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evinda

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We have the space $x^2+y^2 \leq 4$ and we consider the Cauchy problem $y'=\cos x-1-y^2, y(0)=1$.

I want to find for which $b>0$ the Picard theorem ensures the existence of the solution on $(-b,b)$.

I have thought the following:

Since $g(x,y):=\cos x-1-y^2$ is continuous as for $x$ in a space $A$ it remains to be Lipschitz as for $y$ in any closed and bounded subset of $A$, let $B=(b_1,b_2) \times [-c_1,c_1]$.

$|g(x,y_1)-g(x,y_2)|=|y_2^2-y_1^2|=|y_1-y_2| | y_1+y_2| \leq (|y_1|+|y_2|) |y_1-y_2|$.

Thus we can pick $c_1$ as $b$. Am I right?