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

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:

\[\log_{e}y.\]