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:

\(y\left( {{\rm{lo}}{{\rm{g}}_e}2} \right) = \frac{{{\rm{lo}}{{\rm{g}}_e}2}}{4}\)

y(x) is decreasing in \(\left( {\frac{1}{2},1} \right)\)

If y(x) is the solution of the differential equation \(\frac{{dy}}{{d

1.	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:
A.	y(loge 2) = loge 4
B.	\(y\left( {{\rm{lo}}{{\rm{g}}_e}2} \right) = \frac{{{\rm{lo}}{{\rm{g}}_e}2}}{4}\)
C.	y(x) is decreasing in \(\left( {\frac{1}{2},1} \right)\)
D.	y(x) is decreasing in (0,1)
Answer» D. y(x) is decreasing in (0,1)

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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:

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