The integrating factor of the differential eq | vedclass

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Question

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

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$e^{\frac{1}{2}(\log x)^2}$

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

The integrating factor of the differential equation $x \frac{dy}{dx} + y \log x = x e^x \cdot x^{-1/2} \log x$ for $x > 0$ is:

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