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Math Help - Taylor series with two variables?

  1. #1
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    Taylor series with two variables?

    I have this problem:

    y'= x+y , with y(0)=1

    I am asked to estimate y when x=0.4 using Taylor series to the 5th power.
    However, I am in a jam, because I don't understand how I am to do this while having 2 variables to differentiate.

    Can someone please show how to differentiate at least for the 1st step to get y''? All I know is that the Taylor series will start with "1+....", I think.

    Thank you,
    Jen
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  2. #2
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    Quote Originally Posted by Jenberl View Post
    I have this problem:

    y'= x+y , with y(0)=1

    I am asked to estimate y when x=0.4 using Taylor series to the 5th power.
    However, I am in a jam, because I don't understand how I am to do this while having 2 variables to differentiate.

    Can someone please show how to differentiate at least for the 1st step to get y''? All I know is that the Taylor series will start with "1+....", I think.

    Thank you,
    Jen
    Implicit differentiation

    \frac{d}{dx} \frac{dy}{dx} = \frac{d}{dx} (x+y)

     \frac{d^2y}{dx^2} = 1+\frac{dy}{dx}

     \frac{d^2y}{dx^2} = 1+x+y
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  3. #3
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    Quote Originally Posted by Mush View Post
    Implicit differentiation

    \frac{d}{dx} \frac{dy}{dx} = \frac{d}{dx} (x+y)

     \frac{d^2y}{dx^2} = 1+\frac{dy}{dx}

     \frac{d^2y}{dx^2} = 1+x+y

    Ok.

    So, does this mean that:

    y'= x+y , but will y'(0)= y, which is 1? since y(0)=1
    y''= 1+y', which is y"= 1+x+y
    then y'''= y'' ?, a.k.a. 1+x+y (also?) Is it supposed to keep looping to this?
    and y''''= 1+x+y also?

    Oh boy...I am really confused.
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  4. #4
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    Quote Originally Posted by Jenberl View Post
    I have this problem:

    y'= x+y , with y(0)=1

    I am asked to estimate y when x=0.4 using Taylor series to the 5th power.
    However, I am in a jam, because I don't understand how I am to do this while having 2 variables to differentiate.

    Can someone please show how to differentiate at least for the 1st step to get y''? All I know is that the Taylor series will start with "1+....", I think.

    Thank you,
    Jen
    I think a better idea is to solve it as a differential equation.

     \frac{dy}{dx} - y = x

     \phi(x) = e^{\int(-1)dx} = e^{-x}

     \phi(x)y = \int x\phi(x)dx

     e^{-x}y = \int xe^{-x}dx

     e^{-x}y = -xe^{-x}+\int e^{-x}dx

     e^{-x}y = -xe^{-x}-e^{-x}C

     y = -x-1+\frac{C}{e^{-x}}

     y = -x-1+{Ce^x}

    Apply your initial conditions to find C, and the taylor series should be easy from there.

    You should get  y = 2e^x -x-1

    Can you find the taylor series of that?
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  5. #5
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    Got it!!!
    Thank you!!!
    :-)
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