GATE 2016 EE – SET 2 – Question 8

Given equation is: $$y^{“}(t)+2y^{‘}(t)+y(t)=0$$

Applying Laplace Transforms on the above equation

$$L{y^{“}(t)}+2L{y^{‘}(t)}+L{y(t)}=0$$, 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:

$$L[f^{‘}(t)]=s\bar{f}(s) – f(0)$$

$$L[f^{“}(t)]=s^{2}\bar{f}(s)-sf(0)-f^{‘}(0)$$

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

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