June 20th, 2017, 02:22 PM  #11 
Senior Member Joined: Dec 2015 From: Earth Posts: 157 Thanks: 21 
If y is set divisor then $\displaystyle y\neq 0 $

June 20th, 2017, 03:38 PM  #12  
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,158 Thanks: 90  Quote:
Quote:
Last edited by skipjack; June 20th, 2017 at 10:10 PM.  
June 20th, 2017, 10:13 PM  #13 
Global Moderator Joined: Dec 2006 Posts: 18,060 Thanks: 1396  
June 21st, 2017, 05:04 AM  #14 
Math Team Joined: Dec 2013 From: Colombia Posts: 6,973 Thanks: 2296 Math Focus: Mainly analysis and algebra  
June 21st, 2017, 07:35 AM  #15  
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,158 Thanks: 90  Quote:
y=x$\displaystyle {^2}$ BC: y(2)=4 Your turn. I sympathize slightly with the postRomsek trend of the thread because when I took DE I understood nothing and hence remember nothing. It was just one method after the other when I was really looking for a startfromscratch analysis of individual equations. I got an A when I deserved an F and I suspect the people who got an F or D deserved an A because they tried to understand things but under the circumstances couldn't because there was nothing to understand  somebody "just" thought of a method. As for skipjack's claim that conditions for uniqueness don't apply to this equation, see, for example: http://www2.fiu.edu/~aladrog/BasicTheoryLinear.pdf "There exists a unique function f(x) defined on [a, b] such that f(x) is a solution of the linear differential equation and satisfies f(x0) = c0, f’(x0) = c1, f’’(x0) = c2, … , f$\displaystyle ^{n1}$(n1)(x0) = cn1." Last edited by skipjack; June 21st, 2017 at 02:26 PM.  
June 21st, 2017, 08:04 AM  #16  
Math Team Joined: Dec 2013 From: Colombia Posts: 6,973 Thanks: 2296 Math Focus: Mainly analysis and algebra  Quote:
My function meets your boundary condition because $(2)^2=4$. Quote:
Last edited by skipjack; June 21st, 2017 at 02:28 PM.  
June 21st, 2017, 08:25 AM  #17 
Math Team Joined: Dec 2013 From: Colombia Posts: 6,973 Thanks: 2296 Math Focus: Mainly analysis and algebra 
Your theorem (implicitly) assumes that $f(x)$ is analytic everywhere (in the given interval). There's no reason to assume that any derivative past the first exists in $xy'=2y$ and indeed, for my solution, $y''$ doesn't exist at $x=0$. This is confirmed by the corollary which requires $y''(x_0)=0$ in the equation $xy'=2y$ to hold true. You have to understand the theorem before quoting it blindly. Last edited by v8archie; June 21st, 2017 at 08:47 AM. 
June 21st, 2017, 09:01 AM  #18  
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,158 Thanks: 90  Quote:
Obviously the derivatives have to exist to order n. Let's assume everyone else understands whatever you are talking about and there is no point in trying to explain it to me. Last edited by skipjack; June 21st, 2017 at 02:29 PM.  
June 21st, 2017, 09:32 AM  #19 
Math Team Joined: Dec 2013 From: Colombia Posts: 6,973 Thanks: 2296 Math Focus: Mainly analysis and algebra 
Fair enough. But the understanding you have is wrong. In my solution, all derivatives up to $n=1$ exist. So the counterexample contradicts your understanding of the theorem. 
June 21st, 2017, 09:35 AM  #20 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,158 Thanks: 90 
As usual, don't know what you are talking about, so please don't bother explaining  remember, everyone else does so it's entirely unnecessary.
Last edited by skipjack; June 21st, 2017 at 02:31 PM. 

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