Linear (I think?) First Order Diff EQ

[Feldoh]

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?

 

[sutupidmath]

well, i think that deriving an integration factor would work. Like deviding through by x:

y'+\frac{1}{x}y=\frac{cosx}{x}


then r=e^{\int\frac{dx}{x}}=e^{ln|x|}=|x| Now multiplying throught we get

|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

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

(xy)'=cosx=>\int(xy)'dx=\int cosxdx =>xy=sinx+C=>y=\frac{sinx}{x}+x^{-1}C

 

[Pere Callahan]

What did you need the "integration factor" for?

 

[sutupidmath]

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.

 

[epenguin]

xy' + x = (xy)' help?

 

[sutupidmath]

xy' + x = (xy)' help?

well what u did here isn't true..lol...

 

[epenguin]

well what u did here isn't true..lol...

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

 

[sutupidmath]

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

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'.

 

[Pere Callahan]

I'd guess, penguin was not asking this

 

[sutupidmath]

xy' + y = (xy)' help?

I don't know, why would he write this then? I also was surprised..lol..

 

[Pere Callahan]

I woudl interpret this as adressing the OP

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

The mathematical error was most probably just a typo.

 

[sutupidmath]

I woudl 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.

 

[epenguin]

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! As you saw even I 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.

 

[sutupidmath]

Yes I realized it is not stupidmath, but rather sutupidmath. I didn't realize you did!

HAHAHA...Very funny!

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??

 

[epenguin]

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

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?

 

[sutupidmath]

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!

 

[epenguin]

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!

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

And 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!

 

[Feldoh]

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.

 

[sutupidmath]

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 beggining on that problem.