IF `y' = y/x(log y - log x+1),` then the solution of the equation is :

Options

A. `log (x) /( y) = cy`
B. `log (y)/(x) = cy`
C. `log (x)/(y) = cx`
D. `log (y)/(x) =cx`

Answer

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

### Step 1: Rewrite the equation
We start with the equation:
\[ y' = \frac{y}{x} (\log y - \log x + 1) \]
This can be rewritten as:
\[ \frac{dy}{dx} = \frac{y}{x} (\log y - \log x + 1) \]

### Step 2: Substitute \( y = vx \)
We will use the substitution \( y = vx \), where \( v \) is a function of \( x \). Then, we differentiate \( y \):
\[ \frac{dy}{dx} = v + x \frac{dv}{dx} \]

### Step 3: Substitute into the equation
Substituting \( y = vx \) into the equation gives:
\[ v + x \frac{dv}{dx} = \frac{vx}{x} (\log(vx) - \log x + 1) \]
This simplifies to:
\[ v + x \frac{dv}{dx} = v (\log v + \log x - \log x + 1) = v (\log v + 1) \]

### Step 4: Rearranging the terms
Rearranging the equation leads to:
\[ x \frac{dv}{dx} = v (\log v + 1 - 1) = v \log v \]
Thus, we have:
\[ x \frac{dv}{dx} = v \log v \]

### Step 5: Separate variables
Now we can separate the variables:
\[ \frac{dv}{v \log v} = \frac{dx}{x} \]

### Step 6: Integrate both sides
Integrating both sides, we have:
\[ \int \frac{dv}{v \log v} = \int \frac{dx}{x} \]
The right side integrates to:
\[ \log |x| + C \]
For the left side, we can use the substitution \( t = \log v \), so \( dv = v dt = e^t dt \):
\[ \int \frac{e^t}{e^t t} dt = \int \frac{1}{t} dt = \log |\log v| + C_1 \]
Thus, we have:
\[ \log |\log v| = \log |x| + C \]

### Step 7: Solve for \( v \)
Exponentiating both sides gives:
\[ |\log v| = k |x| \quad \text{where } k = e^C \]
This leads to:
\[ \log v = kx \quad \text{or} \quad \log v = -kx \]

### Step 8: Substitute back for \( y \)
Since \( v = \frac{y}{x} \), we have:
\[ \log \left(\frac{y}{x}\right) = kx \]
This simplifies to:
\[ \log y - \log x = kx \]
Thus:
\[ \log y = kx + \log x \]
Exponentiating gives:
\[ y = x e^{kx} \]

### Final Solution
The solution of the differential equation is:
\[ y = Cx e^{kx} \]
where \( C = e^C \) is a constant.