Consider the differential equation 
y' = 

sqrt(y2 − 9)
.
Let 
f(x, y) = 

y2 − 9
.
 Find the partial derivative of f.
df
dy
 = 
 
Determine a region of the xy-plane for which the given differential equation would have a unique solution whose graph passes through a point 
(x0, y0)
 in the region.
A unique solution exits in the region consisting of all points in the x y-plane except (0, 3) and (0, −3).
A unique solution exists in the region −3 < y < 3.    
A unique solution exits in the entire x y-plane.
A unique solution exits in the regions y < −3, −3 < y < 3, and y > 3.
A unique solution exists in the region y < −3 or y > 3.
Determine whether Theorem 1.2.1 guarantees that the differential equation possesses a unique solution through 
(7, 3).
Yes
No    


Let's break down the problem step by step:

### Step 1: Find the partial derivative of \( f(x, y) \)
You are given the function:

\[
f(x, y) = y^2 - 9
\]

To find the partial derivative of \( f(x, y) \) with respect to \( y \), we treat \( x \) as a constant and differentiate \( f(x, y) \) with respect to \( y \).

\[
\frac{\partial f}{\partial y} = \frac{\partial}{\partial y} (y^2 - 9) = 2y
\]

Thus, the partial derivative is:

\[
\frac{\partial f}{\partial y} = 2y
\]

---

### Step 2: Determine the region for a unique solution
The differential equation is:

\[
y' = \sqrt{y^2 - 9}
\]

From the form of the equation, the square root term \( \sqrt{y^2 - 9} \) is only defined when \( y^2 - 9 \geq 0 \), or equivalently when:

\[
y \geq 3 \quad \text{or} \quad y \leq -3
\]

The regions where the differential equation is defined are when \( y > 3 \) or \( y < -3 \). Therefore, we look at the options provided:

- A unique solution exists in the region consisting of all points in the xy-plane except (0, 3) and (0, −3).
- A unique solution exists in the region \( -3 < y < 3 \).
- A unique solution exists in the entire xy-plane.
- A unique solution exists in the regions \( y < -3 \), \( -3 < y < 3 \), and \( y > 3 \).
- A unique solution exists in the region \( y < -3 \) or \( y > 3 \).

Since the differential equation is only valid where \( y > 3 \) or \( y < -3 \), **the correct region is**:

\[
\text{A unique solution exists in the region } y < -3 \text{ or } y > 3.
\]

---

### Step 3: Apply Theorem 1.2.1 (Existence and Uniqueness Theorem)
Theorem 1.2.1 (typically Picard’s Existence and Uniqueness Theorem) states that a first-order differential equation \( y' = f(x, y) \) has a unique solution passing through \( (x_0, y_0) \) if the function \( f(x, y) \) and its partial derivative \( \frac{\partial f}{\partial y} \) are continuous in some region around \( (x_0, y_0) \).

Here, \( f(x, y) = \sqrt{y^2 - 9} \). As discussed earlier, the differential equation is only defined when \( y \geq 3 \) or \( y \leq -3 \).

Now, evaluate at the point \( (7, 3) \):
- At \( y = 3 \), the differential equation \( y' = \sqrt{y^2 - 9} = \sqrt{3^2 - 9} = 0 \).
- The partial derivative of \( f(x, y) \) is \( \frac{\partial}{\partial y} \sqrt{y^2 - 9} = \frac{y}{\sqrt{y^2 - 9}} \). At \( y = 3 \), this becomes undefined because the denominator \( \sqrt{y^2 - 9} = 0 \).

Thus, **Theorem 1.2.1 does not guarantee a unique solution** through the point \( (7, 3) \), since the function is not continuous at \( y = 3 \).

The answer to whether the theorem guarantees a unique solution is:

\[
\boxed{\text{No}}
\]

---

### Questions for Further Understanding:
1. What is the significance of the square root term in the differential equation?
2. How would the solution change if the square root was replaced with a different function?
3. What are the geometric interpretations of the regions where a solution is guaranteed?
4. How does Picard’s theorem apply to more complex systems of differential equations?
5. How would we solve the differential equation \( y' = \sqrt{y^2 - 9} \) explicitly?

### Tip:
Always check the domain of the function and its derivatives when applying existence and uniqueness theorems to differential equations.

Consider the differential equation y' = sqrt(y^2 − 9). Let f(x, y) = y^2 − 9. Find the partial derivative of f, df/dy = ? Determine a region of the xy-plane for which the given differential equation would have a unique solution whose graph passes through a point (x0, y0) in the region. A unique solution exists in the region y < −3 or y > 3. Determine whether Theorem 1.2.1 guarantees that the differential equation possesses a unique solution through (7, 3).

Explore the unique solutions for the differential equation y' = sqrt(y^2 - 9), including determining the region for a solution and applying Picard's Existence and Uniqueness Theorem. Learn how to find the partial derivative of f(x, y) and assess whether a unique solution exists at specific points.

Math Problem Statement

y2 − 9 . Find the partial derivative of f. df dy

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