The integrating factor of the differential eq | vedclass

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(B) The given differential equation is $y \log y \left(\frac{dx}{dy}\right) + x = \log y$. Dividing both sides by $y \log y$,we get: $\frac{dx}{dy} +

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$\log y$

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(B) The given differential equation is $y \log y \left(\frac{dx}{dy}\right) + x = \log y$. Dividing both sides by $y \log y$,we get: $\frac{dx}{dy} + \frac{x}{y \log y} = \frac{\log y}{y \log y} = \frac{1}{y}$. This is a linear differential equation of the form $\frac{dx}{dy} + Px = Q$,where $P = \frac{1}{y \log y}$ and $Q = \frac{1}{y}$. The integrating factor ($I$.$F$.) is given by $e^{\int P dy}$. $I$.$F$. $= e^{\int \frac{1}{y \log y} dy}$. Let $u = \log y$,then $du = \frac{1}{y} dy$. So,$\int \frac{1}{y \log y} dy = \int \frac{1}{u} du = \log u = \log(\log y)$. Therefore,$I$.$F$. $= e^{\log(\log y)} = \log y$.

The integrating factor of the differential equation $y \log y \left(\frac{dx}{dy}\right) + x = \log y$ is

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