The given image contains a partial differential equation and a problem statement. Upon reading and interpreting the text, it says:

$\frac{\partial u}{\partial x} + 3y \frac{\partial u}{\partial y} = 0 \quad (1)$

The task appears to be:
"If $u = x$, then show that (1) holds."

Solution

1. Problem Overview:
   We are given a function $u = x$ and asked to verify whether it satisfies the equation:

   $\frac{\partial u}{\partial x} + 3y \frac{\partial u}{\partial y} = 0.$

2. Partial Derivatives:

   • $u = x$
     • $\frac{\partial u}{\partial x} = \frac{\partial}{\partial x}(x) = 1$.
     • $\frac{\partial u}{\partial y} = \frac{\partial}{\partial y}(x) = 0$ (since $u = x$ does not depend on $y$).

3. Substitute into the Equation: Substituting the partial derivatives into the equation $\frac{\partial u}{\partial x} + 3y \frac{\partial u}{\partial y} = 0$:

   $1 + 3y \cdot 0 = 1 \neq 0.$

4. Conclusion: The equation $\frac{\partial u}{\partial x} + 3y \frac{\partial u}{\partial y} = 0$ does not hold for $u = x$.

Would you like to explore further details or discuss related concepts?

5 Follow-up Questions:

1. How do partial differential equations differ from ordinary differential equations?
2. What are the general methods to solve first-order PDEs like the given one?
3. Could a different form of $u(x, y)$ satisfy the given PDE?
4. What is the significance of boundary conditions in PDEs?
5. How do characteristics help in solving PDEs analytically?

Tip: When dealing with PDEs, always check if the function you're testing satisfies all the terms—sometimes the solution is trivial, but other times it requires transformations.