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differential equation

 
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Dec7-04, 06:03 AM   #1
 

differential equation


somebody slove this differential equations

1/y' = (1/y)+(1/x)

thanx in advance
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Dec7-04, 06:12 AM   #2
 
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Perhaps looking at it like this:

[tex]\frac{1}{\frac{dy}{dx}} = \frac{1}{y} + \frac{1}{x}[/tex]

[tex]\frac{dx}{dy} = \frac{1}{y} + \frac{1}{x}[/tex]

[tex]x\frac{dx}{dy} = \frac{x}{y} + 1[/tex]

lol, I'll stop there because I suddenly realise this is beyond me (but it looks in a 'nicer' form, perhaps it will help you)
Dec7-04, 06:23 AM   #3
 
Your solution is just a peanut compared to where i have gone....there is still more to go...anyhow thanx for trying,do try nmore and figure out the solution.

regards
drdolittle
Dec7-04, 09:54 AM   #4
 
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differential equation


Well can you post what you have done please so others can help.
Dec7-04, 02:29 PM   #5
 
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I ran this through Mathematica: DSolve[1/(y'[x]) == 1/x + 1/y[x], y[x], x]

And it gave me nothing sorry.

Edit: Although I'm not used to using Mathematica and have yet to get it to solve the simplest thing I think I have inputed it right.
Dec7-04, 07:40 PM   #6
 
try seperation of variables...after that iam struggling to cotinue....
Dec7-04, 11:29 PM   #7
 
Even though I just started learning differential equations, I thought I'd give this a try:

[tex]\frac{dx}{dy}=\frac{1}{y}+\frac{1}{x}[/tex]

[tex]\frac{dy}{dx}=\frac{xy}{x+y}[/tex]

[tex]x\frac{dy}{dx}+y\frac{dy}{dx}=xy[/tex]

[tex]y+x\frac{dy}{dx}+y\frac{dy}{dx}=y+xy[/tex]

[tex]\frac{d(xy)}{dx}+\frac{1}{2}\frac{d(y^2)}{dx}=y(1+x)[/tex]

[tex]\frac{1}{y}\,d(xy)+\frac{1}{2y}\,d(y^2)=(1+x)\,dx[/tex]

I don't know what to do now, and I don't know if any of this is right, but I hope it'd be of some use.
Dec8-04, 05:55 AM   #8
 
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Err I still think this is beyond me but I think you made a mistake on the LHS going from the 4th to the 5th line as:

[tex]\frac{d(xy)}{dx} = x\frac{dy}{dx} + y[/tex]
Dec8-04, 11:40 AM   #9
 
I added a y to the LHS in the 4th step.
Dec8-04, 05:48 PM   #10
 
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What I see when I look at that equation is a family of hyperbolas very much like the simple lens equation. There is a change of variables and a rotation that will reduce this equation to something which may be separable. Unfortunately I do not have the time to do all of the algebra for you.

Explore doing a change of variables, perhaps to polar coordinates, see what you get.
Dec13-04, 12:57 PM   #11
 
How do you guys write the nice format of dy/dx and the fractions? Which program do you use, and you post them as photos?

I'll help in solving it, but after knowing how to post a math solution
Dec13-04, 01:24 PM   #12
 
Quote by iSamer
How do you guys write the nice format of dy/dx and the fractions? Which program do you use, and you post them as photos?

I'll help in solving it, but after knowing how to post a math solution
They use LaTeX. See this thread for more info .
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