Use the Laplace transform to solve the given initial-value problem y′′+2y′+y =delta(t-4) y(0)=0 y′(0)=0

Ayaana Buck 2020-11-14 Answered
Use the Laplace transform to solve the given initial-value problem
y+2y+y=δ(t4)
y(0)=0
y(0)=0
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Expert Answer

broliY
Answered 2020-11-15 Author has 97 answers
Step 1
Given a differntial equation,
y+2y+y=δ(t4)
with condition
y(0)=0
y(0)=0
Step 2
Appling laplace transform on both sides,
L{y+2y+y}=L{δ(t4)}
L{y"}+2L{y}+L{y}=L{δ(t4)}
Since,
L{y"}=s2Y(s)sy(0)y(0)
L{y"}=s2Y(s)s(0)(0)=s2Y(s)andL{y}=sY(s)y(0)
L{y}=sY(s)(0)=sY(s)
Step 3
Substituting the values,
s2Y(s)+2sY(s)+Y(s)=e4s
(s2+2s+1)Y(s)=e4sThus ,Y(s)=e4ss2+2s+1
Step 4
Now, applying the inverse Laplace transform,
L1{Y(s)}=L1{e4ss2+2s+1}
y(t)=L1{e4s(s+1)2}Since , L1{1(s+1)2}=tet
Now, if inverse laplace transform of F(s) is f(t) then
L1{eatF(s)}=H(ta)f(ta)
Thus,
y(t)=H(t4)(t4)e(t4)
Where H(t-4) is a Heaviside step function.
Step5
Thus, the solution of the given initial value problem is
y(t)=H(t4)e(t4)(t4)
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