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: Rewrite the equation
We start with the given equation:
\[ (x + \log y) dy + y dx = 0 \]
We can rearrange this into the standard form:
\[ \frac{dy}{dx} = -\frac{y}{x + \log y} \]

### Step 2: Identify \( P \) and \( Q \)
From the rearranged equation, we can identify:
- \( P = \frac{1}{y} \)
- \( Q = -\frac{\log y}{y} \)

### Step 3: Find the integrating factor
The integrating factor \( IF \) is given by:
\[ IF = e^{\int P \, dy} = e^{\int \frac{1}{y} \, dy} = e^{\log y} = y \]

### Step 4: Multiply the entire equation by the integrating factor
Multiplying the entire equation by \( y \):
\[ y(x + \log y) dy + y^2 dx = 0 \]

### Step 5: Rewrite in exact form
This can be rewritten as:
\[ y(x + \log y) dy + y^2 dx = 0 \]
This is now an exact equation.

### Step 6: Integrate
To find the solution, we integrate:
1. Integrate \( y(x + \log y) \) with respect to \( y \):
\[ \int y(x + \log y) \, dy = \frac{xy^2}{2} + \frac{y^2 \log y}{2} - \frac{y^2}{2} + C \]
2. Integrate \( y^2 \) with respect to \( x \):
\[ \int y^2 \, dx = xy^2 \]

### Step 7: Combine results
Combining the results gives us:
\[ xy + \frac{y^2 \log y}{2} - \frac{y^2}{2} + C = 0 \]

### Step 8: Apply initial condition
Now we apply the initial condition \( y(0) = 1 \):
\[ 0 \cdot 1 + \frac{1^2 \log 1}{2} - \frac{1^2}{2} + C = 0 \]
Since \( \log 1 = 0 \):
\[ 0 - \frac{1}{2} + C = 0 \implies C = \frac{1}{2} \]

### Step 9: Final equation
Substituting \( C \) back into the equation:
\[ xy + \frac{y^2 \log y}{2} - \frac{y^2}{2} + \frac{1}{2} = 0 \]
Rearranging gives us the final solution:
\[ xy + \frac{y^2 \log y}{2} - \frac{y^2}{2} + 1 = 0 \]

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