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Differential Equations help?

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Old Oct 26, 2007 | 02:21 PM
  #1  
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Differential Equations help?

I'm having some trouble figuring this problem out and was looking for some help..

I have to solve y'+2xy=0 using a series solution.

So far I have [sum of nCnX^n from n=0 to infinity] + 2 [sum of CnX^(n+1) from n=0 to infinity] = 0

and I don't know where to go from here any ideas?
 
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Old Oct 26, 2007 | 03:33 PM
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If its an odd number, the answer's in the back.
 
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Old Oct 26, 2007 | 03:43 PM
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If you still have time, bust out in tears and run, screaming, to your TA. Nobody without a math degree can help you now. And pray that your professor uses a wicked curve, cause that's the only hope you have of passing the class.

Ok, little bit of sarcasm in there, but seriously-DE is just short of impossible, and your professor or TA are going to be your best bet. This is a class that 95% of people who take it either hate it or have no idea what's going on (I did both) and you can't afford to wait until the last minute for anything. Try to read the notes beforehand and if you don't understand the lectures then go to office hours and ask questions.

To answer your question, no I have no idea. You got farther than I would have.
 
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Old Oct 26, 2007 | 04:59 PM
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I can't help you, and I took the class twice. What I do know, is that this is a first-order problem. Looking in my old Differential Equations book, it mentions that you will need to use the "recurrence formula". Just remember, Google and other search engines are your best friends when it comes to some of this stuff. There are some colleges and universities that have some really good online notes, and examples. What I learned in this class, I used very little after I took it. Some of it I saw again in circuits, fluid mechanics, and environmental engineering classes.
 
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Old Oct 26, 2007 | 05:24 PM
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Originally Posted by captain p4
I'm having some trouble figuring this problem out and was looking for some help..

I have to solve y'+2xy=0 using a series solution.

So far I have [sum of nCnX^n from n=0 to infinity] + 2 [sum of CnX^(n+1) from n=0 to infinity] = 0

and I don't know where to go from here any ideas?
It been a long time since i've had to do any math like this,let lone even know what the heck a series solution is.

what is the ' is that y squared?

if so, wouldn't y=0 and x= +/- 1
how you get to that solution I couldn't tell ya.
 
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Old Oct 26, 2007 | 05:44 PM
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I think that's Y "prime".

I took AP Calc, got way into what you're doing, and now have no idea what the heck it is 23 years later... sorry

But I want to know, so I'm subscribing to this thread...
 
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Old Oct 26, 2007 | 08:51 PM
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Do they give you some initial conditions? You might try this link.

http://www.sosmath.com/diffeq/diffeq.html
 
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Old Oct 26, 2007 | 10:19 PM
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Why do you have to use series to solve this?

Here's the answer without any series used:
y'= -2xy

dy/dx= -2xy

seperate your variables
(1/y)dy = -2xdx

integrate both sides

lny= -x^2 + C

y=Ce^(-x^2)

sorry it would look better on Math Type
 
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Old Oct 26, 2007 | 10:51 PM
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Because thats what the paper says, lol. We're supposed to use the "normal" way as a check to the series solution.
 
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Old Oct 26, 2007 | 11:14 PM
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What you have so far are power series if you use the power series formula for each one don't you get the e^(-x^2) series?

It looks like you should, but I didn't actually calculate it.
 
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Old Oct 26, 2007 | 11:15 PM
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I can't write out the series on this computer without math type it would be too confusing.

Do you know what I'm talking about?

Look up infinate power series rules in your text.
 
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Old Oct 26, 2007 | 11:52 PM
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bczolone you are one smart dude! Since my retirement, I've been revisiting calc, a little at a time, just to make my brain hurt. Maybe I'll get caught up enough to appreciate your analysis

Anyway, when I took diff many years ago, the prof was a really great guy. He was a math junkie, with the little office, and no outside interests, but he was still a great guy.

He was kind enough to tell us that there was no real reason to ever solve diff eqs. He also said that the math world concentrated on classifying the various equations as sovable or not.

One night, he proved his point. He handed out a sample equation, which he maintained was too hard for even the Russians to solve. ( As if I'd know). Then he showed how a relatively simple computer algorithm could approximate a solution good enough for any reasonable use.

I took this to mean study hard, pass the class, and don't worry about it.
I did, and never looked back.

ford2go
 
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Old Oct 27, 2007 | 12:04 AM
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Well the advice came from Mrs. Bczolone!...not me...I have no idea what all this is so I asked her for help and she wrote in my place...sorry to decieve you in that regard, however she wanted to know if she was helpful.
 
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Old Oct 27, 2007 | 12:50 AM
  #14  
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Okay, what the hell are you guys talking about? None of what I just read made ANY sense at all. A math problem classified as unsolveable? Hmm. Maybe there is such thing as global warming...

Kidding. I have no desire to study DE or Calculus. Not my thing. But don't think I'm taking away from what you guys are doing. More power to ya.
 
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Old Oct 28, 2007 | 04:01 PM
  #15  
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I still don't get it. thanks for the attempted help everyone.

If make the n's both = to 0 then my Xs aren't to the same power and if make the X's to the same power my n's are different.. how am I supposed to combine the summations if I can't make these two things equal??
 
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