ODE's: IVP Existence/Uniqueness

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SUMMARY

The discussion centers on the initial value problem (IVP) defined by the differential equation dy/dx = tx^p with the condition y(0) = 0. Participants identified that the existence and uniqueness theorem applies, as both dy/dx and its partial derivative are continuous. The confusion arises from the interpretation of the variable 't' and its role in the equation, leading to the conclusion that different values of 'p' yield distinct differential equations, thus not violating the uniqueness condition.

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Homework Statement



Find a nontrivial solution of the IVP

dy/dx = tx^p (p is any number); y(0) = 0

Does this violate the uniqueness part of the Existence/Uniqueness Theorem? Explain

Homework Equations


The Attempt at a Solution



I've found two solutions: one for p = 1 and one for all other cases.

It seems that having two solutions would violate the uniqueness part of the theorem. Both dy/dx and it's partial derivative of y are continuous, so the Theorem can apply, but not sure other than that.

Any help would be appreciated.
Thanks
 
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No, having different solutions for different values of t does NOT violate the theorem. With different values of p we have different differential equations. But how does 't' come into this equation? Are you sure the left isn't "dx/dt"?
 
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