The integrating factor of the differential eq | vedclass

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Question

(A) The given differential equation is $(x \log x) \frac{dy}{dx} + y = 2 \log x$. Dividing both sides by $(x \log x)$,we get: $\frac{dy}{dx} + \frac{1

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$\log x$

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(A) The given differential equation is $(x \log x) \frac{dy}{dx} + y = 2 \log x$. Dividing both sides by $(x \log x)$,we get: $\frac{dy}{dx} + \frac{1}{x \log x} y = \frac{2 \log x}{x \log x} = \frac{2}{x}$. This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = \frac{1}{x \log x}$ and $Q = \frac{2}{x}$. The integrating factor $(I.F.)$ is given by $e^{\int P dx}$. $I.F. = e^{\int \frac{1}{x \log x} dx}$. Let $u = \log x$,then $du = \frac{1}{x} dx$. So,$\int \frac{1}{x \log x} dx = \int \frac{1}{u} du = \log u = \log(\log x)$. Therefore,$I.F. = e^{\log(\log x)} = \log x$.

The integrating factor of the differential equation $(x \log x) \frac{dy}{dx} + y = 2 \log x$ is

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