# Maximal interval where this IVP has a unique solution

I'm investigating where the IVP $$y'=-y, \enspace y(0)=3$$

has an unique solution. The right hand side satisfies a Lipschitz-condition for the constant $$K$$ with respect to variable $$y$$, and clearly the function $$-y$$ is continuous. This implies there is an interval $$I:=(-\delta, \delta$$) where the solution is unique. According to the proof, we have that $$\delta = \min\{a,\frac{b}{2M}\}$$, where $$M = \sup f(y)_{y\in R}$$. Define $$R$$ as the rectangle $$R : \vert x \vert \leq a, \enspace \enspace \vert y-3\vert\leq b$$

which means $$M=(b+3)$$, the maximal value of $$f(x,y)=-y$$ in R. The interval $$I$$ is maximal, when $$\frac{b}{2M}$$ is maximal, for we can then set $$a$$ as its equal. However, as the function $$q(p)=\frac{b}{2(b+3)}$$

has no extrema and no maximal points, this implies that $$I$$ can be made as large as we wish. Is it now correct to say that the unique solution exists globally?

Your equation is linear, with continuous coefficients on $$\mathbb{R}$$. Therefore, your solutions can be extended to all of $$\mathbb{R}$$ and there is a unique solution through every point.