The integrating factor of the differential equation `(dy)/(dx)(x(log)_e x)+y=2(log)_e x` is given by

Options

A. x
B. `e^(x)`
C. `log_(e)x`
D. `log_(e)(log_(e)x)`

Explanation

To find the integrating factor of the differential equation

\[ \frac{dy}{dx} \cdot x \log_e x + y = 2 \log_e x, \]

we will follow these steps:

### Step 1: Rewrite the Differential Equation
We start by rewriting the given differential equation in the standard form of a linear differential equation:

\[ \frac{dy}{dx} + P(x) y = Q(x). \]

To do this, we will divide the entire equation by \(x \log_e x\):

\[ \frac{dy}{dx} + \frac{y}{x \log_e x} = \frac{2 \log_e x}{x \log_e x}. \]

This simplifies to:

\[ \frac{dy}{dx} + \frac{y}{x \log_e x} = \frac{2}{x}. \]

### Step 2: Identify \(P(x)\) and \(Q(x)\)
From the rewritten equation, we can identify:

\[ P(x) = \frac{1}{x \log_e x}, \quad Q(x) = \frac{2}{x}. \]

### Step 3: Find the Integrating Factor
The integrating factor \(I(x)\) is given by:

\[ I(x) = e^{\int P(x) \, dx}. \]

Substituting \(P(x)\):

\[ I(x) = e^{\int \frac{1}{x \log_e x} \, dx}. \]

### Step 4: Solve the Integral
To solve the integral \(\int \frac{1}{x \log_e x} \, dx\), we can use the substitution:

Let \(t = \log_e x\), then \(dt = \frac{1}{x} dx\), or \(dx = x \, dt = e^t \, dt\).

Thus, the integral becomes:

\[ \int \frac{1}{x \log_e x} \, dx = \int \frac{1}{t} \, dt = \log_e t + C = \log_e (\log_e x) + C. \]

### Step 5: Substitute Back to Find the Integrating Factor
Now substituting back into the expression for the integrating factor:

\[ I(x) = e^{\log_e (\log_e x)} = \log_e x. \]

### Conclusion
The integrating factor of the given differential equation is:

\[ \log_e x. \]