Uniqueness of Solution for y' = y(siny) + x with Initial Value of f(0) = -1

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SUMMARY

The differential equation y' = y(siny) + x has a unique solution with the initial condition f(0) = -1. The continuity of the partial derivative with respect to y, specifically del(x)/del(y) = siny + cosy(y), confirms the uniqueness of the solution within the specified rectangle containing the point (0, -1). The requirement for uniqueness is satisfied as the function f(x,y) is Lipschitz continuous in y, which is implied by the continuous derivative.

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Homework Statement


y' = y(siny) + x does this have a unique solution ,
with the intial value of f(0)= -1

The Attempt at a Solution



the partial dervative with respect to y is
del(x)/del(y) = siny + cosy(y) this is continuous on the rectangle including
(0 , -1 ) so this is unique.

is this right
 
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Yes, that is correct. Strictly speaking, for uniqueness you only requrire that f(x,y), in dy/dx= f(x,y), be "Lischitz" in y but that is implied by continuous derivative so that is sufficient.
 
im just curious do you have your ph.d. in math
 

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