Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

Thanks.

share|cite|improve this question
3  
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.