If `x(dy)/(dx)=y(log y -logx+1),` then the solution of the equation is

Options

A. `log((x)/(y))=Cy`
B. `log((y)/(x))=Cx`
C. `x log((y)/(x))=Cy`
D. `ylog((x)/(y))=Cx`

Explanation

To solve the differential equation \( x \frac{dy}{dx} = y (\log y - \log x + 1) \), we will follow these steps:

### Step 1: Substitute \( y = v(x) \)
Let \( y = v(x) \). Then, we differentiate \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = \frac{dv}{dx} \]

### Step 2: Rewrite the equation
Substituting \( y = v(x) \) into the original equation gives:
\[ x \frac{dv}{dx} = v (\log v - \log x + 1) \]

### Step 3: Simplify the equation
Rearranging the equation, we have:
\[ x \frac{dv}{dx} = v \log v - v \log x + v \]
This can be rewritten as:
\[ x \frac{dv}{dx} = v \log v - v \log x + v \]

### Step 4: Divide both sides by \( v \)
Assuming \( v \neq 0 \), we can divide both sides by \( v \):
\[ \frac{x}{v} \frac{dv}{dx} = \log v - \log x + 1 \]

### Step 5: Rearrange the equation
Rearranging gives:
\[ \frac{dv}{v \log v} = \frac{dx}{x} \]

### Step 6: Integrate both sides
Now, we integrate both sides:
\[ \int \frac{dv}{v \log v} = \int \frac{dx}{x} \]

### Step 7: Use substitution for the left side
Let \( t = \log v \), then \( dv = v dt = e^t dt \):
\[ \int \frac{e^t dt}{e^t t} = \int \frac{dt}{t} \]
The left side becomes:
\[ \int \frac{1}{t} dt = \log |t| + C_1 \]
Thus, we have:
\[ \log |\log v| = \log |x| + C_2 \]

### Step 8: Exponentiate both sides
Exponentiating gives:
\[ |\log v| = C |x| \]
where \( C = e^{C_2} \).

### Step 9: Solve for \( v \)
This implies:
\[ \log v = C x \quad \text{or} \quad \log v = -C x \]
Thus, we have:
\[ v = e^{C x} \quad \text{or} \quad v = e^{-C x} \]

### Step 10: Substitute back for \( y \)
Recall that \( v = \frac{y}{x} \), so:
\[ \frac{y}{x} = e^{C x} \quad \Rightarrow \quad y = x e^{C x} \]
or
\[ \frac{y}{x} = e^{-C x} \quad \Rightarrow \quad y = x e^{-C x} \]

### Final Solution
The general solution of the differential equation is:
\[ y = x e^{C x} \]
or
\[ y = x e^{-C x} \]