Solve y'+y=4 delta(t-2pi). Subject to y(0)=0 , y'(0)=0

Caelan 2021-09-03 Answered
Solve y+y=4δ(t2π)
Subject to
y(0)=0
y'(0)=0
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Expert Answer

timbalemX
Answered 2021-09-04 Author has 108 answers
Step 1
According to the given information, it is required to solve the given differential equation.
y+y=4δ(t2π)
Subject to y(0)=0 , y'(0)=0
Step 2
Use laplace transform to solve the above equation.
y+y=δ(t2π)
L(y)+L(y)=4L(δ(t2π))
sY(s)y(0)+Y(s)=4[e2πs](L(δ(tc))=ecs)
(s+1)Y(s)=4[e2πs]
Y(s)=4[e2πs]s+1
y(t)=4L1([e2πs]s+1)
y(t)=4[u2π(t)e(t2π)](L(uc(t)f(tc))=ecsF(s))
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