If the solution curve of the differential equation ( y - 2 log e x ) d x + ( x log e x 2 ) d y = 0 ,  …

Given,

$(y-2{\log }_{e}x)dx+\left(x{\log }_e{x}^2\right)dy=0, x>1$

$\Rightarrow 2x\ln x\frac{dy}{dx}+y=2\ln x \left\{\text{where} {\log }_{e}x=\ln x\right\}$

$\Rightarrow \frac{dy}{dx}+\frac{y}{2x\ln x}=\frac{1}{x}$

Which is linear differential equation,

So, $I.F=e^{\int \frac{1}{2x\ln x}}=\sqrt{\ln x}$

Now, solution of the equation is given by,

$y·\sqrt{\ln x}=\int \frac{\sqrt{\ln x}}{x}dx$

$\Rightarrow y·\sqrt{\ln x}=\frac{2}{3}{\left(\ln x\right)}^{\frac{3}{2}}+C ...\left(i\right)$

Given, $eq.\left(i\right)$ passes through point $\left(e, \frac{4}{3}\right)$

So, $C=\frac{2}{3}$

Hence, solution of differential equation will be,

$y\sqrt{\ln x}=\frac{2}{3}{\left(\ln x\right)}^{\frac{3}{2}}+\frac{2}{3}$

Also given, above equation passes through point $(e^4, \alpha )$

$\alpha \sqrt{\ln e^4}=\frac{2}{3}{\left(\ln e^4\right)}^{\frac{3}{2}}+\frac{2}{3}$

$\Rightarrow 2\alpha =\frac{2}{3}{\left(4\right)}^{\frac{3}{2}}+\frac{2}{3}$

$\Rightarrow \alpha =\frac{1}{3}\times 8+\frac{1}{3}$

$\Rightarrow \alpha =3$