If y(x) is the solution of the differential equation

$\frac{d y}{d x}+\left(\frac{2 x+1}{x}\right) y=e^{-2 x}$

I.F. $=e^{\int\left(\frac{2 x+1}{x}\right) d x}=e^{\int\left(2+\frac{1}{x}\right) d x}=e^{2 x+\ln x}=e^{2 x} \cdot x$

So, $y\left(x e^{2 x}\right)=\int e^{-2 x} \cdot x e^{2 x}+C$

$\Rightarrow x y e^{2 x}=\int x d x+C$

$\Rightarrow 2 x y e^{2 x}=x^{2}+2 C$

It passess through $\left(1, \frac{1}{2} e^{-2}\right)$ we get $C=0$

$y=\frac{x e^{-2 x}}{2}$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{2} \mathrm{e}^{-2 \mathrm{x}}(-2 \mathrm{x}+1)$

$\Rightarrow f(\mathrm{x})$ is decreasing in $\left(\frac{1}{2}, 1\right)$

$y\left(\log _{\mathrm{e}} 2\right)=\frac{\left(\log _{\mathrm{e}} 2\right) \mathrm{e}^{-2\left(\log _{\mathrm{e}} 2\right)}}{2}$

$=\frac{1}{8} \log _{\mathrm{e}} 2$