Hi Manoj,

"just taking the first and the third part,    dx / (x * y) = df , we get f 
= ln(x) / y"

Now that you explicitly point out difference between dx and dhox, you have 
integrated 1/xy w.r.t x assuming that y is a constant here(This, coupled 
with fact that I was awake for 22 hrs, generated my previous response), 
which is not case as dy/dx = x (as deduced from the first and second part).

>
>>> Hi Manoj, you might want to check this method, because if you directly 
>> substitute Fx = df/dx = c(x,y)/a(x,y) and Fy = df/dy = c(x,y)/b(x,y), you 
>> get LHS = 2*c(x,y). You can take a look at 
>> http://geo.hmg.inpg.fr/loret/enseee/maths/enseee-maths-IBVPs-3.pdf , 
>> which discusses the solution of the pdes of this type based on their 
>> classification.
>>  
>>
>> Thank you for pointing me to a source that, discusses the solution to 
>> pdes, based on their classification. However I'm afraid my question remains 
>> unanswered for the following reasons.
>>
>
>
> Firstly,  I think (I may be wrong) df / dx is equal to (dhof / dhox) (I 
> don't know how to write it here), only if f is a function of x, if not df = 
> (dhof / dhox) * dx + (dhof / dhoy) * dy , since f is a function of x and y, 
> so I suppose df / dx cannot be substituted back in the equation.
>
>  
You are absolutely right. df/dx, in a general case is not equal dhof/dhox. 
They are equal in cases where y and x are independent everywhere i.e. dy/dx 
= 0 and other complicated cases.



> Secondly, I would like to cite this example in the research paper, by Dr. 
> Starrett( Aaron's professor indeed ) Lie 
> Groups<http://www.google.co.in/url?sa=t&rct=j&q=&esrc=s&source=web&cd=1&ved=0CDEQFjAA&url=http%3A%2F%2Feuler.nmt.edu%2F~jstarret%2F05-649LieGroupODEFinalVersion.pdf&ei=WxdcUb65L4OErQeQz4HoAQ&usg=AFQjCNE9ugXh9NmZVLxq9LB8fpYuo6vusA&sig2=HLQVAGHySZ3PndKBIPECdA&bvm=bv.44697112,d.bmk>.
>  On page 16, he mentions clearly that to solve the equation, sx ξ + sy η = 
> 1.
>  
> It can be done by, ds = integral (dx / ξ) , I thought this might be for a 
> case where only ξ is a function of x, but he goes on to say on page 20 that 
> when ξ is equal to y, s would remain (x / y) . Sorry for being a bit 
> bookish here, but I hope my point is being put across.
>
> It would be really helpful, if Aaron or other people here with a much 
> better mathematical background than me over here, could show me the way 
> ahead.
> -- 
> Regards,
> Manoj Kumar,
> Mech Undergrad.
> BPGC
> Blog <http://manojbits.wordpress.com>
>  

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