The solution of ` ( x + log y) dy + y dx=0` when y(0) =1 is :

Options

A. ` y(x-1) + y log y=0`
B. `y(x-1+y log Y ) + 1=0`
C. `xy +y log y-y +1=0`
D. none of these

Answer

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

### Step 1: Rearranging the Equation
We start with the given equation:
\[ (x + \log y) dy + y dx = 0 \]
We can rearrange this to isolate \( dx \) and \( dy \):
\[ y dx = -(x + \log y) dy \]
Dividing both sides by \( y \) gives:
\[ dx = -\frac{x + \log y}{y} dy \]

### Step 2: Identifying p and q
We can rewrite the equation in the standard form:
\[ \frac{dx}{dy} + \frac{x}{y} = -\frac{\log y}{y} \]
Here, we identify:
- \( p = \frac{1}{y} \)
- \( q = -\frac{\log y}{y} \)

### Step 3: Finding the Integrating Factor
The integrating factor \( \mu(y) \) is given by:
\[ \mu(y) = e^{\int p \, dy} = e^{\int \frac{1}{y} \, dy} = e^{\log y} = y \]

### Step 4: Multiplying by the Integrating Factor
We multiply the entire equation by the integrating factor \( y \):
\[ y \frac{dx}{dy} + x = -\log y \]

### Step 5: Integrating Both Sides
The left-hand side can be recognized as the derivative of a product:
\[ \frac{d}{dy}(xy) = -\log y \]
Integrating both sides with respect to \( y \):
\[ xy = \int -\log y \, dy \]
Using integration by parts, let \( u = \log y \) and \( dv = dy \):
\[ du = \frac{1}{y} dy, \quad v = y \]
Then,
\[ \int -\log y \, dy = -y \log y + \int y \cdot \frac{1}{y} dy = -y \log y + y + C \]
Thus, we have:
\[ xy = -y \log y + y + C \]

### Step 6: Applying Initial Conditions
We use the initial condition \( y(0) = 1 \):
\[ 0 \cdot 1 = -1 \cdot \log 1 + 1 + C \]
Since \( \log 1 = 0 \):
\[ 0 = 0 + 1 + C \implies C = -1 \]

### Step 7: Final Equation
Substituting \( C \) back into the equation gives:
\[ xy = -y \log y + y - 1 \]
Rearranging this, we get:
\[ xy + y \log y - y + 1 = 0 \]

### Final Solution
The solution to the differential equation is:
\[ xy + y \log y - y + 1 = 0 \]