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Second order diffy Q

  1. Oct 22, 2013 #1
    1. The problem statement, all variables and given/known data
    Hi,
    I think this is called second order.So I'm working on a problem I haven't had lecture yet. I'm trying to understand.
    Oh wait here is problem. Someone can help me understand this sort of problem in general.
    For what values of k does the function y = coskt satisfy the diffy Q 4y'' = -25y?
    a.) For those values of k, verify that every member of the family of functions y = Asinkt + Bcoskt is also a solution.


    2. Relevant equations



    3. The attempt at a solution



    So, Here is all I know. Which isn't much. I'm used to just doing the first order? The ones that you just mess with then integrate and wahla. So how exactly do I do this and how do I generalize it. Because the book said every second order of ( they give a form) can be represented this way.

    Thanks,
     
  2. jcsd
  3. Oct 22, 2013 #2
    So far I found the second derivavite of y = cos(kt) which is -k^2(cos(kt)) I then put this back into my equation

    4(-k^2(cos(kt)) = -25cos(kt) Not sure why I did that but I did.
     
  4. Oct 22, 2013 #3

    UltrafastPED

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    Just drop in "cos kt" into the ODE and see what conditions have to be true ... in this case you will get a quadratic in k; solve the quadratic for k.
     
  5. Oct 22, 2013 #4

    UltrafastPED

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    Right ... now divide thru by cos(kt) leaving -4k^2 = -25 which is a very simple quadratic equation:
    k = +/- 5/2.
     
  6. Oct 22, 2013 #5
    OK followed. So for
    a.) For those values of k, verify that every member of the family of functions y = Asinkt + Bcoskt is also a solution.

    I would maybe just put in my k values. What is A and B just constants? So will I want to solve for. I really have 0 concept of what is going on here lol.
     
  7. Oct 22, 2013 #6

    UltrafastPED

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    If k=+5/2 is a solution (test it!), and k=-5/2 is a solution (test it!), then you have two solutions:
    cos(5/2t) and cos(-5/2t). You should also check for sin(5/2t), sin(-5/2t).

    but if u is a solution, and v is a solution, then so is Au + Bv ... 2nd order ODE has two independent solutions, and their sum (multiplied by arbitrary constants A,B) is the general solution for the homogeneous case.

    So the general process for an ODE of order n is to find n independent solutions, then multiply each by an arbitrary constant, and sum them all up ... that is the general solution ... you can then apply the boundary conditions provided which will determine the values of A,B, ... for that case.

    There is actually a theory which describes all of this which you will study if you take an advanced course in differential equations - but in the introduction the goal is to learn how to solve them.
     
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