Let `y=f (x) and x/y (dy)/(dx) =(3x ^(2)-y)/(2y-x^(2)),f(1)=1` then the possible value of `1/3 f(3)` equals :

Options

A. 9
B. 4
C. 3
D. 2

Answer

To solve the given differential equation, we start with the equation:

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

### Step 1: Rearranging the Equation
We can rearrange the equation to isolate \(\frac{dy}{dx}\):

\[ \frac{dy}{dx} = \frac{y(3x^2 - y)}{x(2y - x^2)} \]

### Step 2: Cross Multiplying
Cross-multiplying gives us:

\[ x(2y - x^2) dy = y(3x^2 - y) dx \]

### Step 3: Expanding Both Sides
Expanding both sides, we have:

\[ 2xy dy - x^3 dy = 3xy^2 dx - y^2 dx \]

### Step 4: Collecting Terms
Rearranging the terms, we get:

\[ 2xy dy - 3xy^2 dx + y^2 dx = x^3 dy \]

### Step 5: Factoring
Factoring out common terms, we can write:

\[ (2xy - x^3) dy = (3xy^2 - y^2) dx \]

### Step 6: Separating Variables
Now we separate the variables:

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

### Step 7: Integrating Both Sides
Next, we integrate both sides:

\[ \int \frac{dy}{y(3 - 1)} = \int \frac{dx}{x(2 - x)} \]

This gives us:

\[ \ln |y| = \ln |x| + C \]

### Step 8: Exponentiating
Exponentiating both sides leads to:

\[ y = kx \]

### Step 9: Using Initial Condition
Using the initial condition \(f(1) = 1\):

\[ 1 = k \cdot 1 \implies k = 1 \]

Thus, we have:

\[ y = x \]

### Step 10: Finding \(f(3)\)
Now, we need to find \(f(3)\):

\[ f(3) = 3 \]

### Step 11: Finding \(\frac{1}{3} f(3)\)
Finally, we calculate:

\[ \frac{1}{3} f(3) = \frac{1}{3} \cdot 3 = 1 \]

### Final Answer
The possible value of \(\frac{1}{3} f(3)\) is:

\[ \boxed{1} \]