Step 1: Understand the Concept:
The provided differential equation is a first-order linear differential equation. The solution process involves converting it to the standard form $\frac{dy}{dx} + P(x)y = Q(x)$ and then determining the integrating factor (I.F.).
Step 2: Key Formula/Approach:
The standard form for a linear differential equation is $\frac{dy}{dx} + P(x)y = Q(x)$.
The integrating factor is calculated using the formula: I.F. = $e^{\int P(x)dx}$.
Step 3: Detailed Explanation:
The given differential equation is:
\[ (x \log_e x) \frac{dy}{dx} + y = 2\log_e x \]
To transform this into the standard form, divide the entire equation by $(x \log_e x)$:
\[ \frac{dy}{dx} + \frac{1}{x \log_e x} y = \frac{2\log_e x}{x \log_e x} \]
\[ \frac{dy}{dx} + \left(\frac{1}{x \log_e x}\right)y = \frac{2}{x} \]
Comparing this with the standard form $\frac{dy}{dx} + P(x)y = Q(x)$, we identify:
\[ P(x) = \frac{1}{x \log_e x} \]
Next, calculate the integrating factor:
\[ \text{I.F.} = e^{\int P(x)dx} = e^{\int \frac{1}{x \log_e x} dx} \]
To solve the integral $\int \frac{1}{x \log_e x} dx$, employ the substitution method.
Let $t = \log_e x$. Differentiating with respect to $x$, we get $dt = \frac{1}{x} dx$.
The integral transforms to:
\[ \int \frac{1}{t} dt = \ln|t| = \ln(\log_e x) \]
Substitute this result back into the integrating factor formula:
\[ \text{I.F.} = e^{\ln(\log_e x)} \]
Applying the property $e^{\ln u} = u$, we obtain:
\[ \text{I.F.} = \log_e x \]
Step 4: Final Answer:
The integrating factor for the given differential equation is $\log_e x$.