# Linear (I think?) First Order Diff EQ

1. Mar 27, 2008

### 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?

2. Mar 27, 2008

### 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

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

Last edited: Mar 27, 2008
3. Mar 27, 2008

### Pere Callahan

What did you need the "integration factor" for?

4. Mar 27, 2008

### sutupidmath

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.

5. Mar 27, 2008

### epenguin

xy' + x = (xy)' help?

6. Mar 27, 2008

### sutupidmath

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

7. Mar 27, 2008

### epenguin

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

8. Mar 27, 2008

### sutupidmath

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'.

9. Mar 27, 2008

### Pere Callahan

I'd guess, penguin was not asking this

10. Mar 27, 2008

### sutupidmath

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

11. Mar 27, 2008

### 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.

12. Mar 27, 2008

### sutupidmath

Yeah, that makes sens.

13. Mar 28, 2008

### epenguin

Yes I realised it is not stupidmath, but rather sutupidmath. I didn't realise 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.

14. Mar 28, 2008

### sutupidmath

HAHAHA....Very funny!!!!!!:rofl:

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

Last edited: Mar 28, 2008
15. Mar 28, 2008

### epenguin

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

16. Mar 28, 2008

### sutupidmath

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!!

Last edited: Mar 28, 2008
17. Mar 28, 2008

### epenguin

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

:tongue2: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!

18. Mar 28, 2008

### 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.

19. Mar 28, 2008

### sutupidmath

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