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To solve the initial value problem:

\[ \frac{dy}{dx} = x^2 (y - 2), \quad y(0) = 4 \]

we start by separating the variables. First, rewrite the differential equation:

\[ \frac{dy}{y - 2} = x^2 \, dx \]

Now integrate both sides:

\[ \int \frac{1}{y - 2} \, dy = \int x^2 \, dx \]

The integral on the left is:

\[ \int \frac{1}{y - 2} \, dy = \ln|y - 2| \]

The integral on the right is:

\[ \int x^2 \, dx = \frac{x^3}{3} \]

So we have:

\[ \ln|y - 2| = \frac{x^3}{3} + C \]

To solve for \( C \), we use the initial condition \( y(0) = 4 \):

\[ \ln|4 - 2| = \ln 2 = \frac{0^3}{3} + C \]

\[ \ln 2 = C \]

Thus, the solution becomes:

\[ \ln|y - 2| = \frac{x^3}{3} + \ln 2 \]

Exponentiate both sides to solve for \( y \):

\[ |y - 2| = e^{\frac{x^3}{3} + \ln 2} \]

\[ |y - 2| = e^{\frac{x^3}{3}} \cdot e^{\ln 2} \]

\[ |y - 2| = 2 e^{\frac{x^3}{3}} \]

Since \( y = 4 \) when \( x = 0 \), and \( y - 2 = 2 \) is positive, we take the positive root:

\[ y - 2 = 2 e^{\frac{x^3}{3}} \]

Thus, the solution is:

\[ y = 2 + 2 e^{\frac{x^3}{3}} \]

So the implicit solution is:

\[ y - 2 = 2 e^{\frac{x^3}{3}} \]

Would you like more details or have any questions?

Here are 5 questions you might ask:
1. Can you explain the process of separating variables in more detail?
2. What are some techniques for integrating more complex differential equations?
3. Can you show another example of an initial value problem?
4. How do we determine the appropriate root to take in solutions involving absolute values?
5. What are some common pitfalls when solving initial value problems?

**Tip:** Always remember to check initial conditions after finding the general solution to determine the appropriate constant of integration.

Learn how to solve the initial value problem \[ \frac{dy}{dx} = x^2 (y - 2), \quad y(0) = 4 \] using the separation of variables method. This problem involves key concepts in differential equations and integration. Suitable for undergraduate-level students.

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