To solve the exact differential equation $(y^3 - 2x^2y) \, dx + (2xy^2 - x^3) \, dy = 0$, we first rewrite it in the standard form:

$M(x, y) = y^3 - 2x^2y, \quad N(x, y) = 2xy^2 - x^3$

We check if the equation is exact by computing the partial derivatives:

1. $\frac{\partial M}{\partial y}$: $\frac{\partial M}{\partial y} = 3y^2 - 2x^2$

2. $\frac{\partial N}{\partial x}$: $\frac{\partial N}{\partial x} = 2y^2 - 3x^2$

Since $\frac{\partial M}{\partial y} \neq \frac{\partial N}{\partial x}$, the equation is not exact.

To solve the equation, we can attempt to find an integrating factor. A common approach is to look for a function $\mu(x,y)$ that makes the equation exact. However, in this case, let’s examine if a simpler form exists or if we can separate the variables.

We can rewrite the equation as:

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

Next, we will find an integrating factor based on the ratios of the partial derivatives:

The condition for an integrating factor depending solely on $x$ or $y$ might help. The integrating factor can often be derived if:

$\frac{\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}}{M} \quad \text{or} \quad \frac{\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x}}{N}$

In this case, we find:

$\frac{\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}}{M} = \frac{(2y^2 - 3x^2) - (3y^2 - 2x^2)}{y^3 - 2x^2y} = \frac{-y^2 - x^2}{y^3 - 2x^2y}$

This expression depends on both $x$ and $y$, indicating that finding a simple integrating factor may be complex.

Instead, let's go back to the original equation and find potential solutions directly. We can try specific substitutions or look for a function that satisfies the characteristics of the problem.

Finding a potential solution:

After some manipulation, we can integrate $M$ with respect to $x$ and then differentiate with respect to $y$ to find a function $F(x, y)$.

1. Integrate $M$ with respect to $x$:

$F(x, y) = \int (y^3 - 2x^2y) \, dx = xy^3 - \frac{2}{3}x^3y + g(y)$

2. Differentiate $F$ with respect to $y$:

$\frac{\partial F}{\partial y} = 3xy^2 - \frac{2}{3}x^3 + g'(y)$

Set this equal to $N$:

$3xy^2 - \frac{2}{3}x^3 + g'(y) = 2xy^2 - x^3$

This simplifies to:

$xy^2 - \frac{2}{3}x^3 + g'(y) = 0$

From which we can solve for $g'(y)$:

$g'(y) = -xy^2 + \frac{2}{3}x^3$

To solve for $g(y)$, we would need to find a specific relation.

Final Solution

The equation might be quite complex and could involve parameters or a numerical solution depending on the specific form. However, this framework guides toward the complete solution.

If you want, we can continue with a particular integrating factor, or examine numerical methods or specific cases to solve this equation.

Do you have any questions, or would you like more details on any part of this process? Here are some follow-up questions:

1. Would you like to explore specific integrating factors further?
2. Are you interested in numerical methods to solve this equation?
3. Would you like to see examples of similar differential equations?
4. Do you want to delve deeper into the theory of exact differential equations?
5. Are you looking for solutions in a specific context or application?

Tip: Always check if the differential equation can be simplified or transformed into a separable form, as that can often lead to easier solutions.