2
$\begingroup$

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!

$\endgroup$
  • $\begingroup$ it is very hard to read. can you get rid of the shadows. $\endgroup$ – abel Mar 10 '15 at 23:20
  • 2
    $\begingroup$ Is this better? i.imgur.com/GaSTqag.jpg $\endgroup$ – Pat Yates Mar 10 '15 at 23:34
  • $\begingroup$ 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. $\endgroup$ – abel Mar 10 '15 at 23:36
1
$\begingroup$

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 $$(1-x)y' = y. $$

$\endgroup$
  • $\begingroup$ Thank you! I'm going to try to keep my work in sigma notation but this gives me a much better understanding of what I need to do. $\endgroup$ – Pat Yates Mar 11 '15 at 4:44
2
$\begingroup$

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 $$ a_{n+1} = a_{n} $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.