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corey1234
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For the diff eq. (dy/dx)=(1y)/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.
 one year ago
 one year ago
corey1234 Group Title
For the diff eq. (dy/dx)=(1y)/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.
 one year ago
 one year ago

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tornado.adv4 Group TitleBest ResponseYou've already chosen the best response.0
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.
 one year ago

corey1234 Group TitleBest ResponseYou've already chosen the best response.0
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.
 one year ago

corey1234 Group TitleBest ResponseYou've already chosen the best response.0
Excuse me. Tornado.adv4, how do you know what you say is true just from the definition of uniqueness and existence. I cannot tell?
 one year ago
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