An integrating factor of the differential equ | vedclass

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Question

(B) The given differential equation is $x \frac{dy}{dx} + y \log x = x e^x x^{-\frac{1}{2} \log x}$.Dividing by $x$,we get $\frac{dy}{dx} + \left( \fr

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$(\sqrt{x})^{\log x}$

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(B) The given differential equation is $x \frac{dy}{dx} + y \log x = x e^x x^{-\frac{1}{2} \log x}$.Dividing by $x$,we get $\frac{dy}{dx} + \left( \frac{\log x}{x} \right) y = e^x x^{-\frac{1}{2} \log x}$.This is a linear differential equation of the form $\frac{dy}{dx} + P(x) y = Q(x)$,where $P(x) = \frac{\log x}{x}$.The integrating factor $(I.F.)$ is given by $e^{\int P(x) dx} = e^{\int \frac{\log x}{x} dx}$.Let $u = \log x$,then $du = \frac{1}{x} dx$. Thus,$\int \frac{\log x}{x} dx = \int u du = \frac{u^2}{2} = \frac{(\log x)^2}{2}$.Therefore,$I.F. = e^{\frac{1}{2} (\log x)^2} = (e^{\frac{1}{2} \log x})^{\log x} = (e^{\log \sqrt{x}})^{\log x} = (\sqrt{x})^{\log x}$.

An integrating factor of the differential equation $x \frac{dy}{dx} + y \log x = x e^x x^{-\frac{1}{2} \log x}$,$(x > 0)$ is

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