The integrating factor of the differential equation
`ylogy(dx)/(dy)+x-logy=0`, is

Options

A. `log(logy)`
B. `logy`
C. `(1)/(logy)`
D. `(1)/(log(logy))`

Explanation

To find the integrating factor of the differential equation

\[ y \log y \frac{dx}{dy} + x - \log y = 0, \]

we can follow these steps:

### Step 1: Rearranging the Equation
First, we rearrange the equation to isolate the term involving \(\frac{dx}{dy}\):

\[ y \log y \frac{dx}{dy} = \log y - x. \]

### Step 2: Dividing by \(y \log y\)
Next, we divide both sides by \(y \log y\) to express the equation in the standard linear form:

\[ \frac{dx}{dy} + \frac{x}{y \log y} = \frac{\log y}{y \log y}. \]

This simplifies to:

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

### Step 3: Identifying \(p\) and \(q\)
Now, we identify \(p\) and \(q\) from the standard linear form \(\frac{dx}{dy} + p(x) = q(y)\):

- Here, \(p = \frac{1}{y \log y}\) and \(q = \frac{1}{y}\).

### Step 4: Finding the Integrating Factor
The integrating factor \(\mu(y)\) is given by:

\[ \mu(y) = e^{\int p \, dy} = e^{\int \frac{1}{y \log y} \, dy}. \]

### Step 5: Solving the Integral
To solve the integral \(\int \frac{1}{y \log y} \, dy\), we use the substitution \(t = \log y\). Then, we differentiate:

\[ dt = \frac{1}{y} \, dy \quad \Rightarrow \quad dy = y \, dt = e^t \, dt. \]

Substituting into the integral gives:

\[ \int \frac{1}{y \log y} \, dy = \int \frac{1}{t} \, dt = \log t + C = \log(\log y) + C. \]

### Step 6: Exponentiating the Integral
Now, we exponentiate the result to find the integrating factor:

\[ \mu(y) = e^{\log(\log y)} = \log y. \]

### Conclusion
Thus, the integrating factor of the given differential equation is:

\[ \mu(y) = \log y. \]