MHB Partial Differntial problem Cauchy

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Find surface of

$\begin{array}{l}
\text{Problem Cauchy} \\
{a^2} \cdot {x_2} \cdot u \cdot {u_{{x_1}}} + {b^2} \cdot {x_1} \cdot u \cdot {u_{{x_2}}} = 2{c^2}{x_1}{x_2}{\rm{ }} \\
\end{array}$
The partial differntial equation passes through

${\rm{ C: = \{ }}\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1,{x_3} = u({x_1},{x_2}) = 0\} \\
\\ $$a,b,c$ nonzero constants
 
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First. I think you should be consistent with your notation. Either use $$x$$ and $$y$$ or $$x_1$$ and $$x_2$$ but not both. It's confusing.

Second, as you have the boundary of an ellipse, have you thought of introducing new coordinates

$$x = a r \cos \theta, y = b r \sin \theta?$$
 
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Ιn C are x1,y1.
 
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I have the equation ##F^x=m\frac {d}{dt}(\gamma v^x)##, where ##\gamma## is the Lorentz factor, and ##x## is a superscript, not an exponent. In my textbook the solution is given as ##\frac {F^x}{m}t=\frac {v^x}{\sqrt {1-v^{x^2}/c^2}}##. What bothers me is, when I separate the variables I get ##\frac {F^x}{m}dt=d(\gamma v^x)##. Can I simply consider ##d(\gamma v^x)## the variable of integration without any further considerations? Can I simply make the substitution ##\gamma v^x = u## and then...

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