$d y(2 x \ln x)=[(2 \ln x)-y] d x$
$\frac{d y}{d x}=\frac{1}{x}-\frac{y}{2 x \ln x}$
$\frac{d y}{d x}+\frac{y}{2 x \ln x}=\frac{1}{x}$
$\text { I.F }=e^{\int \frac{1}{2 x \ln x} d x}$
$\quad=e^{\frac{1}{2} \int \frac{d f}{t}}=e^{\frac{1}{2} \ln (\ln x)}$
$\Rightarrow I F=(\ln x)^{1 / 2}$
$\therefore y \sqrt{\ln x}=\int \frac{\sqrt{\ln x}}{x} d x$
$=2 \int u^2 d u$
$y \sqrt{\ln x}=\frac{2}{3}(\ln x)^{3 / 2}+c \leftarrow\left(e, \frac{4}{3}\right) \quad\left(\text { Let }, \ln x=u^2\right)$
$\frac{4}{3}=\frac{2}{3}+c \Rightarrow c=\frac{2}{3}$
$y \sqrt{\ln x}=\frac{2}{3}(\ln x)^{3 / 2}+\frac{2}{3} \leftarrow\left(e^4, \alpha\right)$
$\alpha \cdot 2=\frac{2}{3} \times 8+\frac{2}{3}$
$\alpha=3$
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