If y(x) is the solution of the differential equation \(\frac{{dy}}{{dx}} + \left( {\frac{{2x + 1}}{x}} \right)y = {e^{ - 2x}}\), x > 0, where \(y\left( 1 \right) = \frac{1}{2}{e^{ - 2}}\) is:
1
y(loge 2) = loge 4
2
\(y\left( {{\rm{lo}}{{\rm{g}}_e}2} \right) = \frac{{{\rm{lo}}{{\rm{g}}_e}2}}{4}\)
3
y(x) is decreasing in \(\left( {\frac{1}{2},1} \right)\)
4
y(x) is decreasing in (0,1)