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

Answer: (A)

$\log_e y$

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:

$\log_{e}y.$