y' = 3x(y-1)^1/3, y(3) = -7. determine largest open interval that contains xo = 3 where the solution is unique. find the solution and determine the interval

We are given the initial value problem (IVP):

\[
y' = 3x (y - 1)^{1/3}, \quad y(3) = -7
\]

### Step 1: Apply the **Existence and Uniqueness Theorem** (Picard's Theorem)

The Existence and Uniqueness Theorem states that if the function \( f(x, y) = 3x(y - 1)^{1/3} \) and its partial derivative with respect to \( y \) are continuous in some rectangle containing the point \( (3, -7) \), then there exists a unique solution to the differential equation in an interval around \( x_0 = 3 \).

First, let's check for continuity of the function \( f(x, y) = 3x (y - 1)^{1/3} \).

- \( f(x, y) \) is continuous for \( y \neq 1 \), because the term \( (y - 1)^{1/3} \) is undefined or discontinuous at \( y = 1 \).
  
Next, compute the partial derivative of \( f(x, y) \) with respect to \( y \):

\[
\frac{\partial f}{\partial y} = 3x \cdot \frac{1}{3}(y - 1)^{-2/3} = x (y - 1)^{-2/3}
\]

- This derivative is undefined when \( y = 1 \), which means the Existence and Uniqueness Theorem fails at \( y = 1 \).

Thus, the solution exists and is unique in the largest interval where \( y(x) \neq 1 \). We now proceed to find the explicit solution and determine the interval.

### Step 2: Solve the Differential Equation

We solve the differential equation using **separation of variables**.

Given:
\[
y' = 3x (y - 1)^{1/3}
\]

Separate the variables:
\[
\frac{dy}{(y - 1)^{1/3}} = 3x \, dx
\]

Integrate both sides:
\[
\int (y - 1)^{-1/3} \, dy = \int 3x \, dx
\]

Using the standard integral formula \( \int u^n \, du = \frac{u^{n+1}}{n+1} \) for the left-hand side:

\[
\frac{3}{2} (y - 1)^{2/3} = \frac{3x^2}{2} + C
\]

Simplify:
\[
(y - 1)^{2/3} = x^2 + C_1
\]
where \( C_1 = \frac{2}{3}C \).

### Step 3: Apply the Initial Condition \( y(3) = -7 \)

Substitute \( x = 3 \) and \( y = -7 \) into the equation \( (y - 1)^{2/3} = x^2 + C_1 \):

\[
(-7 - 1)^{2/3} = 3^2 + C_1
\]

\[
(-8)^{2/3} = 9 + C_1
\]

Since \( (-8)^{2/3} = 4 \), we have:

\[
4 = 9 + C_1 \quad \Rightarrow \quad C_1 = -5
\]

Thus, the equation becomes:

\[
(y - 1)^{2/3} = x^2 - 5
\]

### Step 4: Solve for \( y \)

Raise both sides to the \( 3/2 \)-power:

\[
y - 1 = (x^2 - 5)^{3/2}
\]

Finally, solve for \( y \):

\[
y(x) = 1 + (x^2 - 5)^{3/2}
\]

### Step 5: Determine the Largest Interval of Uniqueness

For the solution to be real, the term \( (x^2 - 5)^{3/2} \) must be defined, which requires:

\[
x^2 - 5 \geq 0 \quad \Rightarrow \quad x^2 \geq 5 \quad \Rightarrow \quad |x| \geq \sqrt{5}
\]

Therefore, the solution is valid in the intervals \( (-\infty, -\sqrt{5}] \cup [\sqrt{5}, \infty) \).

Since we are asked to find the largest open interval that contains \( x_0 = 3 \), the relevant interval is:

\[
[\sqrt{5}, \infty) \quad \text{or} \quad [\sqrt{5}, 3) \text{ for uniqueness.}
\]

### Final Answer:
- The solution is \( y(x) = 1 + (x^2 - 5)^{3/2} \).
- The largest open interval for the solution's uniqueness that contains \( x_0 = 3 \) is \( (\sqrt{5}, \infty) \).

---

Would you like more details on any of the steps? Here are five related questions:

1. How does the Existence and Uniqueness Theorem work in other cases?
2. Can you explain the separation of variables technique in more detail?
3. Why did we need to check the partial derivative of \( f(x, y) \)?
4. How do you integrate more complex expressions like \( (y - 1)^{-1/3} \)?
5. How do we choose the constant \( C \) from initial conditions?

**Tip:** When solving differential equations, always check both the function and its derivative for continuity to apply the uniqueness theorem correctly.

Discover how to solve the differential equation y' = 3x(y-1)^(1/3) with the initial condition y(3) = -7. Learn how to apply the Existence and Uniqueness Theorem and use separation of variables to find the explicit solution. Suitable for undergraduate calculus students.

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