Concept:
• A differential equation \( \frac{dy}{dx} = F(x, y) \) is homogeneous if \( F(x, y) \) is a homogeneous function of degree zero.
• A function is homogeneous of degree zero if \( F(\lambda x, \lambda y) = F(x, y) \) for any non-zero scalar \( \lambda \).
• This effectively means \( F(x, y) \) can be expressed solely as a function of the ratio \( \frac{y}{x} \).
Step 1: Test function (i)
\( F(x, y) = 3x + 2y \)
\[ F(\lambda x, \lambda y) = 3(\lambda x) + 2(\lambda y) = \lambda(3x + 2y) \neq F(x, y) \]
This is degree 1, not homogeneous of degree 0.
Step 2: Test function (ii)
\( F(x, y) = \sin \frac{y}{x} + \log y - \log x = \sin \frac{y}{x} + \log \left(\frac{y}{x}\right) \)
\[ F(\lambda x, \lambda y) = \sin \left(\frac{\lambda y}{\lambda x}\right) + \log \left(\frac{\lambda y}{\lambda x}\right) = \sin \frac{y}{x} + \log \frac{y}{x} = F(x, y) \]
This is degree 0, so it is a homogeneous differential equation.
Step 3: Test function (iii)
\( F(x, y) = e^{y/x} + 1 \)
\[ F(\lambda x, \lambda y) = e^{\lambda y / \lambda x} + 1 = e^{y/x} + 1 = F(x, y) \]
This is degree 0, so it is a homogeneous differential equation.
Step 4: Test function (iv)
\( F(x, y) = \sqrt{x^2 + y^2} - y \)
\[ F(\lambda x, \lambda y) = \sqrt{(\lambda x)^2 + (\lambda y)^2} - \lambda y = \lambda(\sqrt{x^2 + y^2} - y) \neq F(x, y) \]
This is degree 1, not homogeneous of degree 0.