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For a function $f(x,y)$ of two independent variables we have an incomplete specification of its partial derivatives as follows:

$$\frac {\partial f(x,y)} {\partial x} = \frac {1} {g(x,y) \sqrt {1 - (\frac {k y} {x^{(1/3)}})^2}}$$

$$\frac {\partial f(x,y)} {\partial y} = \left(\frac {3 x} {4}\right)\left (\frac {k} {x^{(1/3)}}\right)^2 (2 y) \frac {1} {g(x,y) \sqrt {1 - (\frac {k y} {x^{(1/3)}})^2}}$$

Problem: finding a suitable $g(x,y)$ that makes the partial derivatives converge to a single function $f(x,y)$ that fulfills the condition $f(x,0) = x$.

I will be grateful if people with many flight hours can offer suggestions for $g(x,y)$. Needless to say, I am not asking that they verify those suggestions, but in case someone would like, these are the inputs to Wolfram integrator:

1 / ( g(x,y as r) sqrt(1 - (k r / x^(1/3))^2) )

(3 t / 4) (k / t^(1/3))^2 (2 x) / ( g(x as t,y as x) sqrt(1 - (k x / t^(1/3))^2) )

Thanks in advance for your help.

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I wonder why it did not process the LaTeX. I know it is correct because I verified it online at latex.informatik.uni-halle.de/latex-online/latex.php –  Alex Feb 16 '11 at 16:37
    
@Alex: 1) Please do not use answers to make comments. The reason you cannot comment on your question or edit is because you are not registered; registering will solve both of these problems. 2) You forgot the dollar signs. In general, it is possible for two different websites to process LaTeX slightly differently. Here on this site it is even possible for the preview and the final product to differ because they do certain things in a different order. –  Qiaochu Yuan Feb 16 '11 at 16:39
    
You need to enclose it in dollar signs. The double dollar signs (you can see if you click on edit) give display mode. Also when you have a multicharacter exponent, like $x^{(1/3)}$ you need to enclose it in brackets, otherwise you get $x^(1/3)$ You can right click on any $LaTeX$ and select Show Source to see how it is done. –  Ross Millikan Feb 16 '11 at 16:40
    
Thank you Qiaochu and Ross for your prompt help. Two points. 1. Sorry to comment again using an answer but I just cannot find a "comment" button, even though I've just registered. 2. I tried to edit the question in order to add the {} to the exponent of the remaining x^(1/3) (outside the square root in the second equation) but the system will not let me do it because it thinks it was posted by a different user. Can someone do it? Thanks. –  Alex Feb 16 '11 at 23:07
    
@Alex: I have an add comment button below the question as well as below the answer. Got the x^(1/3) plus adding \left and \right before parentheses makes them large enough to enclose the inside. –  Ross Millikan Feb 16 '11 at 23:07

1 Answer 1

Hint: From your data one obtains $$-{f_x\over f_y}={-2\over 3 k^2 y x^{1/3}}.$$ It follows that the curves $f(x,y)=$const. satisfy the separable differential equation $$y'={-2\over 3 k^2 y x^{1/3}}.$$

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Thank you very much, Professor. Your observation opened the way to the solution. –  Alex Feb 17 '11 at 2:03

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