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Thread: solution of a differential equation

  1. #1
    Senior Member
    Joined
    Nov 2008
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    solution of a differential equation

    Hi!



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

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

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

    Convert it to a system of first order

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

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

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


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


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

    $\displaystyle y' = f(t,y), y(0) = y_0$ then

    $\displaystyle y^{(0)}(t) = y_0$

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

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

    $\displaystyle 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 $\displaystyle y' = -k^2u_1(t) $ but i dont find the series
    $\displaystyle 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

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    Quote Originally Posted by Rapha View Post
    Hi!



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

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

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

    Convert it to a system of first order

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

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

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


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


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

    $\displaystyle y' = f(t,y), y(0) = y_0$ then

    $\displaystyle y^{(0)}(t) = y_0$

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

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

    $\displaystyle 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 $\displaystyle y' = -k^2u_1(t) $ but i dont find the series
    $\displaystyle 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, $\displaystyle y^{(n+1)}(t) = y_0 + \int^t_0 f(x, y^{(n)}(x)) dx$?

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

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

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

    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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