Solution for the Initial Value Problem:

Problem:

You are asked to analyze the existence and uniqueness of the solution to the initial value problem:

$\frac{dy}{dx} = 1 + y^2, \quad y(0) = 0$

in the rectangle defined by $|x| < 5$ and $|y| < 3$.

Step 1: Statement of the Existence and Uniqueness Theorem

The Existence and Uniqueness Theorem for a first-order differential equation states:

For the equation $\frac{dy}{dx} = f(x, y)$, if:

1. $f(x, y)$ and
2. $\frac{\partial f}{\partial y}$

are continuous in a region $R$ containing the initial point $(x_0, y_0)$, then there exists a unique solution $y(x)$ to the initial value problem in some interval around $x_0$.

Step 2: Verifying Continuity of $f(x, y)$

Here, $f(x, y) = 1 + y^2$.

• $f(x, y) = 1 + y^2$ is a polynomial in $y$ and is therefore continuous for all $x$ and $y$.
• In the given rectangle ($|x| < 5$, $|y| < 3$), $f(x, y)$ is clearly continuous.

Step 3: Verifying Continuity of $\frac{\partial f}{\partial y}$

Next, calculate $\frac{\partial f}{\partial y}$:

$\frac{\partial f}{\partial y} = \frac{\partial}{\partial y}(1 + y^2) = 2y$

• $\frac{\partial f}{\partial y} = 2y$ is also a polynomial and hence continuous for all $x$ and $y$.
• In the rectangle ($|x| < 5$, $|y| < 3$), $\frac{\partial f}{\partial y}$ is continuous.

Step 4: Applying the Theorem

Since both $f(x, y)$ and $\frac{\partial f}{\partial y}$ are continuous in the rectangle ($|x| < 5$, $|y| < 3$), the Existence and Uniqueness Theorem guarantees that:

1. A solution to the initial value problem exists.
2. The solution is unique in some interval containing $x = 0$, as long as the solution stays within the rectangle ($|x| < 5$, $|y| < 3$).

Step 5: Observing Solution Behavior

• The equation $\frac{dy}{dx} = 1 + y^2$ suggests that the slope $dy/dx$ grows rapidly as $y$ increases because of the $y^2$ term.
• The solution might eventually leave the rectangle due to the rapid growth of $y$, but this does not violate the existence and uniqueness result as long as the solution starts at $(x_0, y_0) = (0, 0)$.

Final Answer:

The solution to the initial value problem exists and is unique in the rectangle $|x| < 5$ and $|y| < 3$.

Do you have any questions or want further clarification? Here are some possible expansions:

1. Would you like to solve the differential equation explicitly?
2. Should I demonstrate numerical behavior of the solution near $x = 0$?
3. Should I analyze the behavior of $y(x)$ as $x$ approaches the rectangle’s boundary?
4. Do you need help with other initial value problems of a similar form?
5. Should I explain the significance of the rapid growth term $1 + y^2$?

Tip: In problems involving existence and uniqueness, always confirm continuity of both $f(x, y)$ and $\frac{\partial f}{\partial y}$ in the region of interest.