The solution of the differential equation `y (1 + log x) (dx)/(dy) - x log x = 0` is

Options

A. y log x = y + c
B. x log x = yc
C. y (1 + log x) = c
D. logx - y = c

Explanation

To solve the differential equation \( y (1 + \log x) \frac{dx}{dy} - x \log x = 0 \), we can follow these steps:

### Step 1: Rearrange the equation
We start with the equation:

\[ y (1 + \log x) \frac{dx}{dy} - x \log x = 0 \]

We can rearrange it to isolate \(\frac{dx}{dy}\):

\[ y (1 + \log x) \frac{dx}{dy} = x \log x \]

Dividing both sides by \(y (1 + \log x)\):

\[ \frac{dx}{dy} = \frac{x \log x}{y (1 + \log x)} \]

### Step 2: Separate variables
Now we can separate the variables \(x\) and \(y\):

\[ \frac{dx}{x \log x} = \frac{dy}{y (1 + \log x)} \]

### Step 3: Integrate both sides
Next, we integrate both sides. The left side can be integrated as follows:

\[ \int \frac{dx}{x \log x} \]

This integral can be solved using the substitution \(u = \log x\), which gives \(du = \frac{1}{x} dx\):

\[ \int \frac{1}{u} du = \log |u| = \log |\log x| \]

The right side integrates to:

\[ \int \frac{dy}{y} = \log |y| \]

### Step 4: Combine the results
After integrating, we have:

\[ \log |\log x| = \log |y| + C \]

Where \(C\) is the constant of integration.

### Step 5: Exponentiate both sides
To eliminate the logarithm, we exponentiate both sides:

\[ |\log x| = e^{\log |y| + C} = |y| e^C \]

Let \(k = e^C\), we can rewrite this as:

\[ \log x = k y \]

### Step 6: Solve for \(y\)
Now, we can express \(y\) in terms of \(x\):

\[ x = e^{k y} \]

### Final Step: Write the solution
Thus, the solution of the differential equation is:

\[ x \log x = C y \]

Where \(C\) is a constant.