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Differential equation 
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#1
Dec704, 06:03 AM

P: 27

somebody slove this differential equations
1/y' = (1/y)+(1/x) thanx in advance 


#2
Dec704, 06:12 AM

Sci Advisor
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P: 1,123

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) 


#3
Dec704, 06:23 AM

P: 27

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 


#4
Dec704, 09:54 AM

Sci Advisor
HW Helper
P: 1,123

Differential equation
Well can you post what you have done please so others can help.



#5
Dec704, 02:29 PM

Sci Advisor
HW Helper
P: 1,123

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. 


#6
Dec704, 07:40 PM

P: 27

try seperation of variables...after that iam struggling to cotinue....



#7
Dec704, 11:29 PM

P: n/a

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. 


#8
Dec804, 05:55 AM

Sci Advisor
HW Helper
P: 1,123

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] 


#9
Dec804, 11:40 AM

P: n/a

I added a y to the LHS in the 4th step.



#10
Dec804, 05:48 PM

Mentor
P: 7,314

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. 


#11
Dec1304, 12:57 PM

P: 4

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 


#12
Dec1304, 01:24 PM

P: 553




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