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Find the particalur equation for the given initial conditions

  1. May 7, 2007 #1
    Find the particalur equation for the given initial conditions:-

    1/2y'' + y' + 13y = 0 ; y(0) = 5, y'(0) = 0

    The only method I know how to solve these doesn't seem to work. Any help is much appreciated.
     
  2. jcsd
  3. May 7, 2007 #2
    What method did you try? You know that its homogeneous. What is its characteristic equation?
     
    Last edited: May 7, 2007
  4. May 7, 2007 #3
    Sorry I think I've worked out how to do it. I use Euler's formula?
     
  5. May 7, 2007 #4
    The answer I got was y(x) = e-x(5cosx +5sinx)

    Is that correct?
     
  6. May 7, 2007 #5
    No, you should get that if through your calculation of the characteristic equation the roots are complex then it will be of the form
    [tex]roots = \alpha + i\beta[/tex]
    where
    [tex] \alpha = \frac{-b}{2a}[/tex]
    and
    [tex] \beta = \frac{\sqrt{\Delta}}{2a}[/tex]

    the general solution will be given by (after work with Euler's formula):
    [tex]y(x) = e^{\alpha t} (c_1 cos(\beta t) + c_2 sin(\beta t)[/tex]

    Use the initial conditions to determine c1 and c2.
     
  7. May 7, 2007 #6
    I think that's what I did...:frown:
     
  8. May 7, 2007 #7

    HallsofIvy

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    Do you have a problem with showing your work? In your first post you said you had solved the problem but didn't think your solution was correct- but you didn't show us how you solved it. You didn't even tell us your solution until you were specifically asked and then still wouldn't say how you got it. The problem is that there are several different errors you could have made. We don't know which one you actually made until we see your solution.
     
  9. May 7, 2007 #8
    I'm sorry. I was in rush and didn't know how to put the symbols etc I was using on here.
     
  10. May 7, 2007 #9
  11. May 8, 2007 #10

    HallsofIvy

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    What did you get as the solutions to the characteristic equation?

    Remember that if [itex]a\pm bi[/itex] satisfy the characteristic equation, then the general solution to the differential equation is
    [tex]e^{ax}(C_1cos(bx)+ C_2sin(bx))[/tex]
    In particular, what is the imaginary part of the solution to the characteristic equation?
     
  12. May 8, 2007 #11
    Thanks Mindscape - I will use symbols in future and I'm sorry HallsofIvy, I handed the work in this morning and no longer have it.
     
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