 Solution

Continuity on the compact rectangle gives finite constants $M={\max }_R|F|$ and $L={\max }_R|F_y|$. The real mean value theorem in the second coordinate gives $|F(x,y)-F(x,z)|\le L|y-z|$ within the rectangle. Define the Picard iterates on the entire interval byIf the previous iterate stays in the vertical range of $R$, then $|y_{n+1}(x)-b|\le M|x-a|\le Mh\le k$. Induction therefore makes all these iterates well defined on $|x-a|\le h$.

Repeated use of the Lipschitz bound gives the factorial estimateThe case $n=0$ is the direct integral bound; integrating the estimate for $n-1$ proves the next one, on either side of $a$. The series of uniform bounds converges, so the iterates converge uniformly to a continuous function $y$ taking values within the closed vertical range. The Lipschitz estimate permits passage through the integral to give $y(x)=b+{\int }_a^{x}F(t,y(t))dt$. The fundamental theorem of calculus now gives $y^{\prime }=F(x,y)$ and $y(a)=b$. Thus the solution exists on the full stated interval; no extra restriction $Lh<1$ was used.

For the specified equation take $k=1/4$, so $3/4\le y\le 5/4$. On $|x|\le h$,The condition $Mh\le k$ holds at $h^3=9/64$, namely $\boxed{h=3^{2/3}/4}$. The rectangle avoids $y=0$, so both continuity hypotheses hold there.

Direct separation gives $4y^3{y}^{\prime }=3x^2$, hence $(y^4{)}^{\prime }=(x^3{)}^{\prime }$ and the initial value yields $y^4=1+x^3$. The branch through $y(0)=1$ isIt remains positive on this interval. At $x=-1$ it reaches zero, where the differential equation is undefined, so this is its maximal real interval containing zero. The rectangle theorem supplies a sufficient interval, smaller than that actual interval.