The given partial differential equation is:

$x^2 u_{xx} + 2xy u_{xy} + y^2 u_{yy} + xy u_x + y^2 u_y = 0$

This is a second-order linear partial differential equation with variable coefficients. To find the general solution, we check if it has any resemblance to known types of PDEs. Here, the equation appears to be Euler-Cauchy (homogeneous) in nature, as the powers of $x$ and $y$ in the terms suggest scaling properties.

Step 1: Check for a solution of the form

Let us assume the solution is separable and try a solution of the form: $u(x, y) = x^m y^n$ where $m$ and $n$ are constants to be determined.

Step 2: Compute the derivatives

Using the assumed solution $u(x, y) = x^m y^n$, we calculate the necessary partial derivatives:

• First derivatives: $u_x = \frac{\partial}{\partial x} (x^m y^n) = m x^{m-1} y^n$ $u_y = \frac{\partial}{\partial y} (x^m y^n) = n x^m y^{n-1}$
• Second derivatives: $u_{xx} = \frac{\partial}{\partial x} (m x^{m-1} y^n) = m(m-1) x^{m-2} y^n$ $u_{yy} = \frac{\partial}{\partial y} (n x^m y^{n-1}) = n(n-1) x^m y^{n-2}$ $u_{xy} = \frac{\partial}{\partial y} (m x^{m-1} y^n) = mn x^{m-1} y^{n-1}$

Step 3: Substitute into the PDE

Now, substitute these expressions into the original equation: $x^2 (m(m-1) x^{m-2} y^n) + 2xy (mn x^{m-1} y^{n-1}) + y^2 (n(n-1) x^m y^{n-2}) + xy (m x^{m-1} y^n) + y^2 (n x^m y^{n-1}) = 0$ Simplifying each term: $m(m-1) x^m y^n + 2mn x^m y^n + n(n-1) x^m y^n + m x^m y^n + n x^m y^n = 0$ Combine like terms: $x^m y^n \left[m(m-1) + 2mn + n(n-1) + m + n\right] = 0$ Factor the common terms: $x^m y^n \left[m^2 + n^2 + m + n\right] = 0$ For a non-trivial solution ($x^m y^n \neq 0$), we require: $m^2 + n^2 + m + n = 0$

Step 4: Solve the characteristic equation

The equation $m^2 + n^2 + m + n = 0$ can be written as: $(m + \frac{1}{2})^2 + (n + \frac{1}{2})^2 = \frac{1}{2}$ This is the equation of a circle in the $(m, n)$-plane with center $(-1/2, -1/2)$ and radius $\sqrt{1/2}$.

Step 5: General solution

The general solution to the PDE can be expressed as a linear combination of solutions of the form $x^m y^n$, where $m$ and $n$ satisfy the characteristic equation. In practice, this leads to a more general solution involving arbitrary functions of $x$ and $y$.

Thus, the general solution is: $u(x, y) = F(xy) + G\left(\frac{x}{y}\right)$ where $F$ and $G$ are arbitrary differentiable functions of their respective arguments.

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Would you like further clarification or details on any part of the solution?

Here are 5 related questions to expand your understanding:

1. How does the form of the solution $u(x, y) = x^m y^n$ help in solving PDEs with variable coefficients?
2. What are Euler-Cauchy equations, and how do they relate to the given PDE?
3. Can we always apply the method of separation of variables to such PDEs?
4. What other types of partial differential equations can be solved using similar techniques?
5. How would you solve the characteristic equation if the coefficients were different?

Tip: Understanding the method of separation of variables can greatly simplify the process of solving complex differential equations.