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Math Help - xy''-2y'+3y=0 obtain a power series sol'n

  1. #1
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    xy''-2y'+3y=0 obtain a power series sol'n

    Obtain a power series solution of the equation

    xy'' - 2y' + 3y = 0.

    I'm not sure where to go with this.

    So i just used y= (n=0)SUM(infinity) cnx^(n+r), n got y' and y''.

    subbed it into the equation. and simplified to get:

    r(r-3)=0 and C[n+1] = 3Cn/(n+r-2)(n+r+1) = 0
    r=0 , r=3

    is this Frobenius? with difference of positive integer (3)?
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  2. #2
    MHF Contributor chisigma's Avatar
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    Let's write the differential equation as...

    y=\frac{2}{3}\cdot y^{'} - \frac{1}{3}\cdot x\cdot y^{''} (1)

    Now we suppose that (1) has a solution y(*) that can be expressed as power sum, so that is...

    y(x)= \sum_{n=0}^{\infty} a_{n}\cdot x^{n} (2)

    In this case the identity (1) becomes [the intermediate computations are not reported...] the following...

    a_{0} + \sum_{n=1}^{\infty} a_{n}\cdot x^{n} = \sum_{n=1}^{\infty} (n-\frac{n^{2}}{3})\cdot a_{n}\cdot x^{n-1} (3)

    Now we try to compute the a_{n} from (3) imposing that the coefficients ot the terms x^{n} are identical in both terms. Any solution of (1) is defined unless an arbitrary constant, so that we are allowed to set...

    a_{0}=1

    Equating in (3) the coefficients of the term x^{0} we obtain...

    a_{0}= \frac{2}{3}\cdot a_{1} \rightarrow a_{1}= \frac{3}{2}\cdot a_{0}= \frac{3}{2}

    ... all right!... equating the the coefficients of the term x^{1} we obtain...


    a_{1}= \frac{2}{3}\cdot a_{2} \rightarrow a_{2}= \frac{3}{2}\cdot a_{1}= \frac{9}{4}


    ... all right!... equating the the coefficients of the term x^{2} we obtain...

    a_{2}= 0 \cdot a_{3}

    ... ehmm!... there is some minor difficult in computing a_{3} ...

    ... the conclusion is that our hypothesis is false and doesn't exist any solution of (1) that can be expressed in power series like (2) ...

    Kind regards

    \chi \sigma
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