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Ordinary (Or Partial) Differential Equation Unique Solution

  1. Sep 19, 2009 #1
    1. The problem statement, all variables and given/known data
    (d4y)/(dx4) = y. Find the Unique solution y = y(x).


    2. Relevant equations
    Boundary Conditions: y(0)=0, y'(0) = 2, y''(0) = 0 , y([tex]\pi[/tex]) = 0


    3. The attempt at a solution
    I really dont know where to start. I first started off with the guess that y = c1*sin(Ax) + c2*cos(bx). I got some discrepencies that say that c1 was not equal to c1. I also tried getting the characteristic equation, and then seeing if lambda was an eigenvalue, however I"m not completely sure about how to do this.
     
  2. jcsd
  3. Sep 19, 2009 #2

    rock.freak667

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    that is the same as


    [tex]\frac{d^4y}{dx^4}-y=0[/tex]


    constant coefficients, so try y=emx
     
  4. Sep 20, 2009 #3
    It doesnt work because y(0) is never going to be zero, unless the coeffcient in front of it is 0, this y=0. However, usually this is a trivial solution, but in this case that doesnt even work because y'(0) is not = 2
     
  5. Sep 20, 2009 #4

    Office_Shredder

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    Your equation is satisfied by four functions:

    ex, e-x, cos(x), sin(x)

    What other possible solutions can you construct?
     
  6. Sep 20, 2009 #5
    Isnt there a systematic way of solving these? or is it just going to be a bunch of guesses?
     
  7. Sep 20, 2009 #6
    So far, my only success at this problem has been when y = 2 sin(x). And i believe thats the end of it. I was just looking for some mathematical way to solve the system.
     
  8. Sep 20, 2009 #7

    Office_Shredder

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    Basically, a differential equation of degree n will have n "linearly independent" solutions that form a basis. What that means is that it has n functions that are basically distinct from each other, and every other solution can be constructed by summing multiples of those solutions. ex, e-x, cos(x), sin(x) are all different, so any solution to your differential equation must be of the form

    [tex] y=Ae^x + Be^{-x} + Ccos(x) + Dsin(x)[/tex] where A,B,C and D are arbitrary numbers

    If you use your four initial conditions, you get four equations and four unknowns.
     
  9. Sep 20, 2009 #8
    Ahh, thank you. I think I knew this, I suppose I just need some clarification. I do believe that y = 2sinx is the only thing that is left after applying all of the intial conditions. Thank you much. =)
     
  10. Sep 20, 2009 #9

    HallsofIvy

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    Actually trying what Office Shredder suggested, instead of complaining that you can't gives, with y= emx, m4emx- emx= 0 or, dividing through by emx which is never 0, m4- 1= 0 or m4= 1. That has four solutions. What are they?
     
  11. Sep 20, 2009 #10
    The four solutions are 1, -1 -i and i.
     
  12. Sep 20, 2009 #11

    rock.freak667

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    good so putting these together gives the equation Office_Shredder posted as

    [tex]y(x)=Ae^x + Be^{-x} + Ccos(x) + Dsin(x)[/tex]


    Now use y(0)=0, y'(0) = 2, y''(0) = 0 , y(π) = 0 to get the values of A,B,C and D
     
  13. Sep 20, 2009 #12
    There are 4 equations to solve and they are :
    y(0) = 0 ---> A +B + C = 0
    y(pi) = 0 ---> Ae^pi + Be^(-pi) - C = 0
    y'(0) = 2 - -> A - B + D = 2
    y''(0) = 0 ---> A + B-C = 0
     
  14. Sep 20, 2009 #13

    rock.freak667

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    yes now solve, it shouldn't be too difficult to do it
     
  15. Sep 20, 2009 #14
    After all of the mathematical shenanigans, I get A = B = C = 0 and D = 2. Therefore, y = 2sinx. and I never complained. lol
     
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