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How did this second-order initial value problem do this?

  1. Mar 13, 2009 #1
    1. The problem statement, all variables and given/known data

    Find a second-order initial-value problem whose solution is sin at


    3. The attempt at a solution

    so i know that

    y' = a cos at
    y" = -a^2 sin at

    but the solution says that

    "Then y(t) = sin at is
    a solution to 2 y ''+ a y = 0 . Our initial conditions must be y(0) = 0, y '(0) = a ."

    how do i know that it's 2y'' + a y? how come it couldn't be something else like y''+y'+ y?

    thanks
     
  2. jcsd
  3. Mar 14, 2009 #2

    djeitnstine

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    Gold Member

    This is because y"+y=0 is the only DE that has the solution of the form Acos kt + Bsin kt, other DE's have a solution of the form [tex]Ae^{bt}+...[/tex] such as the one you posted. (y"+y'+y). This stems from the characteristic equation [tex]r^{2}+r=0[/tex] which has complex roots...
     
  4. Mar 14, 2009 #3

    gabbagabbahey

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    That can't be right; [itex]y(t)=\sin(at)[/itex] is not the solution to that DE, [itex]y(t)=\sqrt{2a}\sin(\sqrt{\frac{a}{2}}t)
    [/itex] is.

    Either you've made a typo/error, or the solution manual has a typo/error or some combination of the two.

    Well, as you've found, if [itex]y(t)=\sin(at)[/itex], then [itex]y'(t)=a\cos(at)[/itex] which is orthogonal to both y(t) and y''(t) (you can't add some non-zero multiple of cos(at) to some polynomial of sin(at) and get a constant or zero), so the DE cant contain a y'(t) term....
     
  5. Mar 14, 2009 #4

    HallsofIvy

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    Yes, if y= sin(at), then y'= -a cos(at) and y"= -a2sin(at).

    The simplest second order differential equation y= sin(at) satisfies is y"+ a2y= 0. I don't know where you got that "2".

    (I wouldn't be surprised if this were the answer to "problem 2"!)
     
  6. Mar 14, 2009 #5
    thanks!

    turns out there wasn't a "2" anywhere...
     
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