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

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Aicharya · Accepted Answer

I. $F=\log y$ $\begin{aligned} & \Rightarrow e^{\int P(y) d y}=\log y \ & \Rightarrow \quad \int P(y) d y=\log (\log y)\end{aligned}$ Differentiating both sides, we get $\begin{aligned} & \Rightarrow P(y)=\frac{1}{\log y} \cdot \frac{1}{y} \ & \Rightarrow P(y)=\frac{1}{y \cdot \log y}\end{aligned}$

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

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