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Taylors series for simple ODE

  1. Jul 16, 2012 #1
    1. The problem statement, all variables and given/known data
    Find the Taylor expansion y(x) satisfying: y'(x) = 1 - xy

    2. Relevant equations


    3. The attempt at a solution

    So I need expressions for y''(x), y'''(x), ...etc

    I can find y''(x)=-y-xy' by differentiating implicitly.
    By setting y'(x)=z, then dz/dx = δz/δx+(δz/δy)(dy/dx)

    However when I use the same method to try and find y'''(x) I have problems:
    - partial differential of y'(x) ?
    - is the (dy/dx) from the implicit formula still dy/dx or the second differential?

    [Ans: -2y'-xy'']

    Please tell me the technique, or give me a reference, that is used to find y'''(x) in this example?

    Thanks!
     
  2. jcsd
  3. Jul 16, 2012 #2

    Mute

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    Homework Helper

    I'm slightly confused as to what you are trying to do. If your goal is to find a Taylor series expansion for y(x), you should assume it has the form

    [tex]y(x) = \sum_{n=0}^\infty \frac{a_n}{n!}x^n,[/tex]
    where the factor of 1/n! is just conventional because it makes [itex]a_n = y^{(n)}(0)[/itex].

    You then plug that into the differential equation, and since everything has to be zero you can work out a recurrence relation for an, which you can then try to solve.

    Your approach seems to be to try to compute the [itex]y^{(n)}(0)[/itex] by repeatedly differentiating the differential equation. You could do it this way, but it won't get you the full recurrence relation for the derivatives at x = 0.

    To do it your way, you don't need any implicit differentiation or even the chain rule, just the product rule. [itex]y'(x)[/itex] is just [itex]dy/dx[/itex], so if you differentiate your differential equation, you just have

    [tex]\frac{d^2 y}{dx^2} = -x \frac{dy}{dx} - y.[/tex]

    You could then set x = 0 to get y"(0) = -y(0) (i.e., a2 = -a0).

    To get the third derivative you would just differentiate this equation directly again (with x arbitrary! Don't try to differentiate the equation with x = 0), since d(y''(x))/dx = y'''(x), and so on.
     
  4. Jul 16, 2012 #3

    LCKurtz

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    Putting z in there just confuses things
    $$y'=1-xy$$ $$
    y'' =-y -xy' = -y-x(1-xy)=-y-x+x^2y$$ $$
    y'''=-y'-1+2xy+x^2y' = -(1-xy)-1+2xy- x^2(1-xy)=-1+xy-1+2xy-x^2+x^3y=-2+3xy-x^2+x^3y$$
     
  5. Jul 16, 2012 #4
    Yes thanks, quite straightforward really.
    Just reading my notes from a few year's ago.

    thanks
     
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