The integrating factor of the differential eq | vedclass

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Question

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

Vedclass · Accepted Answer

$\log x$

Vedclass · Answer

(B) The given differential equation is $\frac{dy}{dx}(x \log x) + y = 2 \log x$. Dividing both sides by $(x \log x)$,we get: $\frac{dy}{dx} + \frac{y}{x \log x} = \frac{2 \log x}{x \log x} = \frac{2}{x}$. 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 $(IF)$ is given by $e^{\int P dx}$. $IF = e^{\int \frac{1}{x \log x} dx}$. Let $u = \log x$,then $du = \frac{1}{x} dx$. $IF = e^{\int \frac{1}{u} du} = e^{\log u} = u = \log x$. Thus,the integrating factor is $\log x$.

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

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