y"-8y'+41y=0 y(0)=0 y'(0)=5 First , using Y for the Laplace transform of y(t), i.e., Y=Lleft{y(t)right} find the equation you get by taking the Laplace transform of the differential equation

necessaryh 2021-02-14 Answered
y"8y+41y=0
y(0)=0
y(0)=5First , using Y for the Laplace transform of y(t), i.e., Y=L{y(t)}find the equation you get by taking the Laplace transform of the differential equation
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Expert Answer

Sadie Eaton
Answered 2021-02-15 Author has 104 answers
Step 1
Given
y"8y+41y=0
y(0)=0
y(0)=5
Step 2
Calculation
Apply Laplace transform.
L{y"}=s2Y(s)sy(0)y(0)
L{y}=sY(s)y(0)
L{y}=Y(s)
L{c}=cs
L{y"}8L{y}+41L{y}=L{0}
s2Y(s)sy(0)y(0)8[sY(s)y(0)]+41Y(s)=0s
s2Y(s)058[sY(s)0]+41Y(s)=0
Hence the equation is
s2Y(s)058[sY(s)0]+41Y(s)=0
Now,
s2Y(s)058[sY(s)0]+41Y(s)=0
Y(s)[s28s+41]=5
Y(s)=5(s28s+41)
Hence the value of Y(s)=5(s28s+41)
Apply inverse Laplace transform
Y(s)=5(s28s+41)
=5(s28s+16+25)
=5(s4)2+52
y(t)=L1(5(s4)2+52)
y(t)=e4tsin(5t)
Hence the value of y(t)=e4tsin(5t)
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