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The given differential equation is:

\[
xy \ln \left( \frac{x}{y} \right) dx + \left( y^2 - x^2 \ln \left( \frac{x}{y} \right) \right) dy = 0
\]

This appears to be a first-order differential equation. To solve it, we can attempt to rewrite the equation in a form that allows us to identify an integrating factor or use a substitution that simplifies the logarithmic terms.

### Suggested Steps:
1. **Substitution:**
   Since the logarithmic terms involve \(\frac{x}{y}\), it suggests a substitution of the form \(u = \frac{x}{y}\), which may simplify the equation. Let's introduce this new variable and rewrite the equation accordingly.

2. **Differentiating the substitution:**
   If \( u = \frac{x}{y} \), then:
   \[
   du = \frac{1}{y} dx - \frac{x}{y^2} dy
   \]
   This substitution should allow us to express the equation in terms of \(u\), and potentially solve the resulting simpler differential equation.

Let me work through this substitution and check if it simplifies the equation further.From the substitution \( u = \frac{x}{y} \), we obtained the differentials:

\[
\frac{du}{dx} = \frac{1}{y} \quad \text{and} \quad \frac{du}{dy} = -\frac{x}{y^2}
\]

Now, I will proceed to substitute \( u = \frac{x}{y} \) and its differentials into the original equation and simplify. Let's see if this yields a more tractable form.After substituting \( u = \frac{x}{y} \), the equation does not appear to simplify significantly using this method. However, I will now explore a different approach to find an integrating factor or apply a symmetry transformation to solve the differential equation.

Let's revisit the equation:

\[
xy \ln \left( \frac{x}{y} \right) dx + \left( y^2 - x^2 \ln \left( \frac{x}{y} \right) \right) dy = 0
\]

The equation might require an integrating factor to proceed further, which could involve a functional dependence on \( x \) and \( y \). Let me attempt to find an integrating factor to help solve this.To solve this differential equation more effectively, we should check if it is **exact**, meaning that there exists a potential function \( \phi(x, y) \) such that:

\[
\frac{\partial \phi}{\partial x} = M(x, y) = xy \ln \left( \frac{x}{y} \right)
\quad \text{and} \quad
\frac{\partial \phi}{\partial y} = N(x, y) = y^2 - x^2 \ln \left( \frac{x}{y} \right)
\]

If the equation is exact, then:

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

Let me check this condition to determine if the equation is exact.The partial derivatives are:

\[
\frac{\partial M}{\partial y} = x \left( \ln \left( \frac{x}{y} \right) - 1 \right)
\quad \text{and} \quad
\frac{\partial N}{\partial x} = -x \left( 2 \ln \left( \frac{x}{y} \right) + 1 \right)
\]

Since these two expressions are not equal, the differential equation is not exact.

Given that the equation is not exact, an integrating factor may be required to proceed with solving it. Integrating factors often depend on \(x\) or \(y\), and finding one can simplify the equation. Would you like to explore potential integrating factors, or should I proceed with another method, such as attempting a different substitution? Let me know if you need more details on any part so far.

Here are 5 related questions:
1. How do you determine if a differential equation is exact?
2. What are common methods for finding an integrating factor?
3. What types of substitutions are useful for simplifying logarithmic terms?
4. Can this equation be simplified by converting to polar coordinates?
5. What is the general approach for solving first-order linear differential equations?

**Tip:** Always check the exactness of a differential equation before searching for integrating factors.

Solve xy ln(x/y) dx + (y^2 - x^2 ln(x/y)) dy = 0

Explore the step-by-step solution for solving the differential equation xy ln(x/y) dx + (y^2 - x^2 ln(x/y)) dy = 0. This problem covers concepts such as logarithmic functions, substitution methods, and checking for exactness in differential equations. Suitable for undergraduate calculus students.

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