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 May 10th, 2014, 10:44 AM #1 Senior Member   Joined: Nov 2013 Posts: 137 Thanks: 1 Solutions for diff equation in form of series Given the following diff equation: $(1 - D)y(x)= f(x)$, being D = d/dx, the "implicit" solution is: $y(x)= \frac{1}{1-D}f(x)$, so, for "explicit" the solution is necessary to expand the fraction 1/(1-D) by identity: $\frac{1}{1-x}=\sum_{0}^{\infty}x^n \Delta n$, but this infinity series is true only for |x|<1, for |x|>1 is necessary utilize the following series: $\frac{1}{1-x}= -\sum_{-\infty}^{-1} x^n \Delta n$. Happens that D isn't a number for I say that |D| is less or greater than 1. So, how can I interprate this form of solution correctly? PS, this ideia from Operational Calculus: https://es.wikipedia.org/wiki/Transf...hist.C3.B3rica Heaviside’s Operational Calculus | Rip's Applied Mathematics Blog
 May 10th, 2014, 11:32 AM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,671 Thanks: 2651 Math Focus: Mainly analysis and algebra D is an operator. You can't leave it with nothing to operate on. If I were solving that with a series, I'd say $Dy(x) = y(x) - f(x)$ and then crreate a Taylor series. Of course, this relies on having an initial condition. Then $y(x) = y(0) + Dy(0)x + \cdots = y(0) + (y(0) - f(0)x + \cdots$ We can get $D^2y(x)$ and higher orders by differentiating the equation for $Dy(x)$. Edit: having browsed the blog, I think I've seen something similar before. I would rather understand why it works than the approach seen there, where it seems to for my functions, so lets say it's right. Last edited by v8archie; May 10th, 2014 at 11:49 AM.
May 10th, 2014, 12:59 PM   #3
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 Originally Posted by v8archie D is an operator. You can't leave it with nothing to operate on. If I were solving that with a series, I'd say $Dy(x) = y(x) - f(x)$ and then crreate a Taylor series. Of course, this relies on having an initial condition. Then $y(x) = y(0) + Dy(0)x + \cdots = y(0) + (y(0) - f(0)x + \cdots$ We can get $D^2y(x)$ and higher orders by differentiating the equation for $Dy(x)$. Edit: having browsed the blog, I think I've seen something similar before. I would rather understand why it works than the approach seen there, where it seems to for my functions, so lets say it's right.

 May 10th, 2014, 04:59 PM #4 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,671 Thanks: 2651 Math Focus: Mainly analysis and algebra Shouldn't you write $$(1 -D)y = f(x) \Longrightarrow y = -\frac{1}{D-1}f(x)$$ and then proceed as in the blog?
May 11th, 2014, 04:54 AM   #5
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 Originally Posted by v8archie Shouldn't you write $$(1 -D)y = f(x) \Longrightarrow y = -\frac{1}{D-1}f(x)$$ and then proceed as in the blog?
Now the ideia is expand the fraction $\displaystyle \frac{1}{1-D}$ with $\displaystyle \frac{1}{1-D}=\sum_{0}^{\infty}D^n \Delta n$ or with $\displaystyle \frac{1}{1-D} = -\sum_{-\infty}^{-1} D^n \Delta n$...

http://www.latp.univ-mrs.fr/~chaabi/...20,2006%29.pdf

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