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The given differential equation is:

\[
y(y^2 - 2x^2)dx + x(2y - x^2)dy = 0
\]

To solve this, let’s check if this is an exact differential equation. We can rewrite the equation in the form:

\[
M(x, y)dx + N(x, y)dy = 0
\]

where:
- \( M(x, y) = y(y^2 - 2x^2) \)
- \( N(x, y) = x(2y - x^2) \)

For the equation to be exact, the following condition must hold:

\[
\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}
\]

Let's compute these partial derivatives:

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

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

Since \( \frac{\partial M}{\partial y} \neq \frac{\partial N}{\partial x} \), this is **not an exact equation**.

### Next Step: Try an Integrating Factor
One approach is to find an integrating factor, possibly based on the form of the equation. Since the equation is in a non-exact form, a common method is to find an integrating factor that depends on either \( x \) or \( y \).

Would you like me to try finding an integrating factor for this equation or explore another method of solving it? Let me know if you'd prefer a specific approach.

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### Related Questions:
1. What are the conditions for a differential equation to be exact?
2. How do you compute partial derivatives in exactness checks?
3. What are some common methods to find an integrating factor?
4. When is a differential equation called linear or non-linear?
5. Can we apply separation of variables to this equation?

#### Tip:
Always check for exactness when you encounter a first-order differential equation of the form \( M(x, y)dx + N(x, y)dy = 0 \).

Solve the differential equation: y(y^2 - 2x^2)dx + x(2y - x^2)dy = 0

Learn how to solve the differential equation y(y^2 - 2x^2)dx + x(2y - x^2)dy = 0. This problem involves testing for exactness, partial derivatives, and exploring the use of an integrating factor.

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