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anonymous
 3 years ago
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.
anonymous
 3 years ago
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.

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0The 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
 3 years ago
Best ResponseYou've already chosen the best response.0actually, 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
 3 years ago
Best ResponseYou've already chosen the best response.0Excuse 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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