The solution of the differential equation, for $t 0$, $y''\left(t\right) + 2y'\left( t \right) + y\left( t \right) = 0$ with initial conditions $y' \left(0 \right) = 1\;and\;y\left(0\right) = 0$, is $u\left( t \right)$ denotes the unit step functions).

Question

The solution of the differential equation, for $t > 0$, $y''\left(t\right) + 2y'\left( t \right) + y\left( t \right) = 0$ with initial conditions $y' \left(0 \right) = 1\;and\;y\left(0\right) = 0$, is $u\left( t \right)$ denotes the unit step functions).

TuteeHUB · Accepted Answer

$\left( {{e^{ - t}} - t{e^{ - t}}} \right)u\left( t \right)$

TuteeHUB · Answer

$t{e^{ - t}}u\left( t \right)$

TuteeHUB · Answer

$\left( { - {e^{ - t}} + t{e^{ - t}}u\left( t \right)} \right)$

TuteeHUB · Answer

${e^{ - t}}u\left( t \right)$

1.	The solution of the differential equation, for \(t > 0\), \(y''\left(t\right) + 2y'\left( t \right) + y\left( t \right) = 0\) with initial conditions \(y' \left(0 \right) = 1\;and\;y\left(0\right) = 0\), is \(u\left( t \right)\) denotes the unit step functions).
A.	\(t{e^{ - t}}u\left( t \right)\)
B.	\(\left( {{e^{ - t}} - t{e^{ - t}}} \right)u\left( t \right)\)
C.	\(\left( { - {e^{ - t}} + t{e^{ - t}}u\left( t \right)} \right)\)
D.	\({e^{ - t}}u\left( t \right)\)
Answer» B. \(\left( {{e^{ - t}} - t{e^{ - t}}} \right)u\left( t \right)\)

The solution of the differential equation, for \(t > 0\), \(y''\left(t\right) + 2y'\left( t \right) + y\left( t \right) = 0\) with initial conditions \(y' \left(0 \right) = 1\;and\;y\left(0\right) = 0\), is \(u\left( t \right)\) denotes the unit step functions).

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