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Finding a diff. eq.

  1. Sep 1, 2007 #1
    1. The problem statement, all variables and given/known data

    Find a differential eq. y''+ay'+by=0 with this basis: exp(-1-i)x , exp(-1+i)x

    2. The attempt at a solution

    I see from this that the solutions of the characteristic eq. is two complex conjugates, and that a^2-4b=-4.

    Substitution gives me the eq. -2i-a+ai+b=0, and combining these two gives me

    a=(1/2) v (1/2)-i

    Then I calcutale b when a=(1/2)

    b=(17/16)

    So that the eq. is

    y''+0.5y'+(17/16)y=0

    Looks quite nice, but when I transform the bases (exp(-x)*sinx and exp(-x)*cosx) and substitute, it proves not to be correct.

    What's my mistake?
     
  2. jcsd
  3. Sep 1, 2007 #2

    learningphysics

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    Can you explain how you got the above equation?
     
  4. Sep 1, 2007 #3
    The roots of the characteristic eq. are 0.5(-a (+/-) sqrt(a^2-4c))

    Since the two solutions are complex numbers, a^2-4c must be negative, more precisely -4, since the imaginary part of the complex solutions is (+/-) 1.
     
  5. Sep 1, 2007 #4

    learningphysics

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    I see... you must have made a mistake when you solved the two equations... I'd just take the sum and product of the roots... then you can write out the equation:

    s^2 - qs + p... where q is the sum of the roots and p is the product...

    then you get a = -q and b = p for the differential equation

    or you can just multiply out:

    (s - (-1 - i))(s - (-1 + i))
     
  6. Sep 1, 2007 #5
    Ah, so I use the superposition principle and substitute the sum of Y1 and Y2 into the original eq?
     
  7. Sep 1, 2007 #6

    learningphysics

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    No, I'm not using the superposition principle here. Since you know that the basis is exp(-1-i)x , exp(-1+i)x... you immediately know that the characteristic equation has roots (-1-i) and (-1+i).

    or another way to think about it... you know that the characterstic equation is:
    s^2 + as + b

    This means that any function of the form e^sx that satisfies the equation must have the property that s^2 + as + b =0... ie s=(-1-i) must satisfy this equation, and s = (-1+i) must satisfy this equation... ie these are the two roots of the equation... Once you know the roots of a degree-2 equation r1 and r2... then you immediately know the form of the equation... s^2 - (r1+r2)s + r1r2... which you can also get by (s-r1)(s-r2)

    You can also plug in -1-i into the equation and -1+i into the equation, then get 2 equations in 2 unknowns (a and b) and solve... but this is a little more roundabout way to do the problem I think...
     
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