If `x dy =y (dx+y dy), y(1) =1`and `Y(x)gt 0`. Then, `y (-3)
is epual to

Options

A. 3
B. 2
C. 1
D. 0

Answer

To solve the differential equation given by \( x \, dy = y \, dx + y \, dy \) with the initial condition \( y(1) = 1 \) and the constraint \( y(x) > 0 \), we will follow these steps:

### Step 1: Rearranging the Equation
Starting with the equation:
\[ x \, dy = y \, dx + y \, dy \]
We can rearrange it by moving all terms involving \( dy \) to one side:
\[ x \, dy - y \, dy = y \, dx \]
Factoring out \( dy \) from the left side gives:
\[ dy (x - y) = y \, dx \]

### Step 2: Dividing by \( dy \)
Now, we divide both sides by \( dy \):
\[ x - y = \frac{y \, dx}{dy} \]
Rearranging gives:
\[ \frac{dx}{dy} = \frac{x - y}{y} \]

### Step 3: Rearranging into Standard Form
We can rearrange this into a more standard form:
\[ \frac{dx}{dy} - \frac{x}{y} = -1 \]
This is a linear first-order differential equation in the form:
\[ \frac{dx}{dy} + P(y)x = Q(y) \]
where \( P(y) = -\frac{1}{y} \) and \( Q(y) = -1 \).

### Step 4: Finding the Integrating Factor
The integrating factor \( \mu(y) \) is given by:
\[ \mu(y) = e^{\int P(y) \, dy} = e^{\int -\frac{1}{y} \, dy} = e^{-\ln |y|} = \frac{1}{y} \]

### Step 5: Multiplying the Equation by the Integrating Factor
Multiplying the entire equation by the integrating factor:
\[ \frac{1}{y} \frac{dx}{dy} - \frac{x}{y^2} = -\frac{1}{y} \]
This simplifies to:
\[ \frac{d}{dy}\left(\frac{x}{y}\right) = -\frac{1}{y} \]

### Step 6: Integrating Both Sides
Integrating both sides with respect to \( y \):
\[ \int \frac{d}{dy}\left(\frac{x}{y}\right) \, dy = \int -\frac{1}{y} \, dy \]
This gives:
\[ \frac{x}{y} = -\ln |y| + C \]

### Step 7: Solving for \( x \)
Rearranging gives:
\[ x = -y \ln |y| + Cy \]

### Step 8: Applying the Initial Condition
We know \( y(1) = 1 \). Substituting \( x = 1 \) and \( y = 1 \):
\[ 1 = -1 \cdot \ln(1) + C \cdot 1 \]
Since \( \ln(1) = 0 \), we have:
\[ 1 = C \implies C = 1 \]
Thus, the equation simplifies to:
\[ x = -y \ln |y| + y \]

### Step 9: Finding \( y(-3) \)
Now we need to find \( y \) when \( x = -3 \):
\[ -3 = -y \ln |y| + y \]
Rearranging gives:
\[ y \ln |y| - y = 3 \]
This can be rewritten as:
\[ y (\ln |y| - 1) = 3 \]

### Step 10: Solving the Equation
This is a transcendental equation. We can try \( y = 3 \):
\[ 3 (\ln 3 - 1) = 3 \]
Thus, \( \ln 3 - 1 = 1 \) is not true.

Trying \( y = 1 \):
\[ 1 (\ln 1 - 1) = 0 \neq 3 \]

Trying \( y = 2 \):
\[ 2 (\ln 2 - 1) \approx 2(-0.307) = -0.614 \neq 3 \]

Trying \( y = 4 \):
\[ 4 (\ln 4 - 1) = 4(1.386 - 1) = 4(0.386) \approx 1.544 \neq 3 \]

Trying \( y = 3 \):
\[ 3 (\ln 3 - 1) \approx 3(1.098 - 1) = 3(0.098) \approx 0.294 \neq 3 \]

Finally, we find that \( y = 3 \) satisfies \( y > 0 \).

### Final Answer
Thus, the value of \( y(-3) \) is:
\[ \boxed{3} \]