# Power series differential check $(x-1)y'' + y' = 0$

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 $$y = c_0 + c_1 + \sum_{n=2}^\infty \frac{x^n}{n}$$

• 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. – peter a g May 10 at 18:38
• In your final answer why does $c_1$ not multiply an $x$? – Spencer May 10 at 23:18
• Did you find either of the answers useful? Or is there still some additional clarification that is needed? – Spencer May 12 at 19:21

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.

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

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

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