A function \(y\left( t \right)\) such that y(0) = 1 and y(1) = 3e-1, is a solution of the differential equation\(\frac{{{d^2}y}}{{d{t^2}}} + 2\frac{{dy}}{{dt}} + y = 0\;\). Then \(y\left( 2 \right)\) is

A function \(y\left( t \right)\) such that y(0) = 1 and y(1) = 3e-1,

1.	A function \(y\left( t \right)\) such that y(0) = 1 and y(1) = 3e-1, is a solution of the differential equation\(\frac{{{d^2}y}}{{d{t^2}}} + 2\frac{{dy}}{{dt}} + y = 0\;\). Then \(y\left( 2 \right)\) is
A.	\(5{e^{ - 1}}\)
B.	\(5{e^{ - 2}}\)
C.	\(7{e^{ - 1}}\)
D.	\(7{e^{ - 2}}\)
Answer» C. \(7{e^{ - 1}}\)

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