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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: , being D = d/dx, the "implicit" solution is: , so, for "explicit" the solution is necessary to expand the fraction 1/(1D) by identity: , but this infinity series is true only for x<1, for x>1 is necessary utilize the following series: . 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: 7,685 Thanks: 2666 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  
Senior Member Joined: Nov 2013 Posts: 137 Thanks: 1  Quote:
 
May 10th, 2014, 04:59 PM  #4 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,685 Thanks: 2666 Math Focus: Mainly analysis and algebra 
Shouldn't you write $$(1 D)y = f(x) \Longrightarrow y = \frac{1}{D1}f(x)$$ and then proceed as in the blog? 
May 11th, 2014, 04:54 AM  #5  
Senior Member Joined: Nov 2013 Posts: 137 Thanks: 1  Quote:
PS: look this article from page 13 until 17. http://www.latp.univmrs.fr/~chaabi/...20,2006%29.pdf  

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