Use the Laplace transform to solve the given initial-value problem. displaystyle{y}{''}+{8}{y}'+{41}{y}=delta{left({t}-piright)}+delta{left({t}-{3}piright)}, {y}{left({0}right)}={1}, {y}'{left({0}right)}={0} {y}{left({t}right)}=?

beljuA 2020-12-06 Answered
Use the Laplace transform to solve the given initial-value problem.
y+8y+41y=δ(tπ)+δ(t3π),    y(0)=1,    y(0)=0  y(t)=?
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Expert Answer

sweererlirumeX
Answered 2020-12-07 Author has 91 answers
Step 1
Given initial value problem,
y+8y+41y=δ(tπ)+δ(t3π),
y(0)=1,
y(0)=0
Solve this problem using Laplace transform method.
Step 2
Now
y+8y+41y=δ(tπ)+δ(t3π)
L[y+8y+41y]=L[δ(tπ)+δ(t3π)]
L[y]+8L[y]+41L[y]=L[δ(tπ)]+L[δ(t3π)]
Use the formula such that
L[y]=s2L[y]sy(0)y(0)
L[y]=sL[y]y(0)
L[δ(ta)]=eas
Step 3
Then,
s2L[y]sy(0)y(0)+8[sL[y]y(0)]+41L[y]=eπs+e3πs
s2L[y]s00+8[sL[y]0]+41L[y]=eπs+e3πs
s2L[y]+8sL[y]+41L[y]=eπs+e3πs
[s2+8s+41]L[y]=eπs+e3πs
L[y]=eπs+e3πss2+8s+41
Taking inverse Laplace transform of both sides,
y=L1[eπs+e3πss2+8s+41]

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