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

The given differential equation is, <br/><br/>$$ \begin{aligned} & (y-2 \log x) d x+\left(x \log x^2\right) d y=0 \\\\ & \Rightarrow \frac{d y}{d x}=\frac{(2 \log x-y)}{2 x \log x} \\\\ & \Rightarrow \frac{d y}{d x}+\frac{y}{2 x \log x}=\frac{1}{x} \end{aligned} $$ <br/><br/>It is a linear differential equation. <br/><br/>$\therefore \text { I.F. }=e^{\int \frac{1}{2 x \log x} d x}$ <br/><br/>Put $\log x=t \Rightarrow \frac{1}{x} d x=d t$ <br/><br/>$$ \therefore \text { I.F. }=e^{\int \frac{1}{2 t} d t}=e^{\log (t)^{\frac{1}{2}}}=\sqrt{t}=\sqrt{\log x} $$ <br/><br/>So, required solution is, <br/><br/>$$ \begin{aligned} & y \sqrt{\log x}=\int \frac{\sqrt{\log x}}{x} d x \\\\ & \log x=v \Rightarrow \frac{1}{x} d x=d v \\\\ & \Rightarrow y \sqrt{\log x}=\int \sqrt{v} d v+C \\\\ & \Rightarrow y \sqrt{\log x}=\frac{2 v^{3 / 2}}{3}+C \\\\ & \Rightarrow y \sqrt{\log x}=\frac{2}{3}(\log x)^{3 / 2}+C \end{aligned} $$ <br/><br/>Now, this curve passes through $\left(e, \frac{4}{3}\right)$ and $\left(e^4, \alpha\right)$ <br/><br/>$$ \begin{aligned} & \therefore \frac{4}{3} \sqrt{\log e}=\frac{2}{3}(\log e)^{3 / 2}+C \\\\ & \Rightarrow C=\frac{4}{3}-\frac{2}{3}=\frac{2}{3} \end{aligned} $$ <br/><br/>Also, $\alpha \sqrt{\log e^4}=\frac{2}{3}\left(\log e^4\right)^{3 / 2}+\frac{2}{3}$ <br/><br/>$$ \begin{aligned} & \Rightarrow 2 \alpha=\frac{2}{3} \times(4)^{3 / 2}+\frac{2}{3}=\frac{16}{3}+\frac{2}{3}=\frac{18}{3} \\\\ & \Rightarrow \alpha=3 \end{aligned} $$