with not only the derivatives of v(t) but v(t) itself, and am unsure of how to remove it so I can proceed.

    v(t) is the function I'd like to find with my current known solution, y_1(t)=t^2. I then found the derivatives of all the y_2's, plugged in, but I could not proceed as I still had the v(t) term whereas I'm supposed to be only left with its derivatives.

    y_2(t)=t^2v, y_2'(t)=2tv+t^2v'', y_2''(t)=2vt'+2tv'+2tv'+t^2v''

    Plugging in: t^2(t^2v''+2vt'+4tv')-2(t^2v)=0<br />
=> t^4v''+2t^2vt'+4t^3v'-2t^2v=0
    And this leads me to believe that the terms 2t^2vt' and -2t^2v are the ones that are 'supposed' to cancel out and that I'm adding one too many t' to the equation, but I don't really see how that's wrong 'cause it's what I get when I repeat it, as well.

    Hopefully someone will spot it, thanks!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor chisigma's Avatar
    Joined
    Mar 2009
    From
    near Piacenza (Italy)
    Posts
    2,162
    Thanks
    5
    The DE is of the 'Euler's type' and its solutions are of the form y= t^{\alpha}... imposing that You find a second order algebraic equation in \alpha one solution of which is \alpha=2... what is the other solution?...

    Kind regards

    \chi \sigma
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Mar 2011
    Posts
    4
    Hi!

    Thanks for the quick response, but unfortunately your hint escapes me, as the 'other solution' is exactly what I'm trying to find through reduction of order - which I can't seem to do correctly, hence my problem.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    5
    Awards
    2
    Quote Originally Posted by endreoc View Post
    Hello,

    I have the equation t^2y''-2y=0, where t>0, and I know that
    y_1(t)=t^2.

    I would then like to use reduction of order to find some v(t), but I am getting a bit stuck, so I will just show my workings and hopefully someone will pick up on what I am doing wrong, as I remain with not only the derivatives of v(t) but v(t) itself, and am unsure of how to remove it so I can proceed.

    v(t) is the function I'd like to find with my current known solution, y_1(t)=t^2. I then found the derivatives of all the y_2's, plugged in, but I could not proceed as I still had the v(t) term whereas I'm supposed to be only left with its derivatives.

    y_2(t)=t^2v, y_2'(t)=2tv+t^2v'', y_2''(t)=2vt'+2tv'+2tv'+t^2v''

    Plugging in: t^2(t^2v''+2vt'+4tv')-2(t^2v)=0<br />
=> t^4v''+2t^2vt'+4t^3v'-2t^2v=0
    And this leads me to believe that the terms 2t^2vt' and -2t^2v are the ones that are 'supposed' to cancel out and that I'm adding one too many t' to the equation, but I don't really see how that's wrong 'cause it's what I get when I repeat it, as well.

    Hopefully someone will spot it, thanks!
    I get

    y_{2}(t)=t^{2}v\implies y_{2}'(t)=2tv+t^{2}v'\implies<br />
y_{2}''(t)=2(v+tv')+2tv'+t^{2}v''

    =2v+4tv'+t^{2}v''.

    You'll notice a few differences between this result and yours. In particular, there is no t', because I'm taking the derivative with respect to t.

    Plugging into the DE yields

    t^{2}(2v+4tv'+t^{2}v'')-2t^{2}v=0\implies 4t^{3}v'+t^{4}v''=0.

    And now you can reduce the order, cancel some t's, and proceed on your merry way. Alternatively, you could notice that, right now, the LHS of the above equation is a perfect derivative, allowing you to simply integrate almost immediately. That is, you have

    (t^{4}v')'=0.

    Take your pick.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Mar 2011
    Posts
    4
    Yes, Ackbeet, thank you, this was the mistake I was looking for.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    5
    Awards
    2
    You're welcome!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. 2nd Order, Homog., Reduction of Order
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: November 27th 2011, 07:36 AM
  2. [SOLVED] Reduction of Order
    Posted in the Differential Equations Forum
    Replies: 4
    Last Post: October 17th 2010, 10:07 PM
  3. Reduction to first order
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: September 30th 2009, 06:45 PM
  4. Replies: 4
    Last Post: August 12th 2008, 05:46 AM
  5. reduction of order?
    Posted in the Calculus Forum
    Replies: 1
    Last Post: September 2nd 2006, 12:56 AM

Search Tags


/mathhelpforum @mathhelpforum