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(B) The given differential equation is $(y-2 \ln x) dx + (2x \ln x) dy = 0$.Rearranging the terms,we get $(2x \ln x) dy = (2 \ln x - y) dx$.Dividing b

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$3$

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(B) The given differential equation is $(y-2 \ln x) dx + (2x \ln x) dy = 0$.Rearranging the terms,we get $(2x \ln x) dy = (2 \ln x - y) dx$.Dividing by $dx$ and $(2x \ln x)$,we have $\frac{dy}{dx} = \frac{1}{x} - \frac{y}{2x \ln x}$.This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = \frac{1}{2x \ln x}$ and $Q(x) = \frac{1}{x}$.The Integrating Factor ($I$.$F$.) is $e^{\int P(x) dx} = e^{\int \frac{1}{2x \ln x} dx}$.Let $\ln x = t$,then $\frac{1}{x} dx = dt$. So,$I$.$F$. $= e^{\frac{1}{2} \int \frac{1}{t} dt} = e^{\frac{1}{2} \ln t} = \sqrt{t} = \sqrt{\ln x}$.The general solution is $y \cdot 	ext{I.F.} = \int Q(x) \cdot 	ext{I.F.} dx + C$.$y \sqrt{\ln x} = \int \frac{\sqrt{\ln x}}{x} dx$.Let $\ln x = u^2$,then $\frac{1}{x} dx = 2u du$. The integral becomes $\int u \cdot 2u du = 2 \int u^2 du = \frac{2}{3} u^3 + C = \frac{2}{3} (\ln x)^{3/2} + C$.Using the point $(e, \frac{4}{3})$,we get $\frac{4}{3} \sqrt{\ln e} = \frac{2}{3} (\ln e)^{3/2} + C \Rightarrow \frac{4}{3} = \frac{2}{3} + C \Rightarrow C = \frac{2}{3}$.So,$y \sqrt{\ln x} = \frac{2}{3} (\ln x)^{3/2} + \frac{2}{3}$.For the point $(e^4, \alpha)$,we have $\alpha \sqrt{\ln e^4} = \frac{2}{3} (\ln e^4)^{3/2} + \frac{2}{3}$.$\alpha \sqrt{4} = \frac{2}{3} (4)^{3/2} + \frac{2}{3} \Rightarrow 2\alpha = \frac{2}{3} \cdot 8 + \frac{2}{3} = \frac{16+2}{3} = 6$.Therefore,$\alpha = 3$.

If the solution curve of the differential equation $(y-2 \ln x) dx + (x \ln x^2) dy = 0, x > 1$ passes through the points $(e, \frac{4}{3})$ and $(e^4, \alpha)$,then $\alpha$ is equal to $................$.

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