# Showing the solution is a circle: Why is this true?

• Apr 28th 2012, 12:10 PM
karatekid
Showing the solution is a circle: Why is this true?
Hi guys,

I've been trying this all day, hopefully I'm just being stupid. The full problem is to show that two simultaneous differential equations in x(t) and y(t) have a solution that is a circle, which I believe is the circle $\displaystyle {(x-\mu)}^2 + y^2 = a$ where mu is given and a is a constant depending on initial conditions. I've reduced the problem to showing $\displaystyle x\dot{y} - y\dot{x} - \mu \dot{y} = const.$

I can't show this, so if anyone could help with that I'd be really grateful.

I'm also confused as to where I'm going wrong here:
If $\displaystyle x = a\cos{\omega t} + \mu$ and $\displaystyle y = a\sin{\omega t}$ then the constant in the last expression above is zero (I hope that's not my mistake!). But then we can rearrange this (when x'(t) is non-zero and x isn't mu) to get $\displaystyle \dot{y}/\dot{x} = y/(x-\mu)$ which is false in most cases. Why can't I divide through?

Thanks!!
• Apr 28th 2012, 08:04 PM
Prove It
Re: Showing the solution is a circle: Why is this true?
Quote:

Originally Posted by karatekid
Hi guys,

I've been trying this all day, hopefully I'm just being stupid. The full problem is to show that two simultaneous differential equations in x(t) and y(t) have a solution that is a circle, which I believe is the circle $\displaystyle {(x-\mu)}^2 + y^2 = a$ where mu is given and a is a constant depending on initial conditions. I've reduced the problem to showing $\displaystyle x\dot{y} - y\dot{x} - \mu \dot{y} = const.$

I can't show this, so if anyone could help with that I'd be really grateful.

I'm also confused as to where I'm going wrong here:
If $\displaystyle x = a\cos{\omega t} + \mu$ and $\displaystyle y = a\sin{\omega t}$ then the constant in the last expression above is zero (I hope that's not my mistake!). But then we can rearrange this (when x'(t) is non-zero and x isn't mu) to get $\displaystyle \dot{y}/\dot{x} = y/(x-\mu)$ which is false in most cases. Why can't I divide through?

Thanks!!

My psychic powers are a little off today, so I have no idea if what you have reduced the original problem to is correct. Please state the original problem and what working you have done so far for the first question.
• Apr 29th 2012, 03:32 AM
karatekid
Re: Showing the solution is a circle: Why is this true?
Yeah, sorry. That's the last time I ask a question after midnight.

The original equations are
$\displaystyle \ddot{x} - 2\dot{y} &= \frac{(x-\mu)}{2{r_2}^3} \\ \ddot{y} + 2\dot{x} &= \frac{y}{2{r_2}^3}$

Where $\displaystyle {r_2}^2 = \left( x - \mu \right)}^{2} + y^2$ and in this particular case $\displaystyle \mu = 0.5$. To try and solve these, I divided the second by the first (or if x = mu, the first by the second) and cross multiplied to get
$\displaystyle x \ddot{y} + 2x\dot{x} - 2\mu\dot{x} - \mu \ddot{y} = y \ddot{x} - 2y\dot{y}$

and we can rearrange and integrate this to get
$\displaystyle x^2 + y^2 - 2\mu x = y\dot{x} - x\dot{y} + \mu \dot{y} + const.$

which is of the form $\displaystyle {(x-\mu)}^2 + y^2 = const.$ if and only if $\displaystyle y\dot{x} - x\dot{y} + \mu \dot{y} = const.$

The context is a spaceship orbiting a large planet. I just don't get why this last equation is true (which it must be, since the circle solves the original problem). Sorry for my vagueness before!
• Apr 29th 2012, 03:36 AM
karatekid
Re: Showing the solution is a circle: Why is this true?
Quote:

Originally Posted by karatekid
I'm also confused as to where I'm going wrong here:
If $\displaystyle x = a\cos{\omega t} + \mu$ and $\displaystyle y = a\sin{\omega t}$ then the constant in the last expression above is zero (I hope that's not my mistake!). But then we can rearrange this (when x'(t) is non-zero and x isn't mu) to get $\displaystyle \dot{y}/\dot{x} = y/(x-\mu)$ which is false in most cases. Why can't I divide through?

I've just worked this last bit out. Late night syndrome obviously, since I was claiming sin^2 + cos^2 = 0 not 1....

I'm still unsure of the first part though.