Given: y\'\'+8y\'+4y=u(t) , where u(t) is a unit-step function L[ u(t)]=1/s and

Question

Given: y''+8y'+4y=u(t) , where u(t) is a unit-step function L[ u(t)]=1/s and the initial conditions are y(0) = 0 and y'(0) = 0 . Solve y(t). Can anyone please provide a detailed breakdown. If I have to graph the resultant y(t), how would I do it since the answers I have been getting have y(t)=...+...+u(t) as an example. Do I plug in 1 for u(t)? Do I plug in time increments for u(t)? I have to graph from initial time 0 to final time 10 in step increments of 0.1 (can do on computer). I just need find out the significance of u(t) in the answer to graph the function.

Explanation / Answer

taking laplace on both sides,

s^2 Y(s)-sy(0)-y'(0)+8sY(s)-8y(0)+4Y(s) =1/s

given y(0) and y'(0) =0

hence s^2 Y(s)+8sY(s)+4Y(s) =1/s

Y(s){s^2+8s+4} =1/s

Y(s) = 1/{(s)((s+4)^2-12)}

y(t) =LAPLACE inverse {1/{(s)((s+4)^2-12)}}

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