Question

The integrating factor of the differential equation $ (y\log y)dx=(\log y-x)dy $ is

• A. $ \frac{1}{\log y} $
• B. $ \log (\log y) $
• C. $ 1+\log y $
• D. $ \frac{1}{\log (\log y)} $
• E. $ \log y $

Answer 1

Answer: (E)

Given differential equation can be rewritten as $ \frac{dx}{dy}=\frac{(\log y-x)}{y\log y} $
$ \Rightarrow $ $ \frac{dx}{dy}+\frac{x}{y\log y}=\frac{1}{y} $
$ \therefore $ $ IF={{e}^{\int{\frac{1}{y\log y}}dy}} $
$={{e}^{\log \log y}}=\log y $