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?!
 
Can you state for us:

the theorem of existence and

the theorem of uniqueness

Of solutions of DE?
 
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)
 
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?
 
I need to prove that although the statements of the theorem are not true, the problem has a unique solution
 
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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