Results 1 to 6 of 6

Math Help - Picards Iterates

  1. #1
    Junior Member
    Joined
    May 2006
    Posts
    30

    Picards Iterates

    Needing a little help with this problem

    Calculate the picards equation y' - y = x^2 with the condition y(0) = -1

    First Step:
    (Find Y prime)

    y' = x^2 + y

    y(0) = -1

    then

    Y1(x) = -1 + Integral from {0 to X} (t^2 - 1)dt
    = -1 - x + t^3 / 3


    Dont know if im doing this correct and what to do next.

    Please Help
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    10,054
    Thanks
    368
    Awards
    1
    Quote Originally Posted by mathlg View Post
    Needing a little help with this problem

    Calculate the picards equation y' - y = x^2 with the condition y(0) = -1

    First Step:
    (Find Y prime)

    y' = x^2 + y

    y(0) = -1

    then

    Y1(x) = -1 + Integral from {0 to X} (t^2 - 1)dt
    = -1 - x + t^3 / 3


    Dont know if im doing this correct and what to do next.

    Please Help
    Where did the "t" come from?

    Let's start over.

    y' - y = x^2, y(0) = -1

    Solve the homogeneous equation:
    yh' - yh = 0

    The characteristic equation is:
    m - 1 = 0 ==> m = 1

    Thus
    yh(x) = Ae^{x}

    Now we need a particular solution:
    yp' - yp = x^2

    Try
    yp(x) = Bx^2 + Cx + D
    yp'(x) = 2Bx + C

    So
    [2Bx + C] - [Bx^2 + Cx + D] = x^2

    -Bx^2 + (2B - C)x + (C - D) = x^2

    Thus
    -B = 1
    2B - C = 0
    C - D = 0

    Or B = -1, C = -2, and D = -2

    So yp(x) = -x^2 - 2x - 2

    Thus the solution to
    y' - y = x^2
    is
    y(x) = yh(x) + yp(x) = Ae^{x} - x^2 - 2x - 2

    We know that y(0) = -1, so
    y(0) = Ae^{0} - (0)^2 - 2(0) - 2 = -1

    A - 2 = -1

    A = 1

    Thus
    y(x) = e^{x} - x^2 - 2x - 2
    fits the bill.

    -Dan
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    May 2006
    Posts
    30
    I think I understand how you got that. I think I messed up with the question. I forgot to write

    Calculate the first four Picard Iterates of the equation y' - y = x^2 with the condition y(0) = -1

    and it was given that y'(x) = x^2 +y and y(0)= -1

    I was going off memory.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    May 2006
    Posts
    30
    Not sure if im starting off the problem right.

    Y(1) = -1 + integral from {0, X} (x^2 - 1)
    = -1 + x^3 / 3 - 1x

    Not sure how to get to Y2????
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    May 2006
    Posts
    30
    Calculate the first four Picard Iterates of the equation y' - y = x^2 with the condition y(0) = -1

    and it was given that y'(x) = x^2 +y and y(0)= -1



    Y1(t)= -1+ x^3/3- x

    y'= x^2- 1+ x^3/3 - x.

    For Y2 do I Integrate again Y2(0)= -1 ???

    This correct??
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    May 2006
    Posts
    30

    Checking my answer

    Just want to see if my answer is correct now.


    Y1(x) = y(0) + int(0,x) t^2 + Y0(t) dt
    = -1 + int(0,x) t^2 - 1 dt
    = -1 + x^3/3 - x
    = x^3/3-x-1.

    Y2(x) = y(0) + int(0,x) t^2 + Y1(t) dt
    = -1 + int(0,x) (t^2 + t^3/3 - t - 1) dt
    = x^4/12 + x^3/3 - x^2/2 - x - 1 .


    Y3(x) = y(0) + int(0,x) t^2 + Y2(t) dt
    = -1 + int(0,x) (t^2 + t^4/12 + t^3/3 - t^2/2 - t - 1)dt
    = -1 + int(0,x) (t^4/12 + t^3/3 + t^2/2 - t - 1)dt
    = x^5/60 + x^4/12 + x^3/6 - x^2/2 - x - 1

    Y4(x) = y(0) + int(0,x) t^2 + Y3(t) dt
    = -1 + int(0,x) (t^2 + t^5/60 + t^4/12 + t^3/6 - t^2/2 - t - 1) dt
    = -1 + int(0,x) (t^5/60 + t^4/12 + t^3/6 - t^2/2 - t - 1)dt
    = x^6/360 + x^5/60 + x^4/12 + x^3/6 - x^2/2 - x - 1
    Follow Math Help Forum on Facebook and Google+

Search Tags


/mathhelpforum @mathhelpforum