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I'm a bit stuck on how to solve this:

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

so assuming y is a solution in this form:

$$\sum_{n=0}^\infty C_nx^n$$

$$\sum_{n=1}^\infty nC_nx^{n-1}$$

$$\sum_{n=2}^\infty n(n-1)C_nx^{n-2}$$

subbing in and distributing: $$\sum_{n=2}^\infty n(n-1)C_nx^{n-1} - \sum_{n=2}^\infty n(n-1)C_nx^{n-2} + \sum_{n=1}^\infty nC_nx^{n-1}$$

$$\sum_{n=2}^\infty n(n-1)C_nx^{n-1} - \sum_{n=1}^\infty (n+1)(n)C_{n+1}x^{n-1} \sum_{n=1}^\infty nC_nx^{n-1}$$

start at n=2

$$\sum_{n=2}^\infty n(n-1)C_nx^{n-1} - \sum_{n=2}^\infty (n+1)(n)C_{n+1}x^{n-1} + \sum_{n=2}^\infty nC_nx^{n-1} - 2C_2 + C_1$$

so $-2C_2 + C_1 = 0$

and $$(n(n-1)C_n - (n+1)(n)C_{n + 1} + nC_n = 0$$ and $$(n^2C_n - (n+1)(n)C_{n + 1} = 0$$ and $c_{n+1} = \frac{nC_n}{n+1}$ for $n \ge 2$

so writing out the first few terms:

$c_0 = c_0$ and $c_1 = c_1$ and $c_2 = \frac{c_1}{2}$ and $c_3 = \frac{2c_2}{3}$ and $c_4 = \frac{3c_3}{4} = \frac{3 \cdot 2 \cdot c_1}{4!}$ and so $$c_n = \frac{c_1}{n}$$

so can I write that $$y = c_0 + c_1 + \sum_{n=2}^\infty \frac{x^n}{n}$$

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    $\begingroup$ Not putting this as answer, as it's probably not what you're asking, but: write $z = y'$. Then your equation is $ (x-1)z' + z = 0$. This can be written (product rule on LHS) as $$ {d\over dx}\Big( (x-1)z\Big ) = 0. $$ Hence $$z ={ A\over x-1},$$ and $y= A \ln (x-1) + B$, for $A$ and $B$ constants. $\endgroup$ – peter a g May 10 at 18:38
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    $\begingroup$ In your final answer why does $c_1$ not multiply an $x$? $\endgroup$ – Spencer May 10 at 23:18
  • $\begingroup$ Did you find either of the answers useful? Or is there still some additional clarification that is needed? $\endgroup$ – Spencer May 12 at 19:21
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You initially wrote that,

$$ y = \sum_{n=0}^\infty c_n x^n,$$

and then you determined that for $n \geq 1$ we have $c_n = c_1/n$.

$$ y = c_0 + \sum_{n=1}^\infty c_n x^n,$$

$$ y = c_0 + \sum_{n=1}^\infty \frac{c_1}{n} x^n,$$

$$ y = c_0 + c_1 \sum_{n=1}^\infty \frac{x^n}{n},$$

which is consistent with the answer given by @peter in the comments.

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Calling $Y = y' = \sum_{k=0}^{\infty}a_k x^k$ we have in $(x-1)Y'+Y = 0$

$$ (x-1)\sum_{k=1}^{\infty}k a_k x^{k-1}+\sum_{k=0}^{\infty}a_k x^k = a_0-a_1+\sum_{k=1}^{\infty}(k+1)a_k x^k-\sum_{k=1}^{\infty}(k+1)a_{k+1} x^k = 0 $$

hence $a_k - a_{k+1} = 0$ or $Y = a_0\sum_{k=0}^{\infty}x^k$ but $Y = y'$ so finally

$$ y = \int_0^x Y dx = a_0\sum_{k=1}^{\infty}\frac{x^k}{k}+ C_0 $$

NOTE

$$ \sum_{k=1}^{\infty}k a_k x^{k-1} = a_1+\sum_{k=1}^{\infty}(k+1) a_{k+1} x^k $$

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  • $\begingroup$ I'm a bit lost as to what you're doing. $\endgroup$ – Jwan622 May 10 at 18:44
  • $\begingroup$ I included some additional steps. I hope it helps. $\endgroup$ – Cesareo May 10 at 18:52

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