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Differential equation help

  1. Oct 6, 2008 #1
    d²y/ dx² = - h y / j

    where h, j are constants. What's y?
     
  2. jcsd
  3. Oct 6, 2008 #2
    This on its own is not actually a question in my assignment but it is a starting point I need to get the problem done.
     
  4. Oct 6, 2008 #3

    gabbagabbahey

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    Can you think of a function whose second derivative is proportional to it?
     
  5. Oct 6, 2008 #4
    I know y is an exponential function with a multiplication constant out the front, just not sure what the argument of the exponential is.
     
  6. Oct 6, 2008 #5
    Actually I know it's going to be the sum of two exponentials with different multipliation contants and one will have the negative argument of the other.
     
  7. Oct 6, 2008 #6
    In other words I know it's in the form

    y = Aexp(c) + B exp (-c), just not sure what the c is.
     
  8. Oct 6, 2008 #7

    gabbagabbahey

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    Since, the derivative is respect to x, why not try [itex]kx[/itex] where [itex]k[/itex] is a constant (it may be complex) for your argument? In fact, the general solution has two terms [itex]y(x)=Ae^{kx}+Be^{-kx}[/itex]. When you plug this into your DE, what do you get?
     
  9. Oct 6, 2008 #8
    and k = sqrt (- h/j)?

    so it is a complex argument as I was expecting because a complex exp can be written in terms of sines and cosines, whose 2nd derivative is their own negative.
     
  10. Oct 6, 2008 #9

    gabbagabbahey

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    Yes, exactly so you may as well write [itex]y(x)=Csin(\frac{h}{j}x)+Dcos(\frac{h}{j}x)[/itex]
     
  11. Oct 6, 2008 #10
    Thank you, gabbagabbahey.
     
  12. Oct 6, 2008 #11
    This is a ay''+by'+cy=0 problem which has been discussed to death on every DE book. One should be able to write down the results (when b^2-4ac>0,<0 and = 0) while sleeping.
     
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