Question

The integrating factor of the differential equation \[ (y \log_e y) \frac{dx}{dy} + x = 2 \log_e y \] is:

• A. $\log_e y$
• B. $\frac{1}{y}$
• C. $y$
• D. $\log_e(\log_e y)$

Answer 1

The Correct Option is A

To find the integrating factor of the differential equation \((y \log_e y) \frac{dx}{dy} + x = 2 \log_e y\), we start by comparing it to the standard form of a linear differential equation \(\frac{dx}{dy} + P(y)x = Q(y)\). 

Rewriting the given equation, we have:

\((y \log_e y) \frac{dx}{dy} + x = 2 \log_e y\)

\(\Rightarrow \frac{dx}{dy} + \frac{x}{y \log_e y} = \frac{2}{y}\)

Here, \(\frac{1}{y \log_e y}\) is \(P(y)\) and \(\frac{2}{y}\) is \(Q(y)\).

The integrating factor \( \mu(y) \) is given by:

\(\mu(y) = e^{\int P(y) \, dy} = e^{\int \frac{1}{y \log_e y} \, dy}\)

To solve the integral \(\int \frac{1}{y \log_e y} \, dy\), let \(u = \log_e y\), which gives \(du = \frac{1}{y} dy\).

The integral becomes:

\(\int \frac{1}{y \log_e y} \, dy = \int \frac{1}{u} \, du = \log_e|u| + C\)

Since \(u = \log_e y\), substitute back to get:

\(\log_e|\log_e y|\)

Thus, the integrating factor is:

\(\mu(y) = e^{\log_e(\log_e y)} = \log_e y\)

The correct answer is \(\log_e y\).

Answer 2

The given differential equation is:

\[(y~\log_{e}y)\frac{dx}{dy}+x=2~\log_{e}y.\]

Rewriting it:

\[\frac{dx}{dy}+\frac{x}{y~\log_{e}y}=\frac{2}{y}.\]

This is a linear differential equation of the form:

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

\[P(y)=\frac{1}{y~\log_{e}y},~~~Q(y)=\frac{2}{y}.\]

The integrating factor (IF) is given by:

\[IF=e^{\int P(y)dy}.\]

Substitute \(P(y)=\frac{1}{y~\log_{e}y}\):

\[\int P(y)dy=\int\frac{1}{y~\log_{e}y}dy.\]

Let \(u=\log_{e}y\) so \(du=\frac{1}{y}dy\). The integral becomes:

\[\int\frac{1}{y~\log_{e}y}dy=\int\frac{1}{u}du=\log_{e}u+C.\]

Substituting back \(u=\log_{e}y\):

\[\int P(y)dy=\log_{e}(\log_{e}y).\]

Thus, the integrating factor is:

\[IF=e^{\log_{e}(\log_{e}y)}=\log_{e}y.\]

Final Answer:

\[\log_{e}y.\]