December 17th, 2009, 03:46 AM |
#1 |

Member Joined: Dec 2009 Posts: 34 Thanks: 0 | ODE question 2
Given this ODE: x' = x+y-xy^2 y' = -x-y+x^2y and a function: u(x,y) = x^2+y^2-2ln|xy-1| prove that for each solution ( x(t), y(t) ) of this system, such as: x(t)*y(t) != 1 (doesn't equal...) , there exists a constant C such as: u ( x(t), y(t) ) = C for every t in R. My attempt: It's very clear that we need to look at the derivative of u... If it will be 0, then we'll get what we need...But since I haven't got that much knowledge in 2 variables functions, I can't really see what is the derivative of u, as well as how to solve this ODE... So, I really need your help in: 1. Solving the ODE. 2. What is the derivative of u(t)? TNX a lot! |

December 17th, 2009, 05:01 AM |
#2 |

Global Moderator Joined: Dec 2006 Posts: 20,653 Thanks: 2086 |
For xy ?1, du/dt = 2xx' + 2yy' - (xy' + yx')/(xy - 1) = 2x(x + y - xy²) + 2y(-x - y + x²y) - 2(x(-x - y + x²y) + y(x + y - xy²))/(xy - 1) = 2x² - 2y² - 2(-x² + x³y + y² - xy³)/(xy - 1), = 2x² - 2y² -2(x² - y²)(xy - 1)/(xy - 1) = 0. Hence u(x, y) is a constant. Other solutions are x = y = 0 and (x, y) = (Ae^t, (1/A)e^(-t)), where A is a non-zero constant. |

December 17th, 2009, 06:30 AM |
#3 |

Member Joined: Dec 2009 Posts: 34 Thanks: 0 | Re: ODE question 2
Wow, so we don't even need to solve the the system... Nice one... Tnx a lot! |