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Theorem 1. (Existence theorem): Suppose that $f(x, y)$ is a continuous function in some region

$$R = \{(x, y) : |x − x_0| ≤ a, |y − y_0| ≤ b\}\qquad a, b > 0$$

Since $f$ is continuous in a closed and bounded domain, it is necessarily bounded in $R$, i.e., there exists $K > 0$ such that $|f(x, y)| \leq K$ $\forall (x, y) \in R$. Then the IVP $$ y'=f(x,y), \qquad y(x_0)=y$$ has at least one solution $y = y(x)$ defined in the interval $|x − x_0| \leq \alpha$ where $α = \min(a , b/K )$.

Why is the interval for finding solution $x=\min(a,b/K)$? Particularly why is there $b/k$ and not just $b$?

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closed as off-topic by Namaste, user99914, mau, Ethan Bolker, José Carlos Santos Mar 22 '18 at 23:21

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    $\begingroup$ Have you looked at the proof of the theorem? $\endgroup$ – Alex R. Mar 22 '18 at 17:31
  • $\begingroup$ Take $y' \equiv 2$, $y(0) = 0$, $a = b = 1$, and look what happens. $\endgroup$ – user539887 Mar 22 '18 at 22:03