Question

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

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

Hint:

Recovering $P(y)$ from an Integrating Factor:

• Use I.F. = $e^\int P(y)dy$.
• Take logarithm of I.F. and differentiate.

Answer 1

The Correct Option is D

Given I.F. = $\log y = e^{\int P(y)dy} \Rightarrow \int P(y)dy = \ln(\log y)$
Differentiate: \[ P(y) = \frac{d}{dy}\ln(\log y) = \frac{1}{\log y} \cdot \frac{1}{y} = \frac{1}{y \log y} \]

Answer 2

Step 1: Given differential equation and integrating factor
The differential equation is \(\frac{dx}{dy} + P(y) x = Q(y)\).
We are given that \(\log y\) is an integrating factor.

Step 2: Condition for integrating factor
An integrating factor \(\mu(y)\) for an equation \(\frac{dx}{dy} + P(y) x = Q(y)\) satisfies:
\[ \frac{d\mu}{dy} = \mu \cdot P(y) \]

Step 3: Apply integrating factor \(\mu = \log y\)
Calculate \(\frac{d}{dy}(\log y) = \frac{1}{y}\).
Using the condition: \(\frac{d\mu}{dy} = \mu P(y)\), we get
\[ \frac{1}{y} = (\log y) \cdot P(y) \]

Step 4: Solve for \(P(y)\)
\[ P(y) = \frac{1}{y \log y} \]

Step 5: Conclusion
Therefore, \(P(y) = \frac{1}{y \log y}\).