Prove y' = |y| , y(0)=0 has unique solution, though states of thm's of existence, uniqueness not true

Panos26

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I need to prove that the DE y' = |y| ,y(0)=0 is an example at which the states of the theorem of existence and uniqueness are not true but the problem has unique solution

Can anyone explain this to me please?!
 

Subhotosh Khan

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Can you state for us:

the theorem of existence and

the theorem of uniqueness

Of solutions of DE?
 

Panos26

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That for a DE y'= f(x,y) if :
•f is continuous
•fy is continuous
The problem y'=f(x,y) with initial condition y(x0) = y0 has a unique solution in the (x0-h,x0+h)
 

Subhotosh Khan

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I need to prove that the DE y' = |y| ,y(0)=0 is an example at which the states of the theorem of existence and uniqueness are not true but the problem has unique solution

Can anyone explain this to me please?!
You state:

"... the states of the theorem of existence and uniqueness are not true ...."

Can you defend/refute the statement?
 

Panos26

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I need to prove that although the statements of the theorem are not true, the problem has a unique solution
 

Dr.Peterson

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I would suggest that you first show that f_y is not continuous (that is, one condition of the theorem is not true). That is not hard.

Second, try solving the problem; I would break it into two cases (y >= 0 and y < 0) and see what happens.
 
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