# Finding power series solution to differential equation $(1-x)y'=y$ centered at $x=0$

I'm trying to teach myself how to solve differential equations with power series. I am stuck on working through $$(1-x)y'=y \text{ centered at } x= 0.$$

I've gotten to the point where I must figure out what $a_{k+1}$ is and I'm completely lost on what to do next... can someone please check if I'm on the right track here, and if so, where to go from here? My work: http://i.imgur.com/KodxeCY.jpg (another photo: i.imgur.com/GaSTqag.jpg)

I know from a non-power series method that answer will be of the form $y = \frac{c_1}{1-x}$.

FWIW, I've been following this video: https://www.youtube.com/watch?v=RJJKq7Uc-9I with a different equation. If you know of a better method for this, please let me know!

• it is very hard to read. can you get rid of the shadows. – abel Mar 10 '15 at 23:20
• Is this better? i.imgur.com/GaSTqag.jpg – Pat Yates Mar 10 '15 at 23:34
• it helps, at least at the beginning, not to use the sigma notation for the sum. just write it in longhand as i have done in my answer. the sigma notation may obscure what is going on. at least that is my view. – abel Mar 10 '15 at 23:36

in the mean time here is what i will do. we will try $$y = a_0 + a_1x + a_2x^2 + a_3x^3 + \cdots, y' = a_1 + 2a_2x + 3a_3 x^2 + \cdots$$ subbing in $$(1-x)y' = y$$ we get $$(1-x) \left(a_1 + 2a_2x + 3a_3 x^2 + \cdots \right) = a_0 + a_1x + a_2x^2 + a_3x^3 + \cdots.$$ equating the coefficients of the powers of $x$ gives $$a_1 = a_0, 2a_2 - a_1 = a_1, 3a_3-2a_2 = a_2, \cdots, na_n - (n-1)a_{n-1} = a_{n-1}$$ that is $$a_0 = a_1 = a_2 = a_3 = \cdots = a_n$$ and $$y = a_0(1 + x + x^2 +\cdots = \frac {a_0}{1-x}, y' = \frac {a_0}{(1-x)^2}$$ it verifies that $y$ solves $$(1-x)y' = y.$$
I think you're making a mistake when you get to the point: $$\sum\limits_{n = 1}^\infty na_{n}x^{n-1} - \sum\limits_{n = 1}^\infty na_{n}x^{n} = \sum\limits_{n = 0}^\infty a_{n}x^{n}$$ The second sum on the left is equal to $$\sum\limits_{n = 0}^\infty na_{n}x^{n}$$ because $n = 0$ in the first term, making it 0. The first sum on the left is equal to $$\sum\limits_{n = 0}^\infty (n + 1)a_{n+1}x^{n}$$ This gives: $$(n + 1)a_{n+1} - na_{n} = a_{n}$$ meaning $$a_{n+1} = a_{n}$$