written 5.0 years ago by
aksh_31 ♦ 2.3k


Given
$\dfrac {dy(t)}{dt}+2y=x(t)$
By Laplace transform,
$s\space y(s) + 2 y(s) = x (s) \\ y (s) (s+2) = x (s) \\ \dfrac {y(s)}{x(s)} = \dfrac 1{(s+2)} \\ H(s) = \dfrac 1{(s+2)} \\ (i)\space x(t) = e^{t} u(t) \\ X (s) =\dfrac 1{(s+1)} $
By Laplace transform,
$Y (s) = x (s) \times H (s) \\ =\dfrac 1{(s+2)}\dfrac 1{(s+1) } $
By partial fraction,
$Y (s) = \dfrac A{(s+1)} + \dfrac B{(s+2)} $
$Y (s) = \dfrac 1{(s+1)} \dfrac 1{(s+2)} $
By Inverse Laplace transform,
$ y(t) = e^{t} u(t)e^{2t} u(t). \\ (ii) \space x(t) = u(t) \\ X (s) = \dfrac 1s $
By Laplace transform,
$Y (s) = x (s) \times H (s) \\ = \dfrac 1{(s+2)}\dfrac 1s$
By partial fraction,
$Y (s) =\dfrac As + \dfrac B{(s+2) } $
$Y (s) = \dfrac {1/2}s+ \dfrac {1/2}{(s+2)}$
By Inverse Laplace transform, $\rightarrow y (t) =1/2( u(t)+e^{2t} u(t)).$