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The problem involves a partial differential equation (PDE) of the form \( F(xD_x, yD_y) = f(x, y) \), where \( D_x \) and \( D_y \) are the differential operators with respect to \( x \) and \( y \), respectively.

### Step-by-Step Understanding:

- The equation is written with variable coefficients that contain terms like \( xD_x \) and \( yD_y \), which are common in PDEs, especially when dealing with separable variables or eigenvalue problems.
- You are asked to identify which of the given options simplifies or transforms the PDE.

#### Explanation of Options:
- **(a) \( xD_x = D_u \) and \( yD_y = D_v \)**:
  - This suggests a possible substitution of variables where \( u = \ln(x) \) and \( v = \ln(y) \), transforming the differential operators.

- **(b) \( x^2 D_x^2 = D_u(D_u - 1) \) and \( y^2 D_y^2 = D_v(D_v - 1) \)**:
  - This applies a second-order differential operator to the transformed variables \( u \) and \( v \). This is typical of certain transformations that simplify PDEs.

- **(c) \( D_x = xD_u \) and \( D_y = yD_v \)**:
  - This suggests another form of variable transformation, possibly related to scaling or logarithmic changes in \( u \) and \( v \).

- **(d) \( x = e^u \) and \( y = e^v \)**:
  - This directly transforms the variables \( x \) and \( y \) into exponential terms in \( u \) and \( v \). This type of transformation is often used to simplify equations with multiplicative variables.

### Solution:
The correct approach involves transforming the equation by logarithmic substitution, which leads to simpler differential operators. Based on the given options, **(d) \( x = e^u \) and \( y = e^v \)** seems to be the correct substitution**, as it directly transforms the equation into a more solvable form by making the coefficients constant.

Would you like to go deeper into any part of the solution?

### Related Questions:
1. What are partial differential operators and how are they used in solving PDEs?
2. How do variable transformations help in solving PDEs?
3. What are the typical methods to solve second-order PDEs?
4. Why are logarithmic substitutions common in PDE transformations?
5. Can you explain how eigenvalue problems relate to differential operators?

**Tip**: Transforming variables often simplifies complex PDEs by making coefficients constant, which can help convert difficult equations into standard forms.

If the partial differential equation of the form F(xDx, yDy) = f(x, y) with variable coefficients contains terms like xDx and yDy only, then its solution is obtained by setting?

Learn how to solve a partial differential equation (PDE) of the form F(xDx, yDy) = f(x, y) by applying variable substitution and logarithmic transformations. This example explores concepts in differential operators, second-order transformations, and is suitable for advanced undergraduate students.

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