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The integrating factor of the differential equation $\frac{d y}{d x}(x \log x)+y=2 \log x$ is given by
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Verified Answer
The correct answer is:
$\log x$
$\begin{aligned} & \frac{d y}{d x}(x \log x)+y=2 \log x \\ & \Rightarrow \quad \frac{d y}{d x}+\frac{y}{x \log x}=\frac{2}{x} \end{aligned}$
Here, $\quad P=\frac{1}{x \log x}, Q=\frac{2}{x}$
IF
$$
=e^{\int P d x}=e^{\int \frac{d x}{x \log x}}
$$
$=\log x$
Here, $\quad P=\frac{1}{x \log x}, Q=\frac{2}{x}$
IF
$$
=e^{\int P d x}=e^{\int \frac{d x}{x \log x}}
$$
$=\log x$
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