Which of the following is the integrating factor of `xlogx (dy)/(dx) + y = 2 log x ?`

Options

A. `x`
B. `e^(x)`
C. `logx`
D. `log(logx)`

Answer

To find the integrating factor for the differential equation \( x \log x \frac{dy}{dx} + y = 2 \log x \), we will follow these steps:

### Step 1: Rewrite the equation in standard form
We start with the given equation:
\[ x \log x \frac{dy}{dx} + y = 2 \log x \]
We can rewrite this in the standard form \( \frac{dy}{dx} + P(x)y = Q(x) \) by dividing the entire equation by \( x \log x \):
\[ \frac{dy}{dx} + \frac{y}{x \log x} = \frac{2 \log x}{x \log x} \]
This simplifies to:
\[ \frac{dy}{dx} + \frac{y}{x \log x} = \frac{2}{x} \]

### Step 2: Identify \( P(x) \) and \( Q(x) \)
From the rewritten equation, we can identify:
- \( P(x) = \frac{1}{x \log x} \)
- \( Q(x) = \frac{2}{x} \)

### Step 3: Find the integrating factor
The integrating factor \( \mu(x) \) is given by:
\[ \mu(x) = e^{\int P(x) \, dx} \]
Substituting \( P(x) \):
\[ \mu(x) = e^{\int \frac{1}{x \log x} \, dx} \]

### Step 4: Solve the integral
To solve the integral \( \int \frac{1}{x \log x} \, dx \), we can use the substitution:
Let \( t = \log x \), then \( dt = \frac{1}{x} \, dx \) or \( dx = x \, dt = e^t \, dt \). Thus, the integral becomes:
\[ \int \frac{1}{x \log x} \, dx = \int \frac{1}{\log x} \cdot \frac{1}{x} \, dx = \int \frac{1}{t} \, dt = \log |t| + C = \log |\log x| + C \]

### Step 5: Write the integrating factor
Now substituting back, we have:
\[ \mu(x) = e^{\log |\log x|} = |\log x| \]
Since we are typically interested in the positive integrating factor, we can simplify this to:
\[ \mu(x) = \log x \]

### Conclusion
Thus, the integrating factor for the given differential equation is:
\[ \mu(x) = \log x \]