`x(dy)/(dx)=y(logy-logx+1)`

Options

A. `logx/y=cy`
B. `logy/x=cy`
C. `logy/x=cx`
D. None of these

Explanation

To solve the differential equation \( x \frac{dy}{dx} = y (\log y - \log x + 1) \), we will follow a systematic approach.

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

### Step 2: Simplify using logarithmic properties
Using the property of logarithms, we can express \(\log y - \log x\) as \(\log \left( \frac{y}{x} \right)\). Thus, we have:
\[ \frac{dy}{dx} = \frac{y}{x} \left( \log \left( \frac{y}{x} \right) + 1 \right) \]

### Step 3: Introduce a substitution
Let \( v = \frac{y}{x} \), which implies \( y = vx \). We can differentiate \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = v + x \frac{dv}{dx} \]
Substituting this into our equation gives:
\[ v + x \frac{dv}{dx} = v \left( \log v + 1 \right) \]

### Step 4: Rearranging the equation
Rearranging the equation, we have:
\[ x \frac{dv}{dx} = v \log v \]
This can be separated as:
\[ \frac{dv}{v \log v} = \frac{dx}{x} \]

### Step 5: Integrate both sides
Now we will integrate both sides:
\[ \int \frac{dv}{v \log v} = \int \frac{dx}{x} \]
The right-hand side integrates to:
\[ \log |x| + C \]
For the left-hand side, we can use the substitution \( u = \log v \), which gives \( du = \frac{1}{v} dv \) or \( dv = v du \):
\[ \int \frac{du}{u} = \log |\log v| + C_1 \]

### Step 6: Combine results
Equating both integrals, we have:
\[ \log |\log v| = \log |x| + C \]
Exponentiating both sides gives:
\[ |\log v| = k |x| \quad \text{where } k = e^C \]

### Step 7: Substitute back for \( v \)
Since \( v = \frac{y}{x} \), we substitute back:
\[ |\log \left( \frac{y}{x} \right)| = k |x| \]

### Step 8: Final expression
This can be simplified to:
\[ \log \left( \frac{y}{x} \right) = \log |x| + \log k \]
Thus, we can express it as:
\[ \frac{y}{x} = Cx \]
Where \( C = e^{\log k} \).

### Final Answer
The general solution of the differential equation is:
\[ y = Cx^2 \]