Using Picard's Theorem, show that $$\frac{dy}{dx}=y^2 \qquad \quad y(0)=1$$ has a unique solution in the rectangle $R=[0,0.5] \times [0,2]$.
Picard's Theorem:
The differential equation $$\frac{dy}{dx}=f(x,y) \qquad \quad y(a)=b$$ has a unique solution in the rectangle $R=\{(x,y):|x-a|≤h, \; |y-b|≤k \}$ if:
(i) $f$ is continuous in $R$
(ii) $M := \max _{(x,y) \in R} |f(x,y)|$ exists and satisfies $Mh≤k$; and
(iii) $f$ is Lipshitz in $R$
I have shown that the differential equation satisfies conditions (i) and (iii), and also that $M$ exists.
Clearly, $M=2^2=4$ in this case, $h=0.5$ and $k=1$, but it is not true that $$4 \times 0.5 ≤ 1$$ Have I misunderstood something about Picard's Theorem, or is there something wrong with this question?
$h$ is not 0.5 here, but 0.25. $R$ in this example can be understood as $\{(x,y):(|x-0|\le0.25,|y-1|\le1\}$.