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I have the differential equation:

$$ (1-x)y'+y-x-1=0 $$

I should find solution in the form of power series:

$$ y(x)=\sum_{n=0}^\infty a_nx^n \implies y'(x) = \sum_{n=1}^\infty a_nnx^{n-1} $$

I substituted (1) to (2) and got:

$$ \sum_{n=1}^\infty a_nnx^{n-1}-x\sum_{n=1}^\infty a_nnx^{n-1}+\sum_{n=0}^\infty a_nx^n-x-1 = 0 $$

Then I made a transformation:

$$ \sum_{n=0}^\infty a_{n+1}(n+1)x^n-\sum_{n=0}^\infty a_{n+1}(n+1)x^{n+1}+\sum_{n=0}^\infty a_nx^n-x-1 = 0 $$

Then I grouped components by $x^n$ and $x^{n+1}$

$$ \sum_{n=0}^\infty x^n (a_{n+1}(n+1)) +a_n)-\sum_{n=0}^\infty a_{n+1}(n+1)x^{n+1}-x-1 = 0 $$

I got coefficient equations with different powers of $x$:

$$ x^0: a_1+a_0-1=0. $$

Let $a_0=0,a_1=1$:

$$ x^1: a_2*(1+1)+a_1-a_1*1-1=0 \implies a_2=\frac{1}{2} $$

$$ x^2: a_3*(2+1)+a_2-a_2*2 = a_3 \implies a_3=\frac{a_2}{3}=\frac{1}{2*3} $$

$$ x^3: a_4*(3+1)+a_3-a_3*(2+1)=0 \implies a_4=\frac{2a_3}{4}=\frac{2}{2*3*4} $$

$$ x^4: a_5*(4+1)+a_4-a_4*(3+1)=0 \implies a_5=\frac{3a_4}{5}=\frac{2*3}{2*3*4*5} $$

$$ x^n: a_{n+1}(n+1)+a_n-a_nn=0 \implies a_{n+1}=\frac{a_n(n-1)}{n+1} $$

So I got coefficient $a_n$:

$$ a_n=\frac{a_1(n-2)!}{n!}=\frac{a_1}{n(n-1)} $$

I got stuck at this moment so could you please suggest where to go from there? Where can I substitute $a_n$?

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  • $\begingroup$ @LutzLehmann That was a typo. Now fixed $\endgroup$
    – FoRRestDp
    Commented Oct 7, 2022 at 16:47
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    $\begingroup$ @AnneBauval Yes I do $\endgroup$
    – FoRRestDp
    Commented Oct 7, 2022 at 16:48
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    $\begingroup$ @AnneBauval I cannot find a way to do it. Could you suggest a solution? $\endgroup$
    – FoRRestDp
    Commented Oct 7, 2022 at 17:31
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    $\begingroup$ Now it looks correct, only that the second term in the derivative could be shortly written as $-\sum_{n=0}^\infty na_nx^n$, which simplifies the whole equation to $\sum_{n=0}^\infty ((n+1)a_{n+1}-(n-1)a_n)x^n=x+1$, so $a_0+a_1=1=2a_2$ and then $n(n+1)a_{n+1}=(n-1)na_n=...=2a_2=1$ $\endgroup$ Commented Oct 7, 2022 at 19:20
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    $\begingroup$ A direct solution would start with setting $y=(x-1)u$ so that $1+x=(1-x)[(x-1)u'+u)+(x-1)u=-(x-1)^2u'$. Thus $u'=-\frac1{x-1}-\frac2{(x-1)^2}$, $u=c-\ln(|x-1|)+\frac2{x-1}$, $y=c(x-1)+2-(x-1)\ln(|x-1|)$. $\endgroup$ Commented Oct 7, 2022 at 19:25

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Choosing $a_0=0,a_1=1$, you have got so far the particular solution: $$y=x+\sum_{n\ge2}\frac{x^n}{n(n-1)}.$$ Since $$\frac1{n(n-1)}=\frac1{n-1}-\frac1n\quad\text{and}\quad\sum_{n\ge1}\frac{x^n}n=-\ln(1-x),$$ $$y=x-x\ln(1-x)-(-\ln(1-x)-x)=2x+(1-x)\ln(1-x).$$

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