A system is described by the differential equation d2ydt2 - 5dydt -6yt - xt- Let x t be a rectangular pulse given by - 1- 0 - t - 20- Otherwise-Assuming that y0 -0 and dydt -0 at t - 0- the Laplace transform of yt is

X(t) = u(t) u(t 2) X(s) =1s e^ 2ss=1 e^ 2ss Given that d^2ydt^2+5dydt+6y(t)=x(t) nbsp; Taking Laplace transform on both the sides [ y(0) =0 and dydt = ...

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Question

A system is described by the differential equation $\frac{d^{2} y}{d t^{2}} = 5 \frac{d y}{d t} + 6 y (t) = x (t)$ . Let $x (t)$ be a rectangular pulse given by $= {\begin{matrix} 1, 0 < t < 2 0, Otherwise \end{matrix}$
Assuming that $y (0) = 0$ and $\frac{d y}{d t} = 0$ at $t = 0,$ the Laplace transform of $y (t)$ is

\frac{1 - e^{- 2 s}}{s (s + 2) (s + 3)}

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\frac{1 - e^{- 2 s}}{(s + 2) (s + 3)}

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\frac{e^{- 2 s}}{s (s + 2) (s + 3)}

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\frac{e^{- 2 s}}{(s + 2) (s + 3)}

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Solution

The correct option is A $\frac{1 - e^{- 2 s}}{s (s + 2) (s + 3)}$
$x (t) = u (t) - u (t - 2)$

$X (s) = \frac{1}{s} - \frac{e^{- 2 s}}{s} = \frac{1 - e^{- 2 s}}{s}$

Given that
$\frac{d^{2} y}{d t^{2}} + 5 \frac{d y}{d t} + 6 y (t) = x (t)$

Taking Laplace transform on both the sides
$[y (0) = 0 a n d \frac{d y}{d t} = 0 a t t = 0]$

$s^{2} Y (s) + 5. s Y (s) + 6 Y (s) = X (s)$

$Y (s) [s^{2} + 5 s + 6] = \frac{1 - e^{- 2 s}}{s}$

$Y (s) = \frac{1 - e^{- 2 s}}{s (s^{2} + 5 s + 6)} = \frac{1 - e^{- 2 s}}{s (s + 2) (s + 3)}$

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A system is described by the differential equation d2ydt2=5dydt+6y(t)=x(t). Let x(t) be a rectangular pulse given by ={1, 0<t<20, Otherwise Assuming that y(0)=0 and dydt=0 at t=0, the Laplace transform of y(t) is

The correct option is A 1−e−2ss(s+2)(s+3)x(t)=u(t)−u(t−2) X(s)=1s−e−2ss=1−e−2ss Given that d2ydt2+5dydt+6y(t)=x(t) Taking Laplace transform on both the sides [y(0)=0 anddydt=0 at t=0] s2Y(s)+5.sY(s)+6Y(s)=X(s) Y(s)[s2+5s+6]=1−e−2ss Y(s)=1−e−2ss(s2+5s+6)=1−e−2ss(s+2)(s+3)