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Math Help - quick question, power series method.

  1. #1
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    quick question, power series method.

    y'(x) + x = y(x)

    im told y(x)= sum(C_n)(x^n) power series method

    i understand how to work the method im just confused about what happens to x?

    does it go to 0?does it already have regular a singular point?help,thanks.
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  2. #2
    MHF Contributor Calculus26's Avatar
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    See the attachment for the details but the x doesn't magically disappear

    but the x term is incorporated into determining the coefficient of the x term once the power series and its derivative are plugged back into the differential equation

    You'll notice c1-c0 = 0 considering the contstant term

    then 2c2-c1+1 = 0 in considering the linear term this is where x comes in

    There are no singularities in this DE
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  3. #3
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    This is just y"- y= -x. x= 0 is not a singular point at all!

    Let y= \sum_{n=0}^\infty C_n x^n. Then y"= \sum_{n=2}^\infty n(n-1)x^{n-2} so the equation becomes \sum_{n=2}^\infty n(n-1)x^{n-2}- \sum_{n= 0}^\infty C_n x^n= -x.

    In order to be able to compare "like powers", change the indices. In the first sum, let j= n-2 so that n= j+2. We have \sum_{j= 0}^\infty (j+2)(j+1)C_{j+2}x^j- \sum_{j=0}^\infty C_j x^j= \sum_{j=0}^\infty ((j+2)(j+1)C_{j+2}- C_j)x^j= -x= 0- 1(x)+ 0(x^2)+ ...

    Setting coefficients of like powers equal we have
    j= 0: 2C_2- C_0= 0

    j= 1: 3(2)C_3- C_1= -1
    (That's where the "-x" on the right hand side comes in!)

    j> 1: (j+2)(j+1)C_{j+2}- C_j= 0
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