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Math Help - Second order DE help

  1. #1
    Member Jones's Avatar
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    Second order DE help

    Hi,

    I have the following

    y'' + 2y' +3y = 0

    So the auxilary equation would look like r^2 +2r +3 = 0

    The solution to this equation is on the form \frac{-2}{2}\pm \sqrt{\frac{4-12}{2}}

    which evaluates to  -1 \pm \sqrt{\frac{-8}{2}} \rightarrow \sqrt{-4}

     = -1\pm2i

    Which gives  Ae^{-t}cos(2t) + Be^{-t}sin(2t)

    but apparently thats incorrect, the answer should be -1\pm\sqrt{2i}

    whats wrong?

    Edit: Darn i just remebered there's a forum for DE's. I can't delete my post
    Forgive me
    Last edited by mr fantastic; January 1st 2010 at 05:06 PM. Reason: I moved it - it's taken care of.
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  2. #2
    MHF Contributor
    Jester's Avatar
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    Quote Originally Posted by Jones View Post
    Hi,

    I have the following

    y'' + 2y' +3y = 0

    So the auxilary equation would look like r^2 +2r +3 = 0

    The solution to this equation is on the form \frac{-2}{2}\pm \sqrt{\frac{4-12}{2}} (1)

    which evaluates to  -1 \pm \sqrt{\frac{-8}{2}} \rightarrow \sqrt{-4}

     = -1\pm2i

    Which gives  Ae^{-t}cos(2t) + Be^{-t}sin(2t)

    but apparently thats incorrect, the answer should be -1\pm\sqrt{2i}

    whats wrong?

    Edit: Damn i just remebered there's a forum for DE's. I can't delete my post
    Forgive me
    Look at step (1) in blue. Not sure why the 2 is in the square root.
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  3. #3
    Member Abu-Khalil's Avatar
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    r=\frac{-2\pm\sqrt{4-12}}{2}=\frac{-2\pm2\sqrt{2}i}{2}=-1\pm \sqrt{2}i

    y(t)=Ae^{-t}\cos\left(\sqrt{2}t\right)+Be^{-t}\sin\left(\sqrt{2}t\right)
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  4. #4
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    Dear Jones,

    I have attached the solution to your question. Hope it will help you to find where you went wrong. But I would greatly appreciate your reply in whether you were able to read my attachment clearly. If it is not clear or if you have any doubt regarding this question please don't hesitate to mail me.
    Attached Thumbnails Attached Thumbnails Second order DE help-dsc02495.jpg  
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