Finding general solution to a differential equation using power seriesby s3a Tags: differential, equation, power, series, solution 

#1
Mar312, 11:57 PM

P: 522

1. The problem statement, all variables and given/known data
Find the general solution near x = 0 of y''  xy' + 2y = 0 (using power series). Answer: y = a_0 * y_1(x) + a_1 * y_2(x) where y_1(x) = 1  x^2 and y_2(x) = x  1/6 * x^3  1/120 x^5  1/1680 * x^7  ... 2. Relevant equations Power series. Sigma notation for summations. Polynomial derivatives. 3. The attempt at a solution My attempt is attached. I get y(x) = ke^x and I think I'm wrong but I'm not sure since the answer my book gives doesn't use sigma notation. I'd appreciate if someone could tell me if I'm right or wrong and if I am wrong, where I went wrong. Thanks in advance! 



#2
Mar412, 02:50 AM

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I see two mistakes. First, the nonzero a_{n} of the second solution aren't given by a_{1}/n! for n>1. Second, even if that were the case, your solution wouldn't be e^{x} because the sign of the first term the series is wrong.
Try calculating a_{7} explicitly using the recurrence relation. 



#3
Mar412, 02:40 PM

P: 522

I get a_7 = 1/14 * a_5 = 1/14 * 1/120 * a_1 = 1/1680 * a_1 and I see that the a_n = 1/n! a_1 does not work out for that particular case.
I also see that a_n = 1/n! does not hold for the zero values. As for the sign of the first term for the e^x part, I didn't understand what you meant. Could you rephrase that please? I would appreciate it if you could tell me what the explicit relationship is and what the reasoning was in order to get it because my book doesn't cover the material in this way and uses another method which I dislike (the no summation notation, "dot dot dot" method). 



#4
Mar412, 03:34 PM

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Finding general solution to a differential equation using power series
I see more mistakes now that I'm awake. They're easy to see if you just compare the two series. Your solution was
$$y_2(x) = a_1\left(x  \frac{x^3}{6}  \frac{x^5}{120}  \frac{x^7}{1680}  \cdots\right)$$ but $$a_1 e^x = a_1\left(1  x  \frac{x^2}{2}  \frac{x^3}{6}  \cdots \right)$$ What I meant about the sign of the "first" term was the sign of the linear term. 



#5
Mar412, 05:14 PM

P: 522

Okay, I also redid my work from scratch with neater handwriting and computed up to the a_9 term but I still can't figure out how to get an explicit equation. Namely, how would I take into account that there are zeros for every even subscript from a_4 onward?
I'm currently thinking of something like a_n = [(1)^n  1]/2 * 3/n! which works for a_7 and a_9 but not for a_5. I'm still really stuck :(. 



#6
Mar412, 07:20 PM

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Why are you insisting on having a closed form expression? Sometimes it's not really possible. Just try writing out more terms and see if you can spot a pattern. It usually doesn't help to simplify as you go because you're looking for a pattern.
Anyway, to get only odd powers, you typically write something like $$\sin x = \sum_{n=0}^\infty \frac{x^{2n+1}}{(2n+1)!}$$ 


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