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 April 4th, 2017, 03:28 PM #1 Member   Joined: Feb 2014 Posts: 91 Thanks: 1 Power Series When you take the derivative of this why does n increment by +1? What does equating const x^0 means? What does equating const x^n means? Differential Equations & Linear Algebra (9780136054252), Pg. 641, Ex. 1 :: Homework Help and Answers :: Slader
 April 4th, 2017, 03:41 PM #2 Global Moderator   Joined: Dec 2006 Posts: 17,139 Thanks: 1281 1. It doesn't. As the term for which n = 0 is a constant, its derivative is zero and can be omitted. 2 and 3. Think about it. What is the power series for the function that is zero everywhere?
April 4th, 2017, 05:33 PM   #3
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 Originally Posted by skipjack 2 and 3. Think about it. What is the power series for the function that is zero everywhere?
The power series is a base zero "a0" which equals 0. I still don't know, could you give me the reason why too?

 April 4th, 2017, 09:48 PM #4 Global Moderator   Joined: Dec 2006 Posts: 17,139 Thanks: 1281 As $y\,^\prime - y = 0$, the power series for $y\,^\prime - y$ is also just 0. Hence the $n$th term of that power series is zero.
 April 5th, 2017, 05:33 AM #5 Member   Joined: Feb 2014 Posts: 91 Thanks: 1 That makes sense, it is homogeneous. I asked about equating const x^0 & x^n because there seems to be inconsistency in how to get the power series with the solutions I was provided. I'm really stuck on question 5. I'm confused because this guy did the problem and even got a 5 star rating yet he did not use the same method he used for the previous question number 1. This mainly starts around equating x^0 where he doesn't equate a0, He ignores it. I really wanted to figure this out on my own which is why I only for meaning of equating x^0 & x^n Differential Equations & Linear Algebra (9780136054252), Pg. 641, Ex. 5 :: Homework Help and Answers :: Slader The question I'm really stuck on.
 April 5th, 2017, 07:47 AM #6 Global Moderator   Joined: Dec 2006 Posts: 17,139 Thanks: 1281 The solutions do not contain "equating const x^n". One solution contains "Equating coeff. of ($x^n$) terms to 0; the other contains "equating $x^n$ terms to 0". Those are slightly different wordings, but the method is essentially the same. Note that $x^0 = 1$.
April 5th, 2017, 11:23 AM   #7
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Quote:
 Originally Posted by skipjack The solutions do not contain "equating const x^n". One solution contains "Equating coeff. of ($x^n$) terms to 0; the other contains "equating $x^n$ terms to 0". Those are slightly different wordings, but the method is essentially the same. Note that $x^0 = 1$.
Thank you, I know what it is, I know how to do, now I just need to learn why so that I don't just do work that I don't understand why I am doing it.

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