Linear (I think?) First Order Diff EQ

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Discussion Overview

The discussion revolves around solving the first-order differential equation xy' + y = cos x. Participants explore various methods for solving this equation, including the use of an integrating factor and the product rule. The conversation includes attempts to clarify the reasoning behind different approaches and the implications of mathematical errors.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant expresses uncertainty about how to solve the equation, noting that separation of variables is not applicable.
  • Another participant suggests using an integrating factor and derives a solution, but later questions the necessity of the integrating factor.
  • There is a discussion about the validity of a mathematical expression, with participants correcting each other regarding the product rule.
  • Some participants engage in light-hearted banter about typos and misunderstandings, while also emphasizing the importance of the constant of integration in the solution.
  • One participant asserts that they solved the problem for the original poster (OP), leading to a brief exchange about who contributed what to the discussion.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the necessity of the integrating factor or the clarity of the mathematical expressions used. There are multiple viewpoints on the approach to solving the differential equation, and some misunderstandings about who contributed to the solution remain unresolved.

Contextual Notes

There are indications of missing assumptions regarding the methods discussed, and some participants express confusion over the application of the product rule. The discussion reflects varying levels of understanding and approaches to the problem.

Feldoh
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So this equation came up:

xy' + y = cos x

Now I was just wondering how to solve this, all I've learned how to do is separation of variables, which cannot be used in this case.

Basically I ask this because a solution for y is an infinite series, so basically I'm just wondering if the infinite series converges to some function of x, and if it does which function is it?
 
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well, i think that deriving an integration factor would work. Like deviding through by x:

[tex]y'+\frac{1}{x}y=\frac{cosx}{x}[/tex]


then [tex]r=e^{\int\frac{dx}{x}}=e^{ln|x|}=|x|[/tex] Now multiplying through we get

[tex]|x|y'+\frac{1}{x}|x|y=\frac{cosx}{x}|x|=>sgn(x)xy'+\frac{1}{x}sgn(x)xy=\frac{cosx}{x}sgn(x)x=>xy'+y=cosx[/tex]

Now we notice that on the left side
[tex]xy'+y=(xy)'[/tex] this way

[tex](xy)'=cosx=>\int(xy)'dx=\int cosxdx =>xy=sinx+C=>y=\frac{sinx}{x}+x^{-1}C[/tex]
 
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What did you need the "integration factor" for?
 
Pere Callahan said:
What did you need the "integration factor" for?

Well, i guess i did not need it at all, because it was already in a nice form. I don't know why i did it. Maybe just to show the OP another way of solving diff. eq. since he said that he has learned so far only the method of separation of variables.
 
xy' + x = (xy)' help? :smile:
 
epenguin said:
xy' + x = (xy)' help? :smile:

well what u did here isn't true..lol...
 
sutupidmath said:
well what u did here isn't true..lol...

It's true if you saw I obviously meant to write xy' + y = (xy)' stupidmath :biggrin:
 
epenguin said:
It's true if you saw I obviously meant to write xy' + y = (xy)' stupidmath :biggrin:

Well first, it is not stupidmath, but rather sutupidmath!

Well, if you are asking as why xy'+y=(xy)', it is simply the product rule.
(fg)'=fg'+gf'.
 
I'd guess, penguin was not asking this:smile:
 
  • #10
epenguin said:
xy' + y = (xy)' help? :smile:
I don't know, why would he write this then? I also was surprised..lol..
 
  • #11
I would interpret this as adressing the OP

xy' + y = (xy)'
Does this help?

The mathematical error was most probably just a typo.
 
  • #12
Pere Callahan said:
I would interpret this as adressing the OP

xy' + y = (xy)'
Does this help?

The mathematical error was most probably just a typo.
Yeah, that makes sens.
 
  • #13
sutupidmath said:
Well first, stupidmath, but rather sutupidmath!

Well, if you are asking as why xy'+y=(xy)', it is simply the product rule.
(fg)'=fg'+gf'.

Yes I realized it is not stupidmath, but rather sutupidmath. I didn't realize you did! :biggrin: As you saw even I o:) can make a typo!

I am saying that that formula makes it easy to solve your d.e. Warning, I think it will be paticularly important not to forget the constant of integration.
 
  • #14
epenguin said:
Yes I realized it is not stupidmath, but rather sutupidmath. I didn't realize you did! :biggrin:

HAHAHA...Very funny!:smile:

epenguin said:
I am saying that that formula makes it easy to solve your d.e. Warning, I think it will be paticularly important not to forget the constant of integration.


Isn't this the same thing, in more details ,what i just wrote above??
 
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  • #15
sutupidmath said:
Well, if you are asking as why xy'+y=(xy)', it is simply the product rule.
(fg)'=fg'+gf'.

sutupidmath said:
Isn't this the same thing, in more details ,what i just wrote above??

Yes it is the formula (fg)'=fg'+gf' applied to xy . Using it can you now solve the problem you asked about?
 
  • #16
epenguin said:
Yes it is the formula (fg)'=fg'+gf' applied to xy . Using it can you now solve the problem you asked about?


I did not ask about anything body! Have you at least read the thread at all, or you are just throwing words here without knowing who you are addressing to? It was i who actually solved the problem for the OP!
 
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  • #17
sutupidmath said:
I did not ask about anything body! Have you at least read the thread at all, or you are just throwing words here without knowing who you are addressing to? It was i who actually solved the problem for the OP!

:redface:Sorry I had not noticed who was posting and assumed I was replying to the OP who I now see never came back!

:-pAnd I confess I had not noticed you had solved it as I didn't read beyond the first line or two of your post and immediately thought 'that is unnecessarily complicated'. However congratulations on getting the same right answer as me! :biggrin:
 
  • #18
Sorry, about two minutes after I posted it I realized it was the product rule and slapped myself in the face for not seeing it earlier.
 
  • #19
Feldoh said:
Sorry, about two minutes after I posted it I realized it was the product rule and slapped myself in the face for not seeing it earlier.

Yeah, i complicated a lill bit my answer, but i just wanted to show another perspective from the beginning on that problem.
 

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