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. $\frac{1}{\log(\log y)}$

Answer 1

(5)

Sol. (y log y)dx = (log y - x)dy

$\therefore$ $\frac{dx}{dy} + \frac{x}{y\log y} = \frac{1}{y}$

$\therefore$ I.F. = $e^{\int_{}^{}{\frac{1}{y\log y}.dy}} = e^{\log(\log y)} = \log y.$