The general solution of the differential equation,
`x((dy)/(dx))=y.log((y)/(x))` is

Options

A. `y=xe^(1-c)`
B. `y=xe^(1+c)`
C. `y=ex.e^(Cx)`
D. `y=xe^(Cx)`

Answer

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

### Step 1: Rewrite the Differential Equation
We start by rewriting the given equation in a more manageable form:
\[ \frac{dy}{dx} = \frac{y}{x} \log\left(\frac{y}{x}\right) \]

### Step 2: Substitute \( y = vx \)
To solve this equation, we use the substitution \( y = vx \), where \( v \) is a function of \( x \). Then, we differentiate \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = v + x \frac{dv}{dx} \]

### Step 3: Substitute into the Equation
Now we substitute \( y = vx \) and \( \frac{dy}{dx} \) into the differential equation:
\[ v + x \frac{dv}{dx} = \frac{vx}{x} \log(v) = v \log(v) \]

### Step 4: Rearrange the Equation
Rearranging gives us:
\[ x \frac{dv}{dx} = v \log(v) - v \]
\[ x \frac{dv}{dx} = v (\log(v) - 1) \]

### Step 5: Separate Variables
We can separate the variables:
\[ \frac{dv}{v(\log(v) - 1)} = \frac{dx}{x} \]

### Step 6: Integrate Both Sides
Now we integrate both sides:
\[ \int \frac{dv}{v(\log(v) - 1)} = \int \frac{dx}{x} \]

### Step 7: Use Substitution for Integration
For the left side, we can use the substitution \( T = \log(v) - 1 \), which gives \( dT = \frac{1}{v} dv \). Thus, we have:
\[ \int \frac{1}{T} dT = \int \frac{dx}{x} \]

### Step 8: Solve the Integrals
The integrals yield:
\[ \log|T| = \log|x| + C \]
Substituting back for \( T \):
\[ \log|\log(v) - 1| = \log|x| + C \]

### Step 9: Exponentiate Both Sides
Exponentiating both sides gives:
\[ |\log(v) - 1| = k|x| \quad \text{(where \( k = e^C \))} \]

### Step 10: Solve for \( v \)
This leads to:
\[ \log(v) - 1 = kx \quad \text{or} \quad \log(v) = 1 + kx \]
Exponentiating again:
\[ v = e^{1 + kx} = e \cdot e^{kx} \]

### Step 11: Substitute Back for \( y \)
Recalling that \( v = \frac{y}{x} \):
\[ \frac{y}{x} = e \cdot e^{kx} \implies y = x \cdot e \cdot e^{kx} \]

### Final General Solution
Thus, the general solution of the differential equation is:
\[ y = Cx e^{kx} \quad \text{(where \( C = e \))} \]