Mumbai university > Electronics and telecommunication engineering, Electronics engineering > Sem 3 > Applied mathematics 3

Marks : 06

Years : DEC 2015

$$y"(t) + 4y'(t) + 8y(t) =1 $$

$L[y"(t) ] + 4L[y'(t) ]+ 8L[y(t) ] = L[1] ---(1) \\ Let \space L[y(t)]=y\bar{(s)} \\ L [y'(t)]=sy\bar{(s)} -y(0) =sy\bar {(s)} -0 \\ L[y" (t) ] =s^2y\bar{(s)} -sy(0) -y'(0) =s^2y\bar {(s)} -0-1\\ Put \space in\space (1) \\ s^2y\bar{(s)} -1 + 4sy\bar{(s)} + 8y \bar {(s)} =\dfrac 1s \\ y\bar {(s)}[s^2+ 4s + 8] = \dfrac 1s+1 \\ y\bar {(s)} = \dfrac {(s+1)}{s(s^2+4s+8)} = \dfrac As + \dfrac {Bs+ C}{s^2+4s+8 } \\ os^2+s+1 = A(s^2+4s+8) + (Bs+ C)s\\ os^2+s+1 = (A+B)s^2 + (4A+C)s+ 8A \\ \Rightarrow 1=8A \Rightarrow A=\dfrac 18 \\ 0=A+B \Rightarrow B=-A =-\dfrac 18 \\ 4A+C =1 \\C=1-4A =1-4(1/8) =1-\dfrac 12 =\dfrac 12 \\ y\bar{(s)} =\dfrac {1/8}s =+ \dfrac {(-1/8) s+ 1/2 }{(s^2+4s+8)} =\dfrac 18 \dfrac 1s-\dfrac 18 [\dfrac {s-4}{(s+2)^2+4}] \\ y{(s)} =\dfrac 18\dfrac 1s-\dfrac 18 [\dfrac {(s+2)-6}{(s+2)^2+4}]_{s+2\to S} $

By Inverse L.T.

$L^{-1}[y\bar{(s)}] = y(t) =\dfrac 18(1) -\dfrac 18 e^{-2t} L^{-1} [\dfrac {s-6}{s^2+4}] \\ y(t) =\dfrac 18-\dfrac 18 e^{-2t} [\cos(2t) -\dfrac 62 \sin (2t)] \\ y(t) =\dfrac 18[1-e^{-2t} [\cos 2t-3\sin 2t]]$