Need help with ODE and Existence and Uniqueness Thm

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SUMMARY

The discussion centers on the application of the Existence and Uniqueness Theorem in the context of the initial value problem defined by the differential equation dy/dx = f(x,y), where f(x,y) = (1+x)√y and y(1)=0. It is concluded that the theorem does not guarantee a unique solution due to the partial derivative ∂f/∂y being undefined at the initial condition y=0. Two solutions are identified: y₁(x)=0 and y₂(x)=1/16((1+x)²-4)², demonstrating the theorem's implications.

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Need help with ODE and "Existence and Uniqueness Thm"

I'm currently helping my neice study for her exams and going though last years test there was this one question that I wasn't sure about.

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Considering the initial value problem,

[tex]\frac{dy}{dx} = f(x,y)[/tex]

Where [tex]f(x,y) = (1+x) \sqrt{y}[/tex] and y(1)=0.

Does the "Existence and Uniqueness Theorem" guarantee the existence of a unique solution?
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I'm pretty sure that the theorem they are referring to is this one,
http://mathworld.wolfram.com/PicardsExistenceTheorem.html
See also,
http://www.utpb.edu/scimath/wkfield/mod3/Exuni.htm

I'm thinking that it (the Theorem) doesn't guarantee a unique solution because [tex]\frac{\partial f}{\partial y}[/tex] is not well defined at the initial condition y=0.

Unfortunately I'm not too familiar with this "Existence and Uniqueness" theorem. Also the DE does seem to have a well defined solution of y(x)=0 for x>=1, so I'm really not sure. Can anyone help me out here, thanks.
 
Last edited by a moderator:
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you are quite right; the theorem cannot be used here to guarantee a unique solution.
In this case, you have (at least) two solutions fulfilling the initial value problem:
[itex]y_{1}(x)=0, y_{2}(x)=\frac{1}{16}((1+x)^{2}-4)^{2}[/itex]
 
Last edited:
Thanks for the answer arildno. I can see how you got the two solutions, y=0 by inspection and the other y = 1/16 (x^2 + 2x -3)^2 by seperating and integrating. Yes, now the question makes sense and it nicely shows the theorem in action. Thanks again :)
 
Last edited:

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