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    Taylor Series Solutions of Differential Equations

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    Hi there, I have a question regarding Taylor Series ODE which can be located here http://nullspace8.blogspot.com/2011/10/14.html. can someone please take a look? Full working step by step solution in pdf or word please. If you think the bid is insufficient and you can do it, please respond with a counter bid. Thank you.

    © BrainMass Inc. brainmass.com December 24, 2021, 10:01 pm ad1c9bdddf
    https://brainmass.com/math/calculus-and-analysis/taylor-series-solutions-of-differential-equations-434107

    SOLUTION This solution is FREE courtesy of BrainMass!

    a) We have y' = x^2 y^2 with initial condition y(0) = 1 and we wish to compute the Taylor series of y(x) up to n=2. First we write

    y = c_0 + c_1 x + c_2 x^2 + O(x^3).

    Since y(0) = 1 we have c_0 = 1. Now

    y' = c_1 + 2 c_2 x + O(x^2).

    But this is also equal to

    x^2 y^2
    = x^2(c_0 + c_1 x + c_2 x^2 + O(x^3)
    = c_0 x^2 + c_1 x^3 + c_2 x^4 + O(x^5).

    Equating terms multiplying 1 and x, we see that c_1 = c_2 = 0, whence

    y = 1 + O(x^3).

    (b) We have y' = x^2 + y^2 with initial condition y(0) = 1 and we wish to compute the Taylor series of y(x) up to n=3. We write

    y(x) = c_0 + c_1 x + c_2 x^2 + c_3 x^3 + O(x^4).

    Since y(0) = 1, we have c_0 = 1. We also have

    y' = c_1 + 2 c_2 x + 3 c_3 x^2 + O(x^3).

    But this is also equal to

    x^2 + y^2
    = x^2 + (c_0 + c_1 x + c_2 x^2 + O(x^3))^2
    = x^2 + c_0^2 + 2 c_0 c_1 x + (2 c_0 c_2 + c_1^2) x^2 + O(x^3)
    = c_0^2 + 2 c_0 c_1 x + (1 + 2 c_0 c_2 + c_1^2) x^2 + O(x^3).

    Equating terms multiplying powers of x up to x^2, we find

    c_1 = c_0^2 = 1
    2 c_2 = 2 c_0 c_1 = 2 c_1 = 2, whence c_2 = 1
    3 c_3 = 1 + 2 c_0 c_2 + c_1^2
    = 1 + 2 + 1 = 4, whence c_3 = 4/3.

    Thus we have

    y(x) = 1 + x + x^2 + 4/3 x^3 + O(x^4).

    (c) We have

    y' = sin(xy), where y(0) = 1

    and we wish to compute the Taylor series of y up to n=2.

    We have

    y(x) = c_0 + c_1 x + c_2 x^2 + O(x^3).

    From the initial condition we have c_0 = 1. We also have

    y' = c_1 + 2 c_2 x + O(x^2)

    = sin(xy) = xy + O(x^3)
    = c_0 x + O(x^2).

    Thus we have c_1 = 0 and 2 c_2 = c_0 = 1, whence c_2 = 1/2, so

    y(x) = 1 + 1/2 x^2 + O(x^3).

    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    © BrainMass Inc. brainmass.com December 24, 2021, 10:01 pm ad1c9bdddf>
    https://brainmass.com/math/calculus-and-analysis/taylor-series-solutions-of-differential-equations-434107

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