Find the solution of the initial value problem given below by Laplace transform y'-y=t e^t sin t y(0)=0

generals336 2021-03-06 Answered
Find the solution of the initial value problem given below by Laplace transform
yy=tetsint
y(0)=0
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Expert Answer

stuth1
Answered 2021-03-07 Author has 97 answers
Step 1
The answer is given by using properties of Laplace transform.
Step 2
yy=tetsin(t)
y(0)=0
Apply the laplace transform
L{y}L{y}=L{tetsin(t)}
(sY(s)y(0))Y(s)=2(s1)(s22s+2)2
sY(s)Y(s)=2(s1)(s22s+2)2
(s1)Y(s)=2(s1)(s22s+2)2
Y(s)=2(s22s+2)2
Now apply inverse laplace transform
L1{Y(s)}=L1{2(s22s+2)2}
y(t)=L1{2((s1)2+1)2}
y(t)=etsin(t)tetcos(t)
y(t)=et(sin(t)tcos(t))
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