If log y is an integrating factor of frac{dx} | vedclass

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(D) The integrating factor $(IF)$ for the linear differential equation $\frac{dx}{dy} + P(y)x = Q(y)$ is given by $IF = e^{\int P(y) dy}$.Given that $

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$\frac{1}{y \log y}$

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(D) The integrating factor $(IF)$ for the linear differential equation $\frac{dx}{dy} + P(y)x = Q(y)$ is given by $IF = e^{\int P(y) dy}$.Given that $IF = \log y$,we have:$e^{\int P(y) dy} = \log y$Taking the natural logarithm on both sides:$\int P(y) dy = \log(\log y)$Differentiating both sides with respect to $y$:$\frac{d}{dy} \left( \int P(y) dy \right) = \frac{d}{dy} (\log(\log y))$$P(y) = \frac{1}{\log y} \cdot \frac{d}{dy}(\log y)$$P(y) = \frac{1}{\log y} \cdot \frac{1}{y}$$P(y) = \frac{1}{y \log y}$

If $\log y$ is an integrating factor of $\frac{dx}{dy} + P(y)x = Q(y)$,then $P(y) =$

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