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Math Help - solution of a differential equation

  1. #1
    Senior Member
    Joined
    Nov 2008
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    461

    solution of a differential equation

    Hi!



    Find the Solution of y''(t) = -k^2 y(t)
    that satisfies y(0) = 1, y'(0) = 0, k \not= 0
    --------------

    the equation has the obvious solution y(t) = cos(kt)

    --------------

    Convert it to a system of first order

    --------------

    i substituted u = y' \Rightarrow u'(t) = y''(t)

    \Rightarrow  u'(t) = -k^2 y(t)
    and y(0) = 1, u(0) = 0, k \not= 0


    --------------


    y_1^{(n)}. y_2^{(n)} , n = 0,1,2,... are solutions found by this algorithm:

    y' = f(t,y), y(0) = y_0 then

    y^{(0)}(t) = y_0

    y^{(n+1)}(t) = y_0 + \int^t_0 f(x, y^{(n)}(x)) dx , n=0,1,2,...

    if y_1^{(0)}(t) = 1, then show that

    y_1^{(2n)} =  \sum^n_{i=0}(-1)^i \frac{(kt)^{2i}}{(2i)!}

    --------------

    i do not to have a clue. I tried to use the algorithm and y' = -k^2u_1(t) but i dont find the series
    y_1^{(2n)} =  \sum^n_{i=0}(-1)^i \frac{(kt)^{2i}}{(2i)!}


    any help would be much apprectiated

    Thank you!

    Rapha
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  2. #2
    MHF Contributor

    Joined
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    Quote Originally Posted by Rapha View Post
    Hi!



    Find the Solution of y''(t) = -k^2 y(t)
    that satisfies y(0) = 1, y'(0) = 0, k \not= 0
    --------------

    the equation has the obvious solution y(t) = cos(kt)

    --------------

    Convert it to a system of first order

    --------------

    i substituted u = y' \Rightarrow u'(t) = y''(t)

    \Rightarrow  u'(t) = -k^2 y(t)
    and y(0) = 1, u(0) = 0, k \not= 0


    --------------


    y_1^{(n)}. y_2^{(n)} , n = 0,1,2,... are solutions found by this algorithm:

    y' = f(t,y), y(0) = y_0 then

    y^{(0)}(t) = y_0

    y^{(n+1)}(t) = y_0 + \int^t_0 f(x, y^{(n)}(x)) dx , n=0,1,2,...

    if y_1^{(0)}(t) = 1, then show that

    y_1^{(2n)} =  \sum^n_{i=0}(-1)^i \frac{(kt)^{2i}}{(2i)!}

    --------------

    i do not to have a clue. I tried to use the algorithm and y' = -k^2u_1(t) but i dont find the series
    y_1^{(2n)} =  \sum^n_{i=0}(-1)^i \frac{(kt)^{2i}}{(2i)!}


    any help would be much apprectiated

    Thank you!

    Rapha
    What, exactly, is the question? You have the solution. Do you want to solve it by that specific method, y^{(n+1)}(t) = y_0 + \int^t_0 f(x, y^{(n)}(x)) dx?

    Here your "Y" is really the pair (y, u) so you will need to apply that to both terms. You are told that y_0= 1 and y'(0)= u(0)= 0 so Y (0)= (1, 0) and f(x,t)= (u, -k^2y).

    (y1, u1)= (1, 0)+ (\int_0^t  (0)dt, \int_0^t -k^2(1)dt)= (1, -k^2t)

    (y2, u2)= (1, 0)+ (\int_0^t (-k^2 t)dt, \int_0^t(-k^2)dt)= (1- k^2t, -k^2 t)

    etc. You should be able to see pretty quickly that the first, y, component is the first n terms of the Taylor's series for cos(kt).
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  3. #3
    Senior Member
    Joined
    Nov 2008
    Posts
    461
    Hello HallsofIvy,

    thank you very much, this really helps a lot.

    Rapha
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