Chapter 13.4, Problem 11E

Let $f\left(x,y\right)=y^2$.

1. Solve explicitly for $\varphi \left(x,y\right)$, the solution to

   $y^{\prime }=f\left(x,y\right)$, $y\left(0\right)=1$ and for $\varphi \left(x,1+\epsilon \right)$, the solution to

   $y^{\prime }=f\left(x,y\right)$, $y\left(0\right)=1+\epsilon$, where $\epsilon >0$.

2. Show that $\left|\partial f/\partial y\right|\le 4$ on the rectangle

   $R=\left\{\left(x,y\right):-1<x<1,-2<y<2\right\}$.

3. Using the result of part (b) in inequality $\left(8\right)$ in Theorem $9$, it appears that

   $\left|\varphi \left(x,1\right)-\varphi \left(x,1+\epsilon \right)\right|\le \epsilon e^4$, $x$ in $\left(-1,1\right)$.

   Show that this is not true (even for $\epsilon >0$ small) by using the results of part(a) to verify that $\varphi \left(x,1\right)\to 1+{\epsilon }^{-1}$ as $x\to \frac{1}{1+\epsilon }$, while $\left|\varphi \left(x,1+\epsilon \right)\right|\to +\infty$ as $x\to \frac{1}{1+\epsilon }$.

4. Which assumption in Theorem $9$ is not satisfied so that inequality $\left(8\right)$ is not applicable to this situation?