GATE 2016 EE - SET 2 - Question 8

Given equation is: y^{

Applying Laplace Transforms on the above equation

L{y^{, which results in, 

s^{2}\bar{y}-sy(0)-y^{'}(0)+2[s\bar{y}-y(0)]+\bar{y}=0 where \bar{y}=L[y]

s^{2}\bar{y}-1+2s\bar{y}+\bar{y}=0

(s^{2}+2s+1)\bar{y}=1

\bar{y}=\frac{1}{s^{2}+2s+1}=\frac{1}{(s+1)^{2}}

Applying inverse laplace transform on the above equation, we get 

L^{-1}[\bar{y}]=L^{-1}[\frac{1}{(s+1)^{2}}]

\Rightarrow y=te^{-t} \Rightarrow y=te^{-t}u(t)

Useful formulae:

L[f^{'}(t)]=s\bar{f}(s) - f(0)

L[f^{

L^{-1}[\frac{1}{s-a}]=e^{at}

L^{-1}[\frac{1}{(s-a)^{2}}]=te^{at}