For what value of n is the following a homogeneous differential equation:
`(dy )/(dx ) = ( x^3 - y^(n ))/( x^2 y+ xy^2 )`

Options

A. `2`
B. `3`
C. `1`
D. none of these

Answer

To determine the value of \( n \) for which the given differential equation is homogeneous, we start with the equation:

\[ \frac{dy}{dx} = \frac{x^3 - y^n}{x^2 y + xy^2} \]

### Step 1: Understand the concept of homogeneous differential equations
A differential equation is homogeneous if both the numerator and the denominator can be expressed as functions of \( x \) and \( y \) that have the same degree. The degree of a term is the sum of the powers of \( x \) and \( y \) in that term.

### Step 2: Determine the degree of the denominator
The denominator is:

\[ x^2 y + xy^2 \]

To find the degree, we analyze each term:
- The term \( x^2 y \) has a degree of \( 2 + 1 = 3 \).
- The term \( xy^2 \) has a degree of \( 1 + 2 = 3 \).

Since both terms have degree 3, the degree of the denominator is 3.

### Step 3: Determine the degree of the numerator
The numerator is:

\[ x^3 - y^n \]

To find the degree, we analyze each term:
- The term \( x^3 \) has a degree of 3.
- The term \( -y^n \) has a degree of \( n \).

For the numerator to have the same degree as the denominator, we need:

\[ \max(3, n) = 3 \]

### Step 4: Set the degrees equal
For the degrees to be equal, we must have:

\[ n = 3 \]

### Conclusion
Thus, the value of \( n \) for which the given differential equation is homogeneous is:

\[ \boxed{3} \]