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Y' '+ y = 0

  1. May 5, 2007 #1
    1. The problem statement, all variables and given/known data
    What are the different methods of solving for y?

    Using the series method is one way but I seem to be stuck to get to the correct form for y. My problem is shown in post 3.

    3. The attempt at a solution
    There is guessing that y=Asin(x)+Bcos(x)
    Using a series solution and finding sin and cos series solutions

    What else?
    Last edited: May 5, 2007
  2. jcsd
  3. May 5, 2007 #2
    hmmm... the possibilities are virtually endless.

    you can cast the problem into matrix form, x'=Ax, A is a matrix and the solution is x=e^A

    you can do Laplace transformation

    you can factor the differential operator, and solve for each of them.

    you can apply idea from Banach fixed point theorem and try a solution and then iteratively integrating it. (similar to a series solution)

    you can keep differentiating the equation and find y, y', y'', y'''... at a particular point then smack it together in a Taylor series fashion.

    you can find an integrating factor and integrate.

    you can have a system of ODE's, k=-y', y=k', and look at the complex number n=k+iy and solve a first order ODE.

    Last edited: May 5, 2007
  4. May 5, 2007 #3
    Ok. But if using the series method, I am having problems constructing the solution to y''+ay=0, a>0

    The solution should be y=Asin(a^.5x)+Bcos(a^.5x)

    In the recursion relation I get -aCn/(n+2)(n+1) = Cn+2

    However to construct sin, I need a to have fractional power which I cannot seem to get.

    i.e. for odd n, Cn should have a with a fractional power like n+.5 or a^(n+.5) but I can only see a^n in Cn for all n.
    Last edited: May 5, 2007
  5. May 5, 2007 #4
    Well, the general method for solving this is to assume y = e^mx, and, once complex roots a +/- bi are obtained for m, assume the solution y = A e^ax cos(bx) + B e^ax cos(bx).
    However, as tim_lou stated among others, the Laplace transform may prove helpful in this case-- especially if your initial conditions are given.
    I don't see any reason to go ahead and do a series solution.
  6. May 5, 2007 #5
    The series solution is slower but it should still work. I tried it and seemed to be stuck. Look at post 3.
  7. May 5, 2007 #6

    George Jones

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    Write each [itex]c_n[/itex] as


    multiplied by something.
  8. May 5, 2007 #7
    But ideally there should be something that forces me to write it in the way you suggested while only doing the series method. RIght now, the only reasoning I have writing it out that way is because I knew what the solution should be using another method before hand.
    Last edited: May 5, 2007
  9. May 5, 2007 #8


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    Is it the "a" that is causing the problem then? Are you able to convert the infinite series to sine and cosine with a= 0?
    You recursion relation is [itex]C_{n+2}= (-a/(n+2)(n+1)) a_n[/itex]. What do you get as the "closed form" for [itex]C_n[/itex]? Did you actually calculate the first three or four terms?

    For given [itex]C_0[/itex] and [itex]C_1[/itex] you should get [itex]C_2= (-a/2)C_0[/itex], [itex]C_3= -(a/6)C_1[/itex], [itex]C_4= -(a/12)C_2= (a^2/24)C_0[/itex], [itex]C_5= -(a/20)C_3= -(a^2/120) C_5.

    You should be able to recognize the denominator of [itex]C_n[/itex] as n! but then you are alternating between [itex]C_0[/itex] and [itex]C_1[/itex]. Do you notice that you will always get [itex]C_0[/itex] in the terms with x to an even power and [itex]C_1[/itex] in the terms with x to an odd power? It would make sense to separate the two so that you have one series with even powers only, multiplied by [itex]C_0[/itex] and one series with odd powers only, multiplied by [itex]C_1[/itex], wouldn't it? Now compare the powers of a and x in each of those terms.
  10. May 5, 2007 #9

    I see what you mean. I did separate the series into even and odd powers of x. For both x^(2n) and x^(2n+1), the coefficient is always a^n. With constant C0 and C1 respectively in the two separated series. I guess you could let C1=B/a^.5, B arbitary so that there is an extra a^.5 in the numerator of the odd powers of x hence a^(n+.5)x^(2n+1) = (a^.5 * x)^(2n+1) which is what is recquired to let sin(a^(.5)x) represent the series with odd powers of x. Which is what George Jones suggested.
  11. May 5, 2007 #10
    lol...:rofl: I completely misunderstood your question. I thought it was like a game where we try to figure out as many ways as possible to solve that linear ODE...
  12. May 5, 2007 #11
    It was originally meant to be something like that but when I actually tried to solve y''+ay=0 using the series method I got some difficulties. Things didn't turn out to be as smooth as expected. I had to 'shift a constant' or do a change of constants in the coefficients of odd powers of x in order to get it to match a sin function.
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