Question

Let \( g(x, y) = f(x, y)e^{2x + 3y} \) be defined in \( \mathbb{R}^2 \), where \( f(x, y) \) is a continuously differentiable non-zero homogeneous function of degree 4. Then,

\[ x \frac{\partial g}{\partial x} + y \frac{\partial g}{\partial y} = 0 \text{ holds for} \]

• A. all points \( (x, y) \) in \( \mathbb{R}^2 \)
• B. all points \( (x, y) \) on the line given by \( 2x + 3y + 4 = 0 \)
• C. all points \( (x, y) \) in the region of \( \mathbb{R}^2 \) except on the line given by \( 2x + 3y + 4 = 0 \)
• D. all points \( (x, y) \) on the line given by \( 2x + 3y = 0 \)

Hint:

For homogeneous functions, use Euler's theorem to derive relations involving partial derivatives. This can help identify conditions where certain equations hold.

Answer 1

The Correct Option is B

We are given the function \( g(x, y) = f(x, y)e^{2x + 3y} \), where \( f(x, y) \) is a homogeneous function of degree 4. According to Euler's homogeneous function theorem, for a function \( f(x, y) \) that is homogeneous of degree \( n \), the following holds:
\[ x \frac{\partial f}{\partial x} + y \frac{\partial f}{\partial y} = n f(x, y) \]
Since \( f(x, y) \) is homogeneous of degree 4, we have:
\[ x \frac{\partial f}{\partial x} + y \frac{\partial f}{\partial y} = 4f(x, y) \]
Next, for the function \( g(x, y) = f(x, y) e^{2x + 3y} \), we calculate:
\[ x \frac{\partial g}{\partial x} + y \frac{\partial g}{\partial y} = x \left( \frac{\partial f}{\partial x} e^{2x + 3y} + f(x, y) \frac{\partial}{\partial x} e^{2x + 3y} \right) + y \left( \frac{\partial f}{\partial y} e^{2x + 3y} + f(x, y) \frac{\partial}{\partial y} e^{2x + 3y} \right) \]
Since \( \frac{\partial}{\partial x} e^{2x + 3y} = 2 e^{2x + 3y} \) and \( \frac{\partial}{\partial y} e^{2x + 3y} = 3 e^{2x + 3y} \), we get:
\[ x \frac{\partial g}{\partial x} + y \frac{\partial g}{\partial y} = e^{2x + 3y} \left( 4f(x, y) + f(x, y)(2x + 3y) \right) \]
For the given equation \( x \frac{\partial g}{\partial x} + y \frac{\partial g}{\partial y} = 0 \) to hold, we require:
\[ 4f(x, y) + f(x, y)(2x + 3y) = 0 \]
This simplifies to:
\[ f(x, y)(2x + 3y + 4) = 0 \]
Since \( f(x, y) \) is non-zero, we have the condition:
\[ 2x + 3y + 4 = 0 \]
Thus, the equation holds for points \( (x, y) \) on the line \( 2x + 3y + 4 = 0 \).

Final Answer
\[ \boxed{B} \quad \text{all points } (x, y) \text{ on the line given by } 2x + 3y + 4 = 0. \]