anonymous
  • anonymous
For the diff eq. (dy/dx)=(1-y)/x, I see that it is not continuous or defined at x=0. however, if you graph the solution y=cx+1, then you get solutions passing thru (0,1). This violates the existence part of the existence and uniqueness theorem doesn't it? I mean, I see that uniqueness fails because the partial derivative with respect to y of f is not defined at x=0. But the existence part should fail everywhere on x=0 as well. But it does not at 0,1. Many solution curves pass through there. ? Please help.
MIT 18.03SC Differential Equations
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
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anonymous
  • anonymous
The solution to your differential equation is actually y=1+c/x. The existence and uniqueness theorem doesn't say when there won't be a solution; it just tells you under what conditions can we guarantee a (unique) solution. If the existence and uniqueness theorem does not apply, it tells you nothing. There could be solutions, or a unique solution, or no solution at all.
anonymous
  • anonymous
actually, the existence and the uniqueness fail at x=0 so there should be no unique solution there. in other words, no solution at all.
anonymous
  • anonymous
Excuse me. Tornado.adv4, how do you know what you say is true just from the definition of uniqueness and existence. I cannot tell?

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