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Second order homogenous problem

  1. Dec 1, 2013 #1
    1. The problem statement, all variables and given/known dataSolve the initial value problem y''-y'-2y=0 y(0)=β , y'(0)=2. Then find β so that the solution approaches zero as t→∞



    2. Relevant equations
    R^2-R-2=0

    C1+C2=β, -C1+2C2=2



    3. The attempt at a solutionI solved the equation got the r- values 1 and -2 , then i solved the two equations to find the constants, my first constant is c1=(2β-2)/3 and my second one is c2=(β+2)/3. Now I've got the equations and I dont know how to continue from here, should I try it from the first derivative (solving for minimum value or anything) or from the y-equation( it never gets to zero)
     
    Last edited: Dec 1, 2013
  2. jcsd
  3. Dec 1, 2013 #2

    tiny-tim

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    hi bigu01! :smile:

    hint: if you want Cekt -> 0 as t -> ∞, with k > 0,

    then you need C = … ? :wink:
     
  4. Dec 1, 2013 #3

    HallsofIvy

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    Well, here's your first problem. r= 1 and r= -2 do NOT satisfy the equation:
    [itex](1)^2- 1- 2= -2[/itex], not 0, and [itex](-2)^2- (-2)- 2= 4[/itex], not 0.

     
  5. Dec 1, 2013 #4

    The R-s are inputed to the e^rt not directly in the equation right, so if you input the r=1 and r=-2 ,
    y becomes y=c1e^t+c2e^-2t
     
  6. Dec 1, 2013 #5

    Mark44

    Staff: Mentor

    No.

    Your characteristic equation is r2 - r - 2 = 0. The solutions are NOT r = 1 and r = -2. This is what HallsOfIvy is saying.

    Try again.
     
  7. Dec 1, 2013 #6
    Oh I see r=-1 and r=2 I solved it like that in my notebook,but when putting here looks like I've mixed them.Sorry.Then I have (β+2)/3*e^2t+(2β+2)*e^-t =y and I have to get y=0 as t->∞
     
  8. Dec 1, 2013 #7
    k=e^-t?
     
  9. Dec 1, 2013 #8

    tiny-tim

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    sorry, you're not making any sense :redface:
     
  10. Dec 1, 2013 #9
    β=-2 C should be 0 , β+2=0
     
  11. Dec 1, 2013 #10

    tiny-tim

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    that's better! :biggrin:
     
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