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This has been solved! Thanks to everyone who read and thought about it

Suppose lines of the form $(x_0,y)$ and $(x,y_0)$ for any given $x_0,y_0\in \mathbb R$ are solutions to the system of differential equations

Let $A,B,C,D$ be functions of $(x,y)$ and $x,y$ are functions of $t$

$$\begin{cases} {d\over dt} (A\dot{x} +B\dot{y})={1\over 2} (A_x\dot{x}^2+2B_x\dot{x}\dot{y}+C_x\dot{y}^2)\\ {d\over dt} (B\dot{x} +C\dot{y})={1\over 2} (A_y\dot{x}^2+2B_y\dot{x}\dot{y}+C_y\dot{y}^2) \end{cases}$$

Where subscripts denote partial derivatives.

How then can we conclude that $$\begin{cases}{\partial \over \partial y}\left({B\over \sqrt{C}}\right)={\partial \sqrt{C}\over \partial x}\\{\partial \over \partial x}\left({B\over \sqrt{A}}\right)={\partial \sqrt{A}\over \partial y} \end{cases}$$

I tried to substitute the known solutions into the system but all I get seems to be 

$$\begin{cases}{d\over dt} A={1\over 2}A_x\\ {d\over dt} B={1\over 2}C_x \\ {d\over dt} B={1\over 2}A_y\\ {d\over dt} C={1\over 2}C_y\end{cases}$$ which doesn't seem to be too helpful.

Any ideas?


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Regarding your edit: please feel free to provide your own answer, which I believe is a common practice here. I think you even get a badge. :) – tentaclenorm May 14 '12 at 22:50
@tentaclenorm: thanks for telling me! :) i tried but the system won't let me do so for at least the next 8hrs, it would seem. :( – Jones May 14 '12 at 23:00

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