The integrating factor of the differential eq | vedclass

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Question

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

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) The given differential equation is $x \frac{dy}{dx} + y \log x = x^2$. Dividing both sides by $x$,we get: $\frac{dy}{dx} + \left(\frac{\log x}{x}\right) y = x$. This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = \frac{\log x}{x}$ and $Q = x$. The integrating factor ($I$.$F$.) is given by $e^{\int P dx}$. $I$.$F$. $= e^{\int \frac{\log x}{x} dx}$. 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,$I$.$F$. $= e^{\frac{(\log x)^2}{2}} = (e^{\log x})^{\frac{\log x}{2}} = x^{\frac{\log x}{2}} = x^{\log (\sqrt{x})}$. Thus,the correct option is $D$.

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

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