Given the following C^1 solution (x(t),y(t)) of:

{x' = -x^3 +x*y^2

{y' = -y^3 +y*x^2

How do I show that it will not interesct a circle centered @ (0, 0) more than once?

Also, I cannot find C^1 solution in my book. What is it?

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- Mar 4th 2011, 08:12 AMBorkborkmathC^1 solutions intersecting circles
Given the following C^1 solution (x(t),y(t)) of:

{x' = -x^3 +x*y^2

{y' = -y^3 +y*x^2

How do I show that it will not interesct a circle centered @ (0, 0) more than once?

Also, I cannot find C^1 solution in my book. What is it? - Mar 4th 2011, 09:19 AMTheEmptySet
$\displaystyle C^1$ means that the function is differentiable once and its derivative is continuous.

Note that

$\displaystyle \displaystyle \frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{dy}{dx}= \frac{-y^3 +yx^2}{-x^3 +xy^2}=\frac{-y(y^2-x^2)}{x(y^2-x^2)}=-\frac{y}{x}$

This can be solved by separation of variables

$\displaystyle \displaystyle \frac{dy}{y}=-\frac{dx}{x} \iff \ln|y|=-\ln(x)+\ln(c) =\ln\left( \frac{c}{x}\right) \implies y=\frac{c}{x}$

So unless there is a typo a hyperbola can intersect a circle centered at the origin more than once. - Mar 4th 2011, 11:08 AMBorkborkmath
When I put

{

{

I meant that as a piecewise function, if that changes anything. I think that the professor might have typo'd the question if, as you beautifully explained it, the equation can be separated by variables.

On a side note, how do you type with all the mathematical symbols? - Mar 4th 2011, 11:16 AMTheEmptySet
2nd question first. It is called LaTex

http://www.mathhelpforum.com/math-help/f47/

I assume you mean parametric form. I just eliminated the parameter. If the minus sign was not there you would get a line though the origin. If there were some restriction like

$\displaystyle x > 0 $ or $\displaystyle y > 0$then the statement would be true.