To explain why the initial value problem (IVP)

$\frac{dy}{dx} = \frac{1}{(y - 1)(x - 3)}, \quad y(0) = 0$

admits a unique local solution, we will apply the Existence and Uniqueness Theorem, commonly known as Picard-Lindelöf Theorem. This theorem provides conditions under which an initial value problem has a unique solution in a neighborhood of the initial point.

Step 1: Verify the function and initial condition

The given differential equation is:

$\frac{dy}{dx} = \frac{1}{(y - 1)(x - 3)}$

The initial condition is:

$y(0) = 0$

Step 2: The general form of the Existence and Uniqueness Theorem

The Existence and Uniqueness Theorem states that if the function $f(x, y) = \frac{1}{(y - 1)(x - 3)}$ is continuous in a region containing the point $(x_0, y_0)$ and its partial derivative with respect to $y$ is also continuous in that region, then there exists a unique solution to the differential equation in some interval around $x_0$.

Step 3: Check continuity of $f(x, y)$

We need to check where $f(x, y) = \frac{1}{(y - 1)(x - 3)}$ is continuous. The function $f(x, y)$ is continuous as long as the denominator is non-zero. That is:

• $y - 1 \neq 0$ (i.e., $y \neq 1$)
• $x - 3 \neq 0$ (i.e., $x \neq 3$)

Thus, $f(x, y)$ is continuous in regions where $y \neq 1$ and $x \neq 3$. The point $(x_0, y_0) = (0, 0)$ satisfies this, as neither $x = 3$ nor $y = 1$ occurs at $(0, 0)$.

Step 4: Check the partial derivative with respect to $y$

Now, we compute the partial derivative of $f(x, y)$ with respect to $y$:

$\frac{\partial}{\partial y} \left( \frac{1}{(y - 1)(x - 3)} \right) = \frac{-1}{(y - 1)^2(x - 3)}$

This partial derivative exists and is continuous as long as $y \neq 1$ and $x \neq 3$. Since the point $(0, 0)$ satisfies these conditions, the partial derivative is continuous at this point.

Step 5: Apply the Existence and Uniqueness Theorem

Since the function $f(x, y) = \frac{1}{(y - 1)(x - 3)}$ is continuous in a neighborhood of $(x_0, y_0) = (0, 0)$ and its partial derivative with respect to $y$ is continuous in the same neighborhood, by the Existence and Uniqueness Theorem, the initial value problem has a unique solution in some local neighborhood of $x = 0$.

Conclusion

Thus, the initial value problem admits a unique local solution because the conditions of the Existence and Uniqueness Theorem are satisfied near $(x_0, y_0) = (0, 0)$.

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If you'd like to go into further detail or if you have any questions, let me know!

Here are some related questions to explore:

1. What is the Existence and Uniqueness Theorem in more detail?
2. How do you use the theorem for partial differential equations?
3. Can we extend this result to global solutions?
4. What happens if $y = 1$ or $x = 3$?
5. How would the solution behave if the initial condition was different, say $y(0) = 1$?

Tip: When checking the continuity of a function in a differential equation, always identify the points where the denominator can vanish, as these are the critical points where the function might not be continuous.