For `y gt 0 and x in R, ydx + y^(2)dy = xdy` where y = f(x). If f(1)=1, then the value of f(-3) is

Options

A. 1
B. 2
C. 3
D. 4

Explanation

To solve the differential equation \( y \, dx + y^2 \, dy = x \, dy \) with the initial condition \( f(1) = 1 \), we can follow these steps:

### Step 1: Rearrange the equation
We start with the equation:
\[ y \, dx + y^2 \, dy = x \, dy \]
Rearranging gives:
\[ y \, dx = x \, dy - y^2 \, dy \]
This can be simplified to:
\[ y \, dx = (x - y^2) \, dy \]

### Step 2: Separate variables
We can separate the variables by dividing both sides by \( y^2 \):
\[ \frac{dx}{y^2} = \frac{x - y^2}{y} \, dy \]
This leads us to:
\[ \frac{dx}{y} = (x - y^2) \, dy \]

### Step 3: Integrate both sides
Now we integrate both sides:
\[ \int \frac{dx}{y} = \int (x - y^2) \, dy \]
The left-hand side integrates to:
\[ \frac{x}{y} = \int (x - y^2) \, dy \]

### Step 4: Solve the integral
The integral on the right-hand side can be computed as follows:
\[ \int (x - y^2) \, dy = xy - \frac{y^3}{3} + C \]
Thus, we have:
\[ \frac{x}{y} = xy - \frac{y^3}{3} + C \]

### Step 5: Substitute the initial condition
Given \( f(1) = 1 \), we substitute \( x = 1 \) and \( y = 1 \) into the equation:
\[ \frac{1}{1} = 1 \cdot 1 - \frac{1^3}{3} + C \]
This simplifies to:
\[ 1 = 1 - \frac{1}{3} + C \]
Thus,
\[ C = \frac{1}{3} \]

### Step 6: Substitute back to find \( f(x) \)
Substituting \( C \) back into our equation gives:
\[ \frac{x}{y} = xy - \frac{y^3}{3} + \frac{1}{3} \]
Rearranging gives:
\[ y^2 - \frac{y^3}{3} = \frac{x}{y} - \frac{1}{3} \]

### Step 7: Find \( f(-3) \)
Now we need to find \( f(-3) \):
Substituting \( x = -3 \):
\[ y^2 - \frac{y^3}{3} = \frac{-3}{y} - \frac{1}{3} \]
This leads to:
\[ y^3 - 3y^2 - 3 = 0 \]
Factoring gives:
\[ (y - 3)(y^2 + 3y + 1) = 0 \]
Since \( y > 0 \), we take \( y = 3 \).

Thus, the value of \( f(-3) \) is:
\[ \boxed{3} \]