The integrating factor of the differential equation \( y\log_{e}(10y)\cdot\displaystyle\frac{dx}{dy}+x=\frac{1}{\left[\log_{e}(10y)\right]^{2}} \), is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( \log_{e}(10y) \)
Here x is the dependent variable, so the equation is brought to the form \( \frac{dx}{dy}+Px=Q \) by dividing by \( y\log_{e}(10y) \).
\( \displaystyle \frac{dx}{dy}+\frac{1}{y\log_{e}(10y)}\,x=\frac{1}{y\left[\log_{e}(10y)\right]^{3}} \)
So \( P=\frac{1}{y\log_{e}(10y)} \). The integrating factor is \( e^{\int P\,dy} \). Put \( u=\log_{e}(10y) \), so that \( du=\frac{dy}{y} \).
\( \displaystyle \int\frac{dy}{y\log_{e}(10y)}=\int\frac{du}{u} \)
\( \displaystyle =\log_{e}u \)
Hence
\( \displaystyle \text{I.F.}=e^{\log_{e}(\log_{e}(10y))} \)
\( \displaystyle =\log_{e}(10y) \)