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Linear ODE BV problem

  1. Feb 29, 2012 #1
    1. The problem statement, all variables and given/known data
    I nearly have this problem solved

    x2d2y/dx2 + 3x*dy/dx + 5y = 8x

    y(1) = 2, y(exp(pi/4)) = 2sinh(pi/4)

    I've found the general solution, but I'm not sure how to get the answer with the boundary values


    2. Relevant equations



    3. The attempt at a solution

    My general solution is y(t) = C1 e-t cos(2t) + c2 e-t sin(-2t) + 2t
     
  2. jcsd
  3. Feb 29, 2012 #2
    I understand how to use BV's it just I'm getting weird answers
     
  4. Feb 29, 2012 #3
    C1 - C2 = 6
    C1 - C2 = 2sinh(pi/4) - 8epi/4
     
  5. Feb 29, 2012 #4

    Dick

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    It looks like you did a change of dependent variable like x=e^t to get a solution like that, right? And I'm not sure I agree with the ' + 2t' part. The initial conditions they are giving you are initial conditions for x, not for t. What values of t correspond to x=1 and x=exp(pi/4)?
     
  6. Feb 29, 2012 #5
    Yes, I used e^t. I'll try and work out new boundary values. I think you're right about the 2t part. Thanks
     
  7. Feb 29, 2012 #6
    t = 0 and t = pi/4.
     
  8. Feb 29, 2012 #7

    Dick

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    Sure, now if you correct the '+ 2t' part to the correct expression you actually should be able to work out nice values for C1 and C2. Remember what the definition of sinh is.
     
  9. Apr 10, 2012 #8
    My general solution is y(t) = C1 e-t cos(2t) + c2 e-t sin(-2t) + e^t, but I don't know how to chnage the boundary values from x to t. Having a slow moment.
     
  10. Apr 10, 2012 #9
    can anyone verify that the final solution is y(t) = (cos(pi/2) + 2 + exp(pi/2)) e-t sin(2t) + e-t cos(2t) + e^t please?
     
  11. Apr 10, 2012 #10

    Dick

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    Show how you got that.
     
  12. Apr 10, 2012 #11
    2 = c1*e^0*sin(0) + c2*e^0*cos(0) + e^0

    that works out nicely to
    C2 = 1

    For C1 it's a bit messier. I let 2Sinh(pi/4) = (e^(pi/2) -1)/ e^(pi/4)

    Therefore

    (e^(pi/2) -1)/ e^(pi/4) = C1*exp(-pi/4)*sin(pi/2) + exp(-pi/4)*cos(pi/2) + exp(-pi/4)
    multiplied by exp(pi/4)

    and found c1 to be Cos(pi/2) + 2 -exp(-pi/2)
     
  13. Apr 10, 2012 #12
    No need, I just realised a huge mistake I made and figure it out. Thanks anyway :)
     
  14. Apr 10, 2012 #13

    Dick

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    I'm taking it on trust you've got the right general solution there. But if so, shouldn't the last term be exp(pi/4) instead of exp(-pi/4)?
     
  15. Apr 10, 2012 #14
    Your dead right, it's been a very long night.
     
  16. Apr 10, 2012 #15
    Final solution is e^(-pi/4)*e^(-t)sin(2t) +e^(-t)cos2t + e^t
     
  17. Apr 10, 2012 #16

    Dick

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    Apparently it has been a long night. I just get C1=1 and C2=(-1), though you have been switching C1 and C2 around. And it might be good to express your final answer in terms of x rather than t.
     
  18. Apr 11, 2012 #17
    Just looked at it with fresh eyes.

    y(x) = 1/x * cos(2ln(x)) - 1/x * sin(2ln(x)) + x
     
  19. Apr 11, 2012 #18

    Dick

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    Looks ok. You can always try substituting into the original ODE if you've got any doubts.
     
  20. Apr 11, 2012 #19
    Brilliant, thanks again.
     
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