The given image presents a question regarding the solution of a partial differential equation (PDE) of the form $F(xD_x, yD_y) = f(x, y)$, where $D_x$ and $D_y$ are differential operators with respect to $x$ and $y$, respectively. The question asks how the solution can be obtained when the PDE contains terms like $xD_x$ and $yD_y$ only.

Let's go over the provided answer options:

1. Option (a): $xD_x = D_u \quad \text{and} \quad yD_y = D_v$ This implies that the operators $D_x$ and $D_y$ have been replaced with other operators $D_u$ and $D_v$, which seems to be a transformation of the variables.

2. Option (b): $x^2D_x^2 = D_u(D_u - 1) \quad \text{and} \quad y^2D_y^2 = D_v(D_v - 1)$ This suggests a quadratic transformation involving the second derivatives of $x$ and $y$.

3. Option (c): $D_x = xD_u \quad \text{and} \quad D_y = yD_v$ Here, the differential operators are multiplied by the variables $x$ and $y$, transforming them into new operators $D_u$ and $D_v$.

4. Option (d): $x = e^u \quad \text{and} \quad y = e^v$ This implies an exponential transformation of the variables $x$ and $y$.

Interpretation:

The question focuses on finding a transformation that simplifies the differential operators $xD_x$ and $yD_y$. This typically involves changing the variables to ones where the operators become simpler.

• Option (d) looks like a standard exponential transformation often used in solving such differential equations, as it linearizes the operators in terms of new variables $u$ and $v$.

• Option (a) and Option (c) provide a direct transformation of the differential operators, which might also lead to simplifications.

Thus, Option (d) seems to be the most appropriate, as it provides a classic method for simplifying PDEs with variable coefficients.

Would you like further details on this or another explanation?

Related Questions:

1. What is the method of characteristics in solving partial differential equations?
2. How do exponential transformations help in solving PDEs?
3. Can you explain how to solve second-order PDEs using operator methods?
4. What are the advantages of transforming variables in PDEs?
5. How do separation of variables and transformations differ in solving PDEs?

Tip:

When solving PDEs, changing variables or transforming the equation often simplifies the operators, allowing for more straightforward integration or solution.