$\log_e y$
$y$
The integrating factor for the differential equation \((y \log_e y) \frac{dx}{dy} + x = 2 \log_e y\) is determined by comparing it to the standard linear differential equation form \(\frac{dx}{dy} + P(y)x = Q(y)\).
Rearranging the given equation yields:
\((y \log_e y) \frac{dx}{dy} + x = 2 \log_e y\)
\(\Rightarrow \frac{dx}{dy} + \frac{1}{y \log_e y}x = \frac{2}{y}\)
From this, we identify \(P(y) = \frac{1}{y \log_e y}\) and \(Q(y) = \frac{2}{y}\).
The integrating factor, denoted by \( \mu(y) \), is calculated using the formula:
\(\mu(y) = e^{\int P(y) \, dy} = e^{\int \frac{1}{y \log_e y} \, dy}\)
To evaluate the integral \(\int \frac{1}{y \log_e y} \, dy\), a substitution \(u = \log_e y\) is made, leading to \(du = \frac{1}{y} dy\).
The integral simplifies to:
\(\int \frac{1}{y \log_e y} \, dy = \int \frac{1}{u} \, du = \log_e|u| + C\)
Substituting back \(u = \log_e y\) gives:
\(\log_e|\log_e y|\)
Therefore, the integrating factor is:
\(\mu(y) = e^{\log_e(\log_e y)} = \log_e y\)
The determined integrating factor is \(\log_e y\).